Download Chapter 2 1．Certain gases in the 293K and 9.97 × 104Pa when the

Chapter 2
1．Certain gases in the 293K and 9.97 × 104Pa when the volume of 0.1910dm3
possession of the quality of 0.132g，Examination of this gas demand relative
molecular mass, it may be what kind of gas?
Solution:
So
the
gas
is
NH3
2．A spacious mouth flask at 280K when it is filled with gas,what temperature we
need to heat in order to escape gas third volume?
Solution:
Known by the title is intended to make one third of the gas escaping bottles or 1.5
times the original volume, while the pressure of this process has remained unchanged,
so there is
3．The same temperature, will 101300Pa the N2 2dm3 and 50650Pa of O2 3 dm3 into
6 dm3 of the vacuum container, seek O2 and N2 partial pressure and mixed gas total
pressure.
Solution:
4．Container in the 4.4 g CO2, 14 g N2, 12.8g O2, a total pressure of 202600Pa,
seeking partial pressure of each component.
Solution：
5．In the 300K, 101300Pa when heating a spacious mouth jug to 500K, and then
closed its nematodirus mouth, and cooled to the original temperature, find the
pressure at this condition
Solution：From the meaning of the questions to know the temperature when the bottle
was promoted to 500K, its size will become the original five thirds times. Therefore,
the amount of bottle gases account for only three-fifths of all the gas, the
corresponding pressure 101300Pa, cool bottle of gas before the same amount of
material
6．At 273K and 1.013 × 105Pa, take 1.0 dm3 clean dry air slowly through the
H3C-O-CH3 liquid, in this process, the fluid loss of 0.0335 g, seeking the saturated
vapor pressure of the liquid at 273K.
Solution
The molar mass of such gas is 46g/mol, pass the air before and after the change of its
volume is negligible
PV=nRT P=0.335 × 8.314 × 273/0.001 × 46=1650Pa
7．There is a gas mixture, with a total pressure of 150Pa, in which the fraction
volume of N2 and H2 is 0.25 and 0.75, find the partial pressure of H2 and N2
Solution
8. In the 291K and the total pressure of 1.013 × 105Pa, 2.70 dm3 air which contains
saturated water vapor, through the CaCl2 drying tube, when completely absorbing
water, the dry air is 3.21 g, seeking the water saturation vapor pressure at this
temperature.
Solution:
9. There is a high-pressure gas cylinder, the volume is 30 dm3,which be able to
withstand 2.6 × 107Pa, how many kilograms of O2 can be put into at 293K and avoid
danger?
Solution:
10 In the 273K,compress the same initial pressure of 4.0 dm3 N2 and 1.0dm3 O2
into a vacuum container of 2 dm3, the total gas mixture pressure is 3.26 × 105 Pa, try
seeking
(1) the initial pressure of two kinds of gas;
(2) the partial pressure of each component in the mixture gas;
(3) The amount of two gases.
Solution:
11.273K, when measured a chlorine of methane densities at different pressure has
been showed in the following table:
P/105
1.013
0.675
0.507
0.338
0.253
2.3074
1.5263
1.1401
0.75713
0.56660
ρ /g·dm −3
Extrapolation used to map (p on / p) data obtained from the chlorine demand a
molecular weight of methane
-3
-5
-1
? /P (g穌 m ? 0 pa )
2.4
2.2
2.0
0.0
0.2
0.4
0.6
0.8
1.0
1.2
5
P (10 pa)
Methane can be drawn from a relative molecular weight of chlorine is 50.495
12.(1) with the ideal gas state equation to prove Avogadro's law;
(2) with the mole fraction to prove that xi =
(3) to prove that μ 2 =
νi
V总
3kT
M
Prove that: (1) PV = nRT
When the p and T is constant, the gas is proportional to V and n
Can be expressed as V ∞ n
(2) under certain conditions in the pressure, V = V1 + V2 + V3 +----According to the definition of sub-volume should be relationship
P Vi = nRT
Mixed gas state equation can be written as P V = nRT
ni
Vi
=
n
V总
ni
=x i
n
（3）
xi =
νi
V总
μA
MB
=
μB
MA
1
pV= N0m( μ 2 )2
3
μ2 =
μ2 =
3RT
3 pV
=
N0m
M
3kT
M
13. Known ether steam heat is 25900J • mol-1, its saturated vapor pressure at the
293K is 7.58 × 104Pa, try to seek the saturated vapor pressure at 308K.
Solution
14. Water gasification heat is 40 kJ • mol-1,, seeking the water saturation vapor
pressure at 298K.
Solution:
15. NaCl, as is shown in Figure 1 is a cell belonging to the cell of Cl (with that), and
Na + (with that) the number of months?
Chapter 3
1. What are the characteristics of electrons in atom in motion? What is difference and
contact between probability and probability density?
Answer:
With a wave-particle duality of electron isnot as a fixed macroscopic objects along the
orbital motion .
Probability density: in the volume who appeared in the probability
Probability: E-emerging opportunities in space
The difference is: more likely refers to the opportunities of electronic, unspecified
range;The probability density per unit volume for the emerging opportunities in terms
of electron
Both were linked: both describe the electron probability appear in the space of
physical quantities.
2. What is the effect of shielding effects and drilling? How to interpret the
phenomenon of main floor of the same energy level splitting and different energy
levels of the main layer interleaving?
Answer:
Drilling effect: Due to electronic differences in the role of drill wear which leads to
changes in the phenomenon of energy
The same energy level splitting the main floor and different stop staggering
phenomenon in the energy levels can be drilled by the shielding effect and the effect
can be explained specifically the energy levels for the same main group, their energy
level high or low depends on effective nuclear charge number, The effective nuclear
charge number also decided in its principal quantum number and angular quantum
numbers, leading to the main floor of each level the same split, another for the same
track on the main floor of its electronic effects due to their different capacities of
drilling, but also can lead to energy level splitting, for different energy levels of the
main layer alternation can also be used to explain the above method.
3. To write the name of the elements of atomic number 24, symbols and their ground
state electronic structure of atoms, and use four quantum numbers, respectively, for
each valence electron moves in space.
Answer
24 element chromium, the ground state electronic structure of formula: 1s2 2s22p6
3s23p6 3d5 4s1
4. M2 + ion 3d orbital are known to have five electronics test release: (1) M atoms
extranuclear electron-arranged; (2) M atoms in the outermost and highest-level group,
the number of electrons; (3) M elements position in the periodic table.
Answer
2
3
outermost two electronics, the highest-level group, 5e
position is the fourth cycle, the seventh vice Tribe
5. According to Slater rules , calculate K, Cu, I the outermost electron of the effective
nuclear charge and the corresponding level of energy.
Answer:
6. According to the atomic structure of knowledge, write No. 17, 23, 80, elements of
the electronic ground state of atomic structure.
Answer:
17 elements in the electronic structure of Cl formula: 1s2 2s2 2p6 3s2 3p5
23 elements in the electronic structure of V: 1s2 2s22p6 3s23p63d3 4s2
80 elements Hg, the electronic structure of formula: 1s2 2s22p6 3s23p63d10
4s24p64d104f14 5s25p65d10 6s2
7. Draw the atomic orbital angular distribution map and radial distribution maps of s p
d, and explain the meaning of these graphics.
A. See textbook 65
s electron cloud is spherical symmetric.
p electron cloud is non-stalk-ya bell-shaped.
d electron cloud is Si petals.
8. Describe the motion of atomic electrons in the state of the physical meaning of the
four quantum numbers of all that? What are their possible values?
Answer:
Principal quantum number n used to describe the atomic electrons in the region there
is most likely distance from the nucleus. Or that it determines the electron energy
level of the important factors.
Azimuthal quantum number l determines the atomic orbitals, or the shape of electron
cloud
Magnetic quantum number m determines the atomic orbitals, or electron cloud in
space, stretching the direction of
Spin quantum number m describes the electron spin of the two states
In the case of n determined, l can only take these values from 0 to n-l, and m is able to
take plus or minus 1, plus or minus 1-l. . . 0 This 2l-1 values for the ms just take plus
or minus half
9. Following sets of quantum numbers which are unreasonable, and why?
(1) n = 2, l = 1, m = 0 (2) n = 2, l = 2, m =- 1
(3) n = 3, l = 0, m = 0 (4) n = 3, l = 1, m = 1
(5) n = 2, l = 0, m =- 1 (6) n = 2, l = 3, m = 2
Answer
Unreasonable to have 2,5,6
l values can only be less than n, so unreasonable 2,6
The l = 0 Shi, m can only take 0, and therefore unreasonable to 5
10. The following is correct? How to rectify incorrect?
(1) s e-nuclear movement around its orbit is a circle, while the electron is to take
S-shaped;
(2) The principal quantum number n is 1, there are two opposite spin-orbit;
(3) The principal quantum number n is 4, its orbit a total of 16 electron shell electron
maximum capacity of 32;
(4) The principal quantum number n is 3, there are 3s, 3p, 3d three tracks.
Answer:
1 This argument is wrong, because there is no movement of electrons find out is
different from the general sense of the macroscopic object movement, said
circumference, as well as dual-spindle is just atomic orbital angular distribution map,
does not track the meaning of
2 is not correct, the principal quantum number n is 1, only one atomic orbital
3 principal quantum number n is 4, its orbit a total of 16, electron shell there are 32
electronic proper
11.The extranuclear electron from the ground state of hydrogen atoms excited to the
2s or 2p,is the energy required equal? If it is the helium atom then what will
happen?
Answer :
The extranuclear electron from the ground state of hydrogen atoms excited to the 2s
or 2p level, their consumption of energy is equal, because the hydrogen atom is a
single-electron atom, 2s and 2p electrons on the energy are the same.As for the
electronic energy of nitrogen atoms 2s2p different, so excited to the 2s2p energy
difference trumpet on track
12. By approximate calculations, 12, 16, 25 elements in atoms, 4s, and 3d which the
energy level is high?
13. According to atomic orbital energy level diagram similar to that of each electron
shell in the following table for errors in the electronic and the reasons.
element
K L
M N
19
2
8
9
22
2
10 8
30
2
8
18
33
2
8
20
2
2
O
P
60
2
8
18 3
18 12 2
Answer
14. Note in the same cycle and the same tribe, the changes of atomic radius, and
discuss its causes.
Answer: In the short-cycle, from left to right as the atomic number increases, the
charge number of nuclei increases, enhancing the attractiveness of the extranuclear
electron, so that the trend of smaller atomic radius, while the new fill in the blank
e-increasing electronic repulsion between, so that larger atomic radius of the trend,
which is contradictory factors in the outer electron e-stable structure does not meet the
8 prior to the increase in nuclear charge dominated, so the radius change from left to
right in the same cycle, small, long-period of the main group elements, the radius of
the changes is similar to the atomic radius of the transition elements from left to right
although the increase due to the nuclear charge decreases, but the rate of decrease is
different from the short-cycle situation.
15. Note the following pairs of atoms in which the first atomic ionization energy high,
and why?
S & P Al and Mg Sr and Rb Cu and Zn Cs and An Rn and At
Answer: Each of the elements in question are the same cycle of elements, in general,
in the same cycle, from left to right as the effective nuclear charge increases, the
radius decreases, the first ionization energy the general trend is to increase, but
because of electronic configuration a greater impact on the ionization energy may
cause some anomalies
P> S for P electronic configuration of 3S23P3, 3P orbital semi-full, while the S
electronic structure of 3P4, loss of a half filled electronic
Mg> Al Mg loss is the 3s electron, while Al lost a 3p electron, E3s> E3p, 3p electron
energy is high and more likely to lose, while the electronic configuration of Mg as 3s2,
3s orbital full-filled, Al electronic configuration of 3s23p1, after the loss of an
electron into a stable structure 3s23p0
Sr> Rb Sr nuclear charge more than the Rb, radius smaller than Rb, followed by a
more stable Sr in 5s2
Zn> Cu Zn nuclear charge more than Cu, while the whole is full of Znde 3d orbital, 4s
orbital was completely filled, Cu, only half-filled with the 4s orbital, loss of an e-3d10
stable structure after
Au> Cs Au as the first Vice-family element, the electronic structure of 5d106s1, due
to 5d electrons shielding 6s electrons play a small role, effective nuclear charge large
radius of the small, Cs as the first Vice-group elements, the electronic structure of 6s1,
After the loss of an electron into a stable structure 5s25p6
Rn> At At the halogens, while the Rn for eight electronic structure of inert gas, it is
difficult to lose the outermost electron, and so the first ionization energy big
16.What is the regularity of relationship between Electron affinity and atomic radius?
Why do some non-metallic elements (such as F, O, etc.) have become abnormal?
Answer: In general, the electron affinity and energy with the decrease of atomic
radius increases because the radius of hours, the nuclear charge of the electron cited
large, therefore, electron affinity and energy left to right in the same cycle have an
increasing trend, while the same The tribe was reduced from top to bottom trend, we
can see elements of oxygen and elemental fluorine electron affinity and energy are not
the greatest when compared to the same tribe, the second element is smaller, this
phenomenon is mainly due to oxygen and fluorine atoms Radius of the country is
small, the electron density is too high, so that when the atoms combine to form an
electron negative ion, due to exclusion of electronic interaction between the release of
energy so that the smaller
17. What is the element electronegativity? How about the variation of each element
electro-negativity in the same cycle, with the family?
Answer: Often the atoms in the molecule the ability to attract electrons is called
electronegativity of elements
In the same period, electro-negativity increases from left to right
In the same clan, from the descending to small electro-negativity
But Vice-group elements,have no significant change in electronegativity
18. If the magnetic quantum number m of values change, that is, m desirable 0, 1, 2, ...
l Total l +1 values, the rest remain unchanged. Then what will the periodic table line
up look like? Under the new periodic table of element to write the top 20 most
active of the alkali metal element, the first of a rare gas element, the first over atomic
number element, element symbols and names.
Answer ： Elements of the former 20, the most active of the alkali metals is Rb
The first element is a rare gases He
The first element of atomic number is 21, element symbol is Sc, name of the
scandium
Chapter 5
1. Try to use ionic bond theory to show a single mass of metal potassium and
chlorine reaction, the formation of potassium chloride process? How to understand
the ionic bond non-directional and saturation?
Answer: Atoms derived from the two because of electrostatic attraction and the
attraction between the ionic bond may become a force.
Non-directional ionic bond can be understood: anions and cations in one direction
which is not the only attractive, but have a strong role in any direction, except when
the distance of its force when only smaller.
2. Seeking to use the following data the electron affinity of hydrogen atom and
energy:
K (s) → K (g)
△ H 1 = 83 kJ · mol
K (g) → K + (g) △ H 2 = 419 kJ · mol
H 2 (g) → H (g) △ H 3 = 218kJ · mol
K + (g) + H
K (s) +
(g) → KH (s) △ H 4 =-742kJ · mol
H 2 (g) → KH (s) △ H 5 =-59kJ · mol
Solution ：now design such cycle：
3. ClF dissociation energy is 246kJ · mol -1, ClF formation of heat is - 56 kJ · mol -1
Cl 2 dissociation energy is 238 kJ · mol -1, please calculate dissociation energy of F 2
(g).
Solution Design cycles are as follows：
4. Test according to the crystal structure type and the relationship between the radius
ratio to determine the following ionic compounds AB-type crystal structure type:
BeO, NaBr, CaS, RbI, BeS, CsBr, AgCl.
Answer ：ZnS:BeO,BaS
NaCl:NaBr,CaS,RbI,AgCl
CsCl:CsBr
5. Viewed from electronegativity data to calculate the following compounds in the
ionic nature of the percentage of one-button be?
And determine which are ionic
compounds? which are the covalent compounds?
NaF, AgBr, RbF, HI, CuI, HBr, CrCl.
Answer
6. How to understand the covalent bond with the direction and saturation?
Answer: According to maximum overlap orbital theory, bonding, the atoms are
always the largest possible atomic orbitals overlap along the direction of bond,
whereas for the atomic orbitals in space have a certain orientation, which is why these
types of atomic orbitals whose only along a certain degree of direction in order to
place the greatest degree of overlap, and thus has a direction, the saturation of the
covalent bond can be understood; covalent bond is shared by the atomic orbital
overlap and electronic form, and each electron can provide the number of orbit and
into the single-electron is for certain.
7. BF 3 is a planar triangular geometry, but the NF 3 is triangular
use the trial hybrid orbital theory to illustrate this.
Geometry,try to
Answer: According to the hybrid orbital theory:
BF3 to sp2 hybrid reciprocity, and its molecular structure is a plane triangle
NF3 to sp3 range of hybrid, and its molecular structure is the triangular shaped
8. Pointed out that which structure of the following compounds is rational structure?
What is irrational structure error?
(A)
(B)
(C)
Answer: a. rational structure for the N2O is: because its structure has produced
relatively large charge the situation is not conducive .
b. For the SCN structure is reasonable: the wrong reasons Ibid
c. contains a double bond structure is more reasonable, and reached the eight e-stable
structure
9. In the following groups which compound of the bond angle is big?
Explain why.
(A) CH 4 and NH 3 (b) OF 2 and Cl 2 O
(C) NH 3 and NF 3 (d) PH 3 and NH3
Answer: a bond angle CH4> NH3
CH4 molecules central atom such as use of sp3 Hybridization, bond angle 109
degrees 28 minutes, while the NH3 molecules central atom to sp3 hybrid inequality,
there is a lone pair of electrons, lone pairs of low-energy electrons from the nucleus
than the near, so lone pairs of electrons bonding electron repulsion larger, so that NH3
molecules bond angles become smaller, for 18 minutes at 107 degrees.
b bond angle Cl2O> OF2
Cl2O and the center of OF2 molecule oxygen atoms are sp3 hybrid inequality, there
are two pairs of lone pair electrons, molecular model for the V-type, but the
coordination of the electronegative F atom is much larger than Cl electrical negative,
OF2 molecule molecule in both pairs of electrons tend to atoms F, ClO2 molecule
bonding electron pairs tend to the central atom O, both in the OF2 molecule bonding
electron pair repulsion between the small molecule, both bonding electron pairs
repulsion and thus is Cl2O molecules bond angle large.
c NH3> NF3 NH3 and NF3 molecules, H and F are the radius of small molecules H,
HF, and F between the repulsion can be ignored, decided to bond angle is a major
factor in the size of lone pairs of electron pairs and bonding electron pair repulsion
size of the .
NF3 molecule bonding electron pairs tend to F atoms, then N's lone pair electrons
closer to the nucleus, energy less, so lone pairs of electrons bonding electron pair
repulsion greater. In the NH3 molecules, bonding electron bias N atoms bonding
electron pairs repulsion between the increase, while N's lone pair electrons by the
nuclear gravity than NF3 molecule large. Lone pairs of electrons energy and the
energy of bonding electron pairs or less, resulting in lone pairs of electrons and
bonding electron pairs of pressure and repulsion or less, so NH3 molecules in the
bond angle and 109 degrees 28 minutes or less.
d bond angle NH3> PH3
NH3 molecules and PH3 molecule configuration are triangular shaped, with the same
body, but a different central atom, N large negative electrical radius of the small, P
electrical radius of the small negative large radius of the small electronegativity large
N atoms around the lone pair electrons and bonding electron pairs between the
maximum angle in order to keep balance, so NH3 bond angle close to 109 degrees for
28 minutes, in addition, PH3 bond angle close to 90 ° may also be P have sp3 orbital
energies close to the 3d orbit.
10. Try to use valence shell electron Repulsion Theory to determine the following
molecules or ions in space configuration. Explain.
HgCl 2 BCl 3 SnCl 2 NH 3 H 2 O PCl 5 TeCl 4 ClF 3
ICl
SF 6 IF 5 FCl 4 CO 2 COCl 2 SO 2 NOCl
SO 2 Cl 2 POCl 3 SO
ClO
IO 2 F
Answer: HgCl2; Hg of the electron number is two for the sp hybrid linear reciprocity
BCl3: B E-3 for sp2 on the number of sexually Hybrid planar triangular
SnCl2 Sn e-3 for the right number of sp2 hybrid linear inequality
NH3 N number of electrons ranging from 4 to sp3 hybrid triangular shaped like
H2O O number of the electronic range of 4 to sp3 hybrid V PCl5 P E-5 for the right number of sexually Hybrid sp3d triangle pairs of cone-shaped
ClF3 F the number of electrons ranging from 5 to sp3d of hybrid T TeCl4 Te number of the electronic sp3d ranging from five to four corners of Hybrid
Cone
ICl2 I the electron number ranging from 5 to sp3d of hybrid linear
SF6 S E-6 for the right number of sp3d2 etc. are octahedral shape of Hybrid
IF5 I of the electron number ranging from 6 to sp3d2 corner of Hybrid Cone
FCl4 F number of the electronic range of 5.5 to sp3d2 Hybridization square-planar
CO2 C e-2 for the right number of sp hybrid linear reciprocity
COCl2 C E-3 for the right number of sexually sp2 hybrid planar triangular
SO2 S e 3 for the right number of sp2 hybrid V-ranging nature of
NOCl2 N e of the number of ranges of 3.5 to sp3 hybrid triangular cone-shaped
SOCl2 S e-4 on the number of inequality for the sp3 hybrid triangular cone-shaped
POCl3 P E-4 for the right number of sexually hybrid sp3 tetrahedral
SO3 S the number of electrons ranging from 4 to sp3 Hybridization triangular cone
ClO2 Cl E-4 for the right number of ranges of sp3 hybrid V IO2F2 I of the electron number ranging from 5 to sp3d four cone of Hybrid
11. Try to use valence bond and molecular orbital method to shows O 2 and F 2
molecular structure. What is difference between these two methods?
Answer
See book 177.
Valence bond theory suggests that the formation of covalent bond electron confined
between two adjacent atoms within cell movement, lack of opposition activists have
fully taken into account as a whole, so it some polyatomic molecules, especially
organic molecular structure can not be Note, while its hydrogen molecular ion in the
single-electron bond, oxygen molecules in the three-electron bond, as well as
magnetic properties of molecules were also unable to explain.
Molecular orbital theory, focusing on the molecular integrity, it is the molecule as a
whole to deal with, a more comprehensive reflection of the molecular movements of
electrons within the state, it can not only explain the existence of molecules in the
electronic key, single-electron bond, Three-electron bond formation, but also the
structure of multi-atomic molecules in order to give a better description.
12. This has the following diatomic molecules or ions
Li 2, Be 2, B 2, N 2, HF, F 2, CO +
① write their molecular orbital-style.
② calculation of their bond, to determine which of the most stable? Which
is the most unstable?
③ determine which molecules or ions are paramagnetic. What is
anti-magnetic?
Answer:（1）Molecular orbital is as follows
13. Write O , O 2, O , O Molecules or ions of the molecular orbital type.
And compare their stability?
Answer:
14. Known NO 2, CO 2, SO 2 molecules whose bond angles were 132 °, 180 °, 120 °,
to determine which hybrid type of the center of their atomic orbitals are?
Answer: NO2 molecules bond angles is between 120-180 degrees, so the central
atom N of the hybrid type is sp2, CO2 its bond angle is linear, so the central atom of
the hybrid type sp, and the SO3 bond angle is 120 °. So is the flat triangular and its
central atom of the hybrid type sp2
15. Write NO +, NO, NO Molecules or ions of the molecular orbital type, pointed
out that the key to their level, which one is magnetic?
Answer:
16. What's the difference about metal conductors, semiconductors and insulators the
energy band structure?
Answer: From the band theory point of view, half of the energy band structure of
solids have, for conductor valence electron energy band is half full, or valence
electrons over the whole band though, but the empty energy band, and can bring the
energy spacing small, overlap one another can occur, for an insulator, starting all over
e-zone, the conduction band is empty, and full band and the conduction band energy
spacing between the large, for the semiconductor, electronic full band has been filled
conduction band is empty, but the energy band structure of a very narrow band gap
18. Compare the following two compounds, the polarization ability of CKS-ion Size:
1.
2.
ZnCl 2, Fe Cl 2, CaCl 2, KCl.
SiCl 4, AlCl 3, PCl 5, MgCl 2, NaCl.
Answer （1) The polarization capacity: Zn2 +> Fe2 +> Ca2 +> K +
Ca, K, Fe, Zn four elements in the same cycle, K + is only one positive
charge, polarization capacity of the weakest, Zn2 +, Fe2 +, Ca2 + stay the
same positive charge, Ca2 + ion charge of 8 electron configuration, Fe2 + for
9-17 electron configuration, Zn2 + for the 18-electron configuration, the
effective charge cation Zn2 +> Fe2 +> Ca2 + Therefore, polarization ability
Zn2 +> Fe2 +> Ca2 +
(2) The polarization ability : P5 +> Si4 +> Al3 +> Mg2 +> Na +
19.Try to use ion polarization point of view to explain the following
phenomena:
AgF soluble in water, AgCl, AgF, AgI is insoluble in water, solubility
in turn reduced by AgF to AgI.
2. AgCl, AgBr, AgI in turn deepen the color.
1.
Answer: As the F-and Ag + mutual polarization is weak, and AgCl, AgBr, AgI exists
a certain degree of polarization effect, it is AgCl, AgBr, AgI which is insoluble in
water, for AgCl, AgBr, AgI from Cl to I of its deformation getting stronger, leading to
their interactions has been strong, and therefore lead to smaller solubility of the
gradual weakening of the color gradually
20. Let us compare the following substances in the key class of size.
NaF, HF, HCl, HI, I 2
Answer: Bond order of：NaF>HF>HCL>I2
21. What is a bond?
Hydrogen bond to the compound nature of the impact?
Answer: Bond is usually available X-H. . . Y said that X and Y on behalf of N, O, F
and other electro-negativity large and small atomic radius of atoms, hydrogen bond in
the X and Y can be both kinds of the same elements, it can be both kinds of different
elements, the formation of hydrogen key produced a strong intermolecular binding
force of the compounds of the boiling point and melting point significantly increased.
22. Which of the following compounds has hydrogen bonding? And point that they
are intermolecular hydrogen bonds or intramolecular hydrogen bond?
C 6 H 6, NH 3, C 2 H 6,
,
,
, H 3 BO 3 (solid)
Answer: There is no hydrogen bonds: C6H6 C2H6
Existence of intermolecular hydrogen bonds also:
Existence of intramolecular hydrogen bonds are: H3BO3 NH3
23. To determine the existence of the following groups of molecules What is the form
of intermolecular forces?
① benzene and CCl 4;
② Helium and water;
gas;
⑤ Methanol and water.
③ CO 2 gases;
④ HBr
Answer: 1 dispersion force
2 dispersion-induced power
3 dispersion-induced power
4 dispersion forces dispersion forces induced by force-oriented force
5 dispersion-induced power-oriented power of hydrogen bond st
24. Try to determine what kind of Si and I 2 crystals a high melting point, and why?
Answer: As the Si crystal is a typical atomic crystals, while the I2 crystal is a typical
molecular crystal, so the melting point of Si crystals higher than the I2 crystals
1. Say BaH 2, SiH 4, NH 3, AsH 3, PdH 0.9 and HI the name and classification?
What was the status of room temperature? Hydride which is good conductor
of electricity?
Answer Ionic hydrides: BaH2
Metal hydrides: PdH0-9
Molecular hydrides SiH4 NH3 AsH3 HI
2.How to use the Lewis structures and valence shell electron pair repulsion theory,
to determine H 2 Se, P 2 H 4, H 3 O + structure?
Answer H2Se Broken line
P2H4 cones
H3O + cones
3 To write the three major chemical industrial hydrogen production laboratory
preparation of hydrogen gas equation and the easiest method?
Answer
4.
He abundance in the universe ranked second, why He concentration in the
atmosphere is very low?
Answer： In space, due to light exposure, is likely to have a series of nuclear fusion,
nuclear fission took place alive, this time He is likely to have generated in the
atmosphere, due to temperature, pressure, light a number of conditions such as lack of
reaction can not occur。 Therefore, in the universe, He's a lot of content, while in the
atmosphere, He content is low,
5 Which noble gas can be used as low-temperature refrigerant? Which noble gas ion
potential is low, do the security needs of gas-discharge light sources? What is the
cheapest kind of noble gases?
Answer: Low-temperature refrigerant: liquid He
Low-ion potential: Xe
The cheapest: He
6 What is salt hydrides? What kind of elements can form salt hydrides?
to prove that salt hydrides exist in H Anion?
How
Answer Salt hydride is ionic hydride hydrogen with less electronegativity of
alkali metals and alkaline earth metals, when the direct chemical
combination, tend to get an electronic form a H-ions of this nature, similar to the
halogens, ionic hydride a strong reaction with water, emit hydrogen
gas
7.
Why is synthetic metal hydride always use the dry method? how many dm 3
of hydrogen gas can produce under 38kg for the role of aluminum hydride
with water (298K, 1.03 × 10 5 Pa)?
Solution
8.
How to purify the hydrogen which obtained from the reaction of zinc with
the acid? Write reaction equation.
LaNi 5 + 3H 2 == LaNi 5 H 6
Pressure in the (2-3) × 10 5 Pa, temperature is warm.
9.
Reaction equation for a trial of xenon fluoride XeF 6 and oxide XeO 3 of the
synthesis methods and conditions?
Answer
10. Write XeO 3 in acidic medium is I
Xe-ion reduction are the reaction equations.
Answer
11. De Batlle PtF 6 with Xe gas and the role of Xe obtained the first compound. In a
second experiment, PtF 6 of the initial pressure of 9.1 × 10 Pa, adding Xe until the
pressure of 1.98 × 10 Pa, reaction pressure of the remaining Xe is 1.68 × 10 Pa,
the product of the chemical formula calculation.
Solution
12. XeO 3 aqueous solution with Ba (OH) 2 solution, the role of generating a
white solid. This white solid mass fraction of each component are as follows:
71.75% of BaO, 20.60% of the Xe, and 7.05% of the O. Seeking chemical
formula of this compound.
Solution
71.75%
20.6%
:
153
131
:
7.05%
16
== 3: 1: 3
BaO: Xe: O == 3: 1: 3
The chemical formula of this compound is Ba 3 XeO 4.
13. Comparison of VB and MO France France molecular structure of XeF 2
treatment.
A VB method, see the book 232;
MO method, see the book 234.
14. Completion and balancing of the following reaction equation:
1. XeF 4 + ClO →
2. XeF 4 + Xe →
3.
4.
5.
6.
Answer
Na 4 XeO 6 + MnSO 4 + H 2 SO 4 →
XeF 4 + H 2 O →
XeO 3 + Ba (OH) 2 →
XeF 6 + SiO 2 →
① XeF 4 +2 ClO
+2 H 2 O = Xe + 2ClO
+ 4HF
③ 5Na 4 XeO 6 + 2MnSO 4 +7 H 2 SO 4 = 5XeO 3 +2 NaMnO4 + 7 H 2 O +
9Na 2 SO 4
⑤ 2XeO 3 +2 Ba (OH) 2 = Ba 2 XeO 6 + Xe + O 2 + 2H 2 O
Chapter 6
1. Thermodynamics of the ideal gas constant expansion can be the same, whether
it meant that the ideal gas constant expansion process does not do work?
Answer U=Q-W
Thermodynamics of the ideal gas constant expansion can be the same, but Q, W has
changed, so they are canceling each other out, that is, to absorb all the heat for doing
work.
2 Thermodynamic calculation of the system can change, is known:
1.
2.
System absorbs heat 1000J, do 540J of the power on the environment;
System absorbs heat 250J, the 635J environment on the system of
power;
Solution
3.
At 298K and 100kPa constant pressure, the 0.5 mol OF 2 react with the
water, release 161.5kJ heat, seeking
OF 2 (g) + H 2 O (g) → O 2 (g) + 2HF (g)
The △ rH
, And △ rU
.
Solution
4.
Reaction N 2 (g) + 3H 2 (g) → 2NH 3 (g) In the constant-volume calorimeter
not carried out within the release of energy generating 2molNH 3 release heat
82.7kJ, seeking Reaction △ rU And 298K when the reaction of △ rH .
Solution
5.
298K, look-up table when seeking the following reaction heat
1. 3NO 2 (g) + H 2 O (l) → 2HNO 3 (l) + NO (g)
2. CuO (s) + H 2 (g) → Cu (s) + H 2 O (g)
Solution
(1)
(2)
6.
N 2 O 4 thermal decomposition in the reactor, when the product generated
when there are 1mol NO 2, respectively, according to the following reaction
formula, the reaction progress were in the number of following formula:
1. N 2 O 4 → 2NO 2
2. N 2 O 4 → NO2
Solution
(1) 0.5mol (2) 1mol
7.
In a known bomb-type calorimeter burning 0.20molH 2 (g) generated H 2 O
(l), so that calorimeter temperature rise 0.88K, when 0.010mol toluene in this
calorimeter in the combustion calorimeter temperature rise of 0.615K, the
combustion reaction of toluene
C 7 H 8 (l) + 9 O 2 (g) → 7CO 2 (g) + 4H 2 O (l)
Seek the reaction of the △ rH
. Known △
H
(H 2 O, l) == - 285.8 kJ/ mol
Solution
△ rV
= △ rH
- △v RT = - 285.8 +1.5 × RT = - 282.1kJ/mol
8.
The following thermo-chemical reaction is known
Fe 2 O 3 (s) + 3CO (g) → 2Fe (s) + 3CO 2 (g) △ rH
= - 27.61 kJ · mol
3Fe 2 O 3 (s) + 3CO (g) → 2 Fe 3 O 4 (s) + CO 2 (g) △ rH
Fe 3 O 4 (s) + CO (g) → 3FeO (s) + CO 2 (g) △ rH
= - 27.61 kJ · mol
= + 38.07 kJ · mol
Solution
9.
Why is the standard state entropy of stable single quality non-zero?
Answer:
Crystal at 0K, any integrity in the arrangement of atoms or molecules is only
one way, that there is only a micro-state, its entropy value is 0, the standard in
the standard state entropy are a difference.
It is not 0
10 Analysis of the following reaction spontaneous temperature conditions
(1) 2N 2 (g) + O 2 (g) → 2N 2 O (g) △ rH
(2) Ag (s) +
Cl 2 (g) → AgCl (s)
(3) HgO (s) → Hg (l) +
O 2 (g)
(4) H 2 O 2 (l) → H 2 O (l) +
= + 38.07 kJ · mol
△ rH
△ rH
= - 127 kJ · mol
= +91 KJ · mol
O 2 (g) △ rH
= - 98kJ · mol
Solution
自发进行
11. The commonly used method of high purity nickel is thick nickel 323K and CO
reaction, the resulting Ni (CO) 4 by the purified decomposition at about 473K of pure
nickel
Ni (s) + CO (g)
Ni (CO) 4 (l)
Known reaction △ rH = - 161 kJ · mol , △ rS = 420J · k ·Mol .
Examination by the thermodynamic method of data analysis to discuss the
purification of nickel is reasonable.
Solution
Rough Nickel 323K and the CO reaction of Ni (CO) 4, to a liquid, it is easily
separated reactants, in the 473K decomposition of available high-purity nickel,
therefore, the above system, the method is reasonable nickel
12 The following are known to bond energy data
Bond N
N
N – F N – Cl
Bond energy / kJ · mol
F - F Cl - Cl
942 272,201,155,243
Examination of data derived from the bond energy standard heat of formation
to illustrate the NF 3 at room temperature, the more stable but explosive NCl 3.
Solution
The results show that the reaction is exothermic reaction, so NF3 stable at room
temperature
Because NCl3 for the endothermic reaction and a larger enthalpy of formation
is positive, indicating NCl3 very unstable, explosive
13. The following data are known to
△
△
H
(CO 2, g) = - 393.5 kJ · mol
△
H
(Fe 2 O 3, s) = - 822.2 kJ · mol
△
G
(CO 2, g) = - 394.4kJ · mol
G
(Fe 2 O 3, s) = - 741.0kJ · mol
Demand response to Fe 2 O 3 (s) + C (s) → 2Fe (s) +
temperature can occur spontaneously.
CO 2 (g) Under what
Solution
14. Look-up table seeking reaction CaCO3 (s) → CaO (s) + CO 2 (g) to
spontaneously carried out by the lowest temperature.
Solution
15. The following data are known to
△
H
(Sn, white) = 0
△
H
(Sn, Gray) = - 2.1kJ · mol
S
(Sn, white) = 51.5J · k
S
(Sn, gray) = 44.3J · k
· Mol
· Mol
Seeking Sn (white) and Sn (gray) of the phase transition temperature.
Solution
Chapter 7
1. What is the chemical reaction of the average rate, instantaneous rate? What's
the difference between the two kinds of reaction rates and contact?
Answer ：The average rate is defined as the time interval t, the concentration of
reactants or the resultant change in volume, while the instantaneous velocity is the
reaction rate at some point, the former is an average concept, as opposed to a period
of time is concerned, while the latter is a instantaneous concept, is relative to a certain
point in terms of
Respectively reactant concentration and the resultant concentration of the
following changes that the average rate and instantaneous reaction rate and
expressed with different concentration of material as shown in the relationship
between the reaction rate. This relationship is the average rate and
instantaneous rate will apply?
1. N 2 + 3H 2 → 2NH 3
2. 2SO 2 + O 2 → 2SO 3
3. aA + Bb → gG + hH
2.
Solution
V
=
V
=
(1) V
=
△[ NH 3 ]
△[ N 2 ]
△[ H 2 ]
=
=
△t
△t
△t
△[ NH 3 ]
lim
lim
△[ N 2 ]
△[ H 2 ]
=
=
△t → 0 △t
△t → 0 △t
△t → 0 △t
lim
1
V
3
H2
=
1
V
2
NH 3
Two kinds of rates apply.
(2) (3) (same with 1).
3.
Brief reaction rate theory, the theory of collision points.
Answer Collision theory that:
① reactant molecules inter-reaction of collisions are a prerequisite for the higher
collision frequency, the greater the reaction rate
② collisions can occur in response to a group of molecules must first have enough
energy to overcome the molecular infinitely close, the repulsion between the
electron cloud, resulting in rearrangement of atoms in molecules, that is, a
chemical reaction.
③ effective collision energy is a necessary condition but not sufficient, only when
the active molecule in the group to take the appropriate orientation of the various
elements for the collision, the reaction later.
④ activation energy: minimum energy efficient collision occurred
4 Brief reaction rate theory of the transition state theory points.
Answer
Transition state theory that:
① When both a sufficient average energy of the reactant molecules close to each
other when the chemical bonds in molecules to undergo rearrangement, the energy to
re-allocate, in the reaction process, go through the middle of a transition state, that is,
reactant molecules must first the formation of activated complex
② reaction rate and the following three factors: the concentration of activated
complexes, activated complex decomposition of the probability of activated complex
decomposition rate of
③ Application of transition state theory to discuss chemical reaction rate, the system
of reaction process can be expressed in response to potential changes in potential
energy - Course map
④ activation energy: reactant complexes and the activation energy difference between
the
5.
Reaction C 2 H 6 → C 2 H 4 + H 2, the initial stage of reaction is
approximately Class, 910K when the rate constant for the 1.13dm 1.5 · mol
· S . Tried calculating C 2 H 6 (g) pressure of 1.33 × 10 4 Pa when the
initial decomposition rate of (In [C 2 H 6] of the changes that have).
Solution
6.
295K, the reaction 2NO + Cl 2 →2NOCL ,its response relationship between
concentration and reaction rate are as follows:
[NO] / mol · dm
[Cl 2] / mol · dm
Cl2
/ mol · dm
0.100
0.100
8.0 × 10
0.500
0.100
2.0 × 10
0.100
0.500
4.0 × 10
·S
Q: (1) to different reaction order of reactants are?
(2) write a reaction rate equation;
(3) The reaction rate constants are?
Solution
7.
Reaction 2NO (g) + 2 H 2 (g) → N 2 (g) +2 H 2 O the rate equation for NO
(g) is quadratic, the H 2 (g) is a linear equation.
1. Write N 2 generation rate equation;
2. If the concentration of mol · dm - 3 that the reaction rate constant k,
the number of units?
3. Write NO concentration reduced rate equation, where the rate
constant k and (1) k-value is the same as the relationship between the
two k values is like?
Solution
8.
Imagine a response to aA + bB + cC →
Product, if the experiments show
that A, B and C increased the concentration of 1-fold, the entire reaction rate
increased to 64 times the original reaction rate; and if [A] and [B] remain
unchanged, only the [C]-fold increase in a , then the reaction rate increased to
four times the original; and [A], [B] was increased to four times in the
individual, its effect on the rate of the same. Seeking a, b, c values. This
may be a response to elementary reactions?
Solution
The reaction may be elementary reactions, it may not be elementary reactions
9. Carbon monoxide and chlorine at a high temperature role of phosgene (COCl 2),
the measured reaction rate equation is:
d [COCl 2 ]
== K [CO] [Cl 2]
dt
It was suggested that the reaction mechanism as follows:
Cl 2
2Cl
Cl + CO
COCl
COCl + Cl 2
1.
COCl 2 + Cl
Test illustrate this mechanism in line with the rate equation;
2.
Pointed out that the reaction rate equation in the k and reaction mechanism of
the rate constant (k 1, k , K 2, k ) relations.
Answer:
10. How to correctly understand the variety of reaction rate theory, the significance
of activation energy?
Answer In the collision theory, the activation energy is an effective collision can
occur by the activation of molecular groups with the minimum energy Na times, Na is
constant, in the transition state theory, the activation energy and embodies a kind of
energy difference, the reactants and activated complexes the energy difference .
11. High temperature decomposition of NO 2 to NO and O 2, the reaction rate
equation for -- v (NO 2) == k [NO 2] 2
In the 592K, rate constant is 4.98 × 10 dm 3 ·mol ·s , At 656K, its value
becomes 4.74dm 3 · mol s , Calculate the reaction activation energy.
Solution According to Allen, the formula you are
selfless
12. If a reaction activation energy is the 117.15kJ · mol , Asked at what
temperature the reaction rate constant k `value is the value of the 400K rate
constants 2 times.
Solution
13. Reaction N 2 O 5 → 2NO 2 + O 2, its temperature and rate constants of the data
columns in the table below, find the activation energy.
T/K
K/s
T/K
K/s
338
4.87 × 10
308
1.35 × 10
328
1.50 × 10
298
3.46 × 10
318
4.98 × 10
273
7.87 × 10
Solution
Ea=103.1KJ/mol
14. CO (CH 2 COOH) 2 decomposition in aqueous solution of acetone and carbon
dioxide, decomposition reaction rate constant of 283K when 1.08 × 10 - 4 mol · dm - 3 · s
- 1,
333K when 5.48 × 10 - 2 mol · dm - 3 · s - 1, trial calculation of 303K, the
decomposition reaction rate constant.
Solution First aim of the activation
energy
15. Are known to HCl (g) at 1.013 × 10 5 Pa and 298K when the heat is
generated - 92.3kJ · mol - 1, generate the activation energy for 1135kJ · mol - 1,
try to calculate the activation energy of reverse reaction.
16. The following statement you believe is right or not?
Reasons.
Reaction and reaction series is synonymous with the number of molecules.
The reaction course of the reaction rate constant speed step is the slowest
step.
3. The size of the reaction rate constant is the size of the reaction rate.
4. From the reaction rate constant of the unit can determine the reaction series.
1.
2.
Answer
(1) This is not correct, the reaction of the series is a macro concept, can be a fraction,
integer or even negative, while the number of molecules in response to a
micro-concept, can only be an integer
(2) This is not correct ,in a typical complex reaction, as reflected in constant speed
parallel to the steps for quick steps, while the opposite reaction of the control steps is
the slow reaction rate
(3) This is not correct ,reaction rate constant is only related to temperature, the
reaction rate constant is not only related to temperature stress with the concentration
of catalyst.
(4) correctly
17. Reaction 2NO (g) + 2 H 2 (g) → N 2 (g) + 2H 2 O (g) of the reaction
rate expression is v== K [NO 2] 2 [H 2], try to discuss the following
conditions on the initial rate of change in question.
1.
2.
3.
4.
5.
Doubling the concentration of NO;
With catalyst to participate;
Increase the volume of the reactor twice;
Increase the volume of the reactor twice;
To the reaction system by adding a certain amount of N 2.
Answer
(1) rate into the original four-fold
(2) reaction rate accelerated
(3) The reaction rate decreases
(4) rate of 0.125 times the original
(5) had no effect (constant volume); decreased (CV)
Chapter 8 Chemical equilibrium
1. How to understand equilibrium state in chemical reaction correctly?
Solution：Chemical equilibrium state is concentration of reactants and products not
follow to the change of time when the rates of the forward and reverse reactions
are equal
2．How to write the expression of experience equilibrium constant and standard
equilibrium constant correctly ?
Solution ： Experience equilibrium constant is the reversible reaction reached
equilibrium, the resultant concentration for power product of exponential for
stoichiometric coefficient in reaction equation and the concentration of reactants for
power product of exponential for stoichiometric coefficient in reaction equation of
the ratio is a constant at a certain temperature.
The experience equilibrium constant K has units generally, only when the sum of
the stoichiometric coefficient of reactants equal to the sum of the stoichiometric
coefficient of resultant, K is the dimensionless quantity.
a A + bB = gG + hH
[G ]g [ H ]h
K=
[ A]a [ B]b
The concentration and pressure of standard equilibrium constant are all a relative
value, are all relative to a standard value of C0 and the standard pressure P°,
whether gas, liquid or a heterogeneous reaction, K°are all dimensionless quantities.
3 ． Write out the expression of
following reversible reaction.
（1）2NOCl(g)
(2) Zn(s) + CO2(g)
(3) MgSO4(s)
(4) Zn(s) + 2H+(aq)
(5) NH4Cl(s)
solution
equilibrium
2NO(g) + Cl2(g)
ZnO(s) + CO(g)
MgO(s) + SO3(g)
Zn2+(aq) + H2(g)
NH3(g) + HCl(g)
constant Kc 、 Kp and
K in
4、Given equilibrium constant in following reaction.
HCN
NH3 + H2O
H2O
H+ + CN −
NH +4 + OH −
H+ + OH −
Calculate equilibrium constant in following reaction .)
solution
5 ． Can equilibrium constant represent conversion rate? How to know the
relationship of each other correctly ?
solution
(Equilibrium constant is the relationship of all substances,s concentration
when the system reach to equilibrium at definite temperature while conversion
rate is oneself some substance and its equal transformed divide primary
quantity .)
6. At 699K the equilibrium constant Kp for the reaction H2(g) + I2(g)
2HI(g)
is 55.3 , if
2.00mol H2 and 2.00mol I2 is placed in a 4.00dm3 container at 699K,How many HI
formation when the reaction reach to equilibrium at 699K?
solution：
(at the same system,V、R
andT are equal. )
7．H2 + CO2
H2O + CO reach to equilibrium at 1259K,
Equilibrium: [H2]=[CO2]=0.44mol·dm −3 ，[H2O]=[CO]=0.56mol·dm −3 .
Calculate experience equilibrium constant and Initial concentrations of H2 and CO2
at 1259K.)
solution：
（Initial）
8. For the reversible reaction H2O + CO
H2 + CO2 found equilibrium in a
closed container and the equilibrium constant Kc is 2.6 at 749K.
(1) Calculate : n（H2O）／n（CO）=1, equilibrium conversion rate of CO;
(2) Calculate : n（H2O）／n（CO）=3, equilibrium conversion rate of CO;
(3)From the calculated results show that the effects of concentration on the balance of
movement.)solution： （1）H2O + CO
a-x
b-x
x
H2 + CO2
x
x
=0.617
a
Therefore, equilibrium conversion rate of CO is 61.7%.
x2=2.6(a-x)2
⇒
（2）H2O + CO
n
n-x
3n
3n-x
x2
=2.6
(n − x)(3n − x)
H2 + CO2
0
0
x
x
⇒
x
=0.865
n
Therefore, equilibrium conversion rate of CO is 86.5%.
9. Decomposition reaction of HI is 2HI
H2 + I2..At the beginning,there is
1mol HI while Decomposed HI have 24.4% at equilibrium, Now we will drop
decomposition percentage to 10%, Computing system should be added to this
balance the number of mol I2. )
solution：
2HI
H2 + I2
(Initial) 起始
(Change ) 转化
1
0.244
0
0.122
0.122 2 (0.05 + x)0.05
=
0.92
0.756 2
0
0.122
⇒ x=0.37mol
10. SO3 is dissociated partly and produce SO2 and O2 at 900K and 1.013×105Pa:
1
SO2（g）+ O2（g）
SO3（g）
2
If density of mixture is 0.925g·dm-3 at equilibrium, Calculate the degree of
dissociation of SO3 .)
solution： PV=nRT
M=
ρRT
P
=
P
m
ρ
PM= ρ RT
=nRT
0.925 × 10 3 × 8.314 × 900
=68.33
1.013 × 10 5
SO3（g）
SO2（g）+
x
0
a-x
x
80(a − x) + 64 x + 32 ×
a−x+x+
1
x
2
1
O2（g）
2
0
1
x
2
1
x
2 = 68.33
x
= 0.34
a
Therefore, the degree of dissociation of SO3 is 34%.
11. N2O4 has 27.2% are decomposed into NO2 at 308K and 1.013×105Pa.
(1) Calculate the K θ for the following reaction
N2O4（g）
2 NO2（g）
(2) Calculate the rate of dissociation of N2O4 at 308K and 1.013×105Pa.
(3) From the calculated results show that the effects of Pressure on the balance of
movement)
（1） N2O4（g）
solution：
0.272
Kθ =
2 NO2（g）
0.544
0.544 2
=0.32
1 − 0.272
n P
=
, Similarly, the rate of dissociation of N2O4 is 19.6%.
V RT
(3) Increasing pressure，Balance to move to the direction of the volume decrease
Reduced pressure，Balance to move to the direction of increasing the volume)
（2） PV=nRT
12. For the reaction PCl5
PCl3（g）+ Cl2（g）
PCl5（g）decomposed up to a balance at 523K.
Equilibrium concentration: [PCl5]=1 mol·dm −3 ,[ PCl3]=[ Cl2]=0.204 mol·dm −3 .
If the temperature constant and the pressure reduced by half, what is the concentration
of various substances in the new balance system?
solution： K θ =
PCl5
0.204 2
=0.041616
1
PCl3（g）+ Cl2（g）
1-x
0.204+x
0.204+x
(0.204 + x) 2
0.204 =
=0.041616
1− x
2
x=0.135mol·dm −3
13.At 698K the equilibrium constant (Kc) for the reaction
H2（g） + I2（g）is 1.82×10 −2 .If HI（g）is placed in a reaction
2HI（g）
flask,question:
(1)If the concentration of HI is 0.0100mol·dm −3 , calculate the concentrations of H2
and I2.
(2) what will the initial concentration be for HI（g）?
(3) what is the conversion rate of HI at equilibrium?
solution： (Solution:)
（1）2HI（g）
H2（g） + I2（g）
2x
x
x2
=1.82×10 −2
0.01 − 2 x
x
⇒ x=1.35×10 −3 mol·dm −3
[H2]和[ I2]各是 1.35×10 −3 mol·dm −3 ，1.35×10 −3 mol·dm −3 。
（2）[HI]=0.01+2×1.35×10 −3 =0.0127mol·dm −3
（3） α =
2 × 1.35 × 10 −3
×100%=21.3%
0.0127
14. SO2Cl2（g）
SO2（g）+ Cl2（g） K θ =2.4
at 375K
7.6g SO2Cl2 and 1.013×105Pa Cl2 are placed in a 1.0 dm—3 flask ,
Calculate the partial pressure of SO2Cl2、SO2 and Cl2 at equilibrium.)
solution： K θ =K p （P θ ) −△n
P so2 cL2 ×1=
7.6
RT
135
P so2 cL2 =6.8×104Pa
P SO2 =8.8×104Pa P Cl 2 =1.90×105Pa
15. At a certain temperature the equilibrium constant (K θ ) for the reaction
PCl（
5 g）
PCl（
+ Cl（
is 2.25. A quantity of PCl5 get into a airless bottle.
3 g）
2 g）
The partial pressure of PCl5 is 2.533×104Pa When up to equilibrium.Question:
(1) What is the partial pressure of PCl3 and Cl2 at equilibrium?(2) What is the pressure
of PCl5 before dissociation?(3) What is dissociation percentage of PCl5 at
equilibrium?
solution： （1）
PPCl3 .PCl2
θ
(P )
2
=2.25×
PPCl5
Pθ
P PCl3 =P Cl2 =7.60×104Pa
（2）P PCl5 =(2.533+7.6)×105Pa
7.6 × 10 4
（3）P PCl5 =
×100% = 75.25%
10.1 × 10 4
16. How to express chemical reaction isotherm?What is the relations between
standard equilibrium constant of chemical reaction and △rG θm ?
solution △rG m =△rG θm +RTlnQ
△rG m =0， △rG θm = —RTlnk
17. What is the relations between△ f G θ and△rG θ ?)
Solution: △rG θ = ∑ ν △ f G θ (生成物）—
∑ν △
f
G θ （反应物）
18. At 523K put 0.110 mol PCl5（g）into a 1dm—3 flask ,put up the following
equilibrium
PCl（
5 g）
PCl（
+ Cl（
the concentration of PCl（
is 0.050 mol·dm −3 .
3 g）
2 g）
3 g）
(1)What will the equilibrium concentrations be for PCl5 and Cl2?(2)What is Kc and
K θ at 523K?
solution：(Solution:)
（1） PCl5（g）
起始(Initial )
PCl3（g）+ Cl2（g）
0.11
0
0
平衡(Equil)
0.06
0.05
0.05
平衡时 PCl5 是 0.06mol·dm −3 ，Cl2 是 0.05mol·dm −3
(PCl5 is 0.06mol·dm −3 and Cl2 is 0.05mol·dm −3 at equilibrium)
（2）K C =
0.05 2
=0.042
0.06
由 K θ =K p （P θ ) −△n 得出 K θ =1.80
19. Look up chemical thermodynamics datasheet and calculate the K θ in following
reaction at 298K.
H2(g) + I2(g)
2HI(g))
solution： 由公式—△rG θm =RTln K θ
得出 K θ =627
( By the formula—△rG θm =RTln K θ draw a conclusion K θ =627)
20. (1) Calculate the K θ for the reaction NiSO4·6H2O(s)
NiSO4(s) +
6H2O(g)
(2)What is the equilibrium vapor tension of H2O to solid NiSO4 • 6H2O?
solution： （1） △rG θm =—773.6—6×228.4+2221.7
=77.7kJ·mol −1
由公式-△rG θm =RTln K θ
得出 K θ =2.4×10 −14
(By the formula-△rG θm =RTln K θ draw a conclusion K θ =2.4×10 −14 )
（2） (from)由 K p =P
H 2O
△rG θm =—RTln K θ
K θ =K p （P θ ) −△n
得出 H2O 在固体 NiSO4·6H2O 上的平衡蒸气压为 544Pa。
( Conclude the equilibrium vapor tension of H2O to solid NiSO4 • 6H2O is
544Pa).
21. What is the reaction quotient of chemical reaction？How to apply the relationship
between reaction quotient and equilibrium constant to judge the direction of reaction
and the direction of movement of chemical equilibrium?)
solution △rG m =△rG θm +RTlnQ
△rG θm =—RTln K θ
当 Q〈 K θ 时 △rG m 〈 0
(△rG m 〈0
When Q〈 K θ
当 Q = K θ 时 △rG m = 0
(△rG m = 0
when Q = K θ
当 Q 〉K θ 时 △rG m 〉0
(△rG m 〉0
when Q 〉K θ
正反应自发进行
the forward reaction is spontaneous〉
反应达到平衡，以可逆方式进行
take place in the way reversed at equilibrium)
逆反应自发进行
reverse reaction is spontaneous)
22. At a temperature the equilibrium constant (Kp) for the reaction
H2O(g) + CO(g)
H2(g) + CO2(g) is 1, at this temperature put 2 dm3
3.04×104Pa CO 、3dm3 2.02×105Pa CO2、6dm3 2.02×105Pa H2O（g）and 1dm3
2.02×105Pa H2 into 6dm3 container.
Which direction is the net reaction to be carried out?
Solution: Reaction is going to the reverse direction by calculating )
23. At a certain temperature and pressure there are quantities of PCl5 and Cl2 ,judging
degree of dissociation of PCl5 is to increase or decrease in the following
circumstances.(1)the volume of PCl5 become to 2 dm—3by decompression;(2)
Adding nitrogen ,so that the volume increased to 2dm 3 in the condition of keeping
pressure constant .(3) Adding nitrogen ,so that the pressure increased to double in the
condition of keeping volume constant(4) Adding nitrogen ,so that the volume
become to 2dm 3 in the condition of keeping pressure constant(5) Adding
nitrogen ,so that the pressure increased to double in the condition of keeping pressure
constant.
solution：
（1）increase
（2）increase
（3）constant
（4）reduce
（5）reduce
24. At 973K the equilibrium constant (K θ ) for the reaction
CO2(g) + H2(g)
CO(g) + H2O(g) is 0.64 , calculate the change in standard
free energy△rG θm at 973K. When partial pressure of various gases at system have
following seven groups.
Identified for the Gibbs free energy change of each component of partial pressure.)
Ⅰ
0.253
p CO = p H 2O
Ⅱ
Ⅲ
Ⅳ
Ⅴ
Ⅵ
Ⅶ
0.507 0.760 1.013 1.266 1.520 1.773
1.773 1.520
p CO2 = p H 2
1.266 1.013
0.760 0.507 0.253
(Table units is 105Pa .△G will be mapping the composition of the reaction mixture.
Which regions may occur and mark on the diagram in positive reaction .Which
regions may occur and mark on the diagram in reverse reaction)
0.253 2
=0.194
solution： Ⅰ K =
1.733 2
θ
△rG m =△rG θm + RTln K θ =kJ·mol −1
Ⅱ Kθ =
0.507 2
=0.111
1.52 2
△rG m =△rG θm + RTln K θ =3610.249—17782.590=—14.2kJ·mol −1
0.76 2
=0.360
Ⅲ K =
1.266 2
θ
△rG m =△rG θm + RTln K θ =3610.249+（—8264.670）=—4.7kJ·mol −1
Ⅳ
K θ =1
△rG m =△rG θm + RTln K θ =3610.249=3.6kJ·mol −1
Ⅴ Kθ =
1.266 2
=2.775
0.76 2
△rG m =△rG θm + RTln K θ =11.9kJ·mol −1
Ⅵ Kθ =
1.52 2
=8.988
0.507 2
△rG m =△rG θm + RTln K θ =21.4kJ·mol −1
Ⅶ Kθ =
1.773 2
=49.111
0.253 2
△rG m =△rG θm + RTln K θ =35.1kJ·mol −1
Chapter 9 electrolyte solution
1. Convert the following hydrogen ion concentration and hydroxide ion
concentration into pH and pOH.）
（1）[H+]=3.2×10 −5 mol·dm −3 ；
（2）[H+]=6.7×10 −9 mol·dm −3 ；
（3）[OH − ]=2.0×10 −6 mol·dm −3 ； （4）[OH − ]=4.0×10 −12 mol·dm −3 。
solution:
2. Convert the following pH and pOH into hydrogen ion concentration and hydroxide
ion concentration.)
（1）pH=0.24；
（2）pH=7.5；
（2）pOH=4.6；
（4）pOH=10.2
solution:Similar to the above subject.
3. Given the concentration of a weak acid is 0.010 mol·dm −3 at 298K. Measuring the
pH is 4.0.
Calculate K θa 、 α and K θa 、 α and pH of diluting to volume into a two-fold.）
solution: K θa is unchanged and is still 10-6when volume diluted into a two-fold.( K θa is
only related to temperature).）
4. How much dilution to volume for 1.0dm30.20 mol·dm −3 HAc solution in order
to make Hac dissociation degree of one-fold increase compared with the original
solution?)
Solution:
(1L diluted to 4L, that is ,diluted fourfold , α increases double )
5. Calculate the concentrations of C2O 24− and HC2O −4 in mixed Solution of 0.10
mol·dm −3 hydrochloric acid and 0.10 mol·dm −3 H2C2O4.
solution：
6．Known K θ2 of H2SO4 is 1.2×10 −2 . Calculate the concentrations of all ions in
0.010 mol·dm −3 H2SO4. )
solution：
7 ． The mixed acid solution has 1.0 mol·dm −3 HF and 0.10 mol·dm −3 HAc.
Calculate the concentrations of H+，F − ，Ac − ,HF and Hac in solution.)
solution：
8．Calculate the range of effective pH in the following buffer solution
（1）HCO 3− —CO 32− ；
（3）H2PO −4 —HPO 24− ；
（2） HC2O −4 —C2O 24− ；
（4）HPO 24− —PO 34− ；
（5）H3PO4—H2PO −4
solution：(The range of effective pH is pKa Plus or minus 1)
( pKa must correspond to acid.)
9． Mix up 0.10 dm30.20 mol·dm −3 HAc with 0.050 dm30.20 mol·dm −3 NaOH.
Calculate pH of the Mixture
solution：
10．For the preparation of 0.50 dm3 、pH=9 and [NH] = 1.0 mol • dm buffer
solution.How much required the volume of concentrated ammonia water of
0.904g·cm −3 density and ammonia mass fraction of 26.0% ?How many grams of solid
ammonium chloride?
solution：
11．Mix up 1.0 mol·dm −3 Na3PO4 with 2.0 mol·dm −3 hydrochloric acid at equal
volume.
Calculate pH of the solution.)
solution：
12．Mix up 0.10mol·dm −3 a weak monoacid with 0.050 dm31.10 mol·dm −3 KOH. The
mixture was diluted to 0.10 dm3. Measured pH = 5.25 then calculate the K
θ
a
of the
weak monoacid.)
solution：
13 ． Calculate pH for 0.10 mol·dm −3 in the following concentration of all the
solutions)
（1）NaHCO3
solution：
(2) Na2S
(3)NH4Cl
(4) NaH2PO4
14. Write the conjugate acid of the following molecules or ions.)
SO 24− ， S 2− ， H2PO −4 ， NH3， HNO3 ，H2O
solution：(The conjugate acid, ,respectively.)
15. Write the conjugate alkali of the following molecules or ions.)
HAc，
H2O， NH3， HPO 24− ， HS −
（The conjugate alkali ,respectively.)
solution：
16．Giving examples to illustrate What kinds of common Reaction at the electronic
theory of acid and alkali.
solution：(KEY: The electronic theory of acid and alkali think that the essence of
acid-base reaction is the process of the formation of coordination bond and ligand
complex of acid and alkali.
It contains reactions between acid and alkali and substitution reactions which also
divide into acid substitution reaction, alkali substitution reaction and double
substitution reaction.)
17．Known Zn(OH)2 solubility product is 1.2 × 10.Calculate its solubility.）
solution：
18．Known the solubility of all the following salt at room temperature(expressed in
mol·dm −3 .) Calculate K θsp of all the salt.)
（1） AgBr(8.8×10 −7 mol·dm −3 )； （2）Mg(NH4)PO4(6.3×10 −5 mol·dm −3 )；
(3)Pb(IO3)2(3.1×10 −5 mol·dm −3 )
solution：
19．Put the solution of 0.10dm3 contains 0.040 mol·dm −3 I − into the solution of
0.10dm3 contains
2.0×10 −3 mol·dm −3 Pb2+ .Can generate PbI2 precipitate?)
solution：
Q=
( Therefore, it is generated precipitate.)
20 ． Mix
up
5.0×10
−3
dm
−3
0.20mol·dm
−3
MgCl2
solution
with
5.0×10 −3 dm −3 0.10mol·dm −3
NH3·H2O solution. Has the precipitation to produce? So that solution does not
precipitate
Mg (OH) 2 precipitation. How many grams adding solid NH4Cl in the solution at
least?
( Neglecte the changes of the volume of solution after adding solid NH4Cl).)
solution：
21． The 0.010mol CuS dissolved in 1.0 dm −3 hydrochloric acid. Calculate the
concentration of hydrochloric acid. Can hydrochloric acid dissolve CuS from the
results?)
solution： (if CuS dissolved in hydrochloric acid.)
(But the concentration of hydrochloric acid can not be achieved. Therefore,
hydrochloric acid can not dissolved in CuS)
22．A 0.10 dm −3 solution contains 0.0010mol NaCl and 0.0010mol K2CrO4 , what
the first precipitate by drops of AgNO3 solution?)
solution：
(AgCl precipitation need a few Ag+ so that it precipitated firstly.)
23．AgI can be dealt with by Na2CO3 to make it into Ag2CO3. What is the
conditions for transformation carried out in the end?According to the calculation
results to predict whether the conversion reaction carried out in the end.)
solution：
The results show that AgI is dissolved a small amount and needed the
concentration of Na2CO3 should greater than 9.1x1020[I-] 2+ 0.5[I-].This is
impossible to achieve virtually.So 要用 Na2CO3 处理 AgI，不能用沉淀转化的
方法，使 AgI 转化为 Ag2CO3。
24．If BaCO3 precipitation still contains 0.010mol BaSO4. In order to make BaSO4
fully into BaCO3, how many Na2CO3 Saturated solution of this precipitation
should be added to?)
solution：
25．The formation of insoluble salt MA of a weak monoacid alkali, the solubility
in pure water (without regard to hydrolysis) is 1.0×10 −3 mol·dm −3 , weak acid
K θa is 10 −6 and calculate the solubility of the salt solution in the[H+] maintained at
2.4×10 −6 mol·dm −3 .)
solution：
Chapter 11．oxidation-reduction reaction
1．Balance the following reaction formula for ion-electron method.
（1） PbO2+
Cl − ⎯
⎯→ Pb 2+ + Cl2
（2） Br2 ⎯
⎯→
BrO 3− +Br −
（酸性介质）(Acidic medium)
（酸性介质）(Acidic medium)
（ 3 ） HgS+2NO 3− +Cl − ⎯
⎯→ HgCl 24− +2NO 2
（碱性介质） (Alkaline
medium)
（4） CrO 24− +HSnO −2 ⎯
⎯→ HSnO 3− +CrO −2
（碱性介质）(Alkaline
medium)
（5） CuS +CN − +OH − ⎯
⎯→ Cu(CN) 34− +NCO − +S
（碱性介质）(Alkaline
medium)
Solution:
（5）2CuS +9CN − +2OH − =2Cu(CN) 34− +NCO − +2S 2− +H2O
2．Balance the following electrode reaction for ion-electron method.)
（1）MnO −4 ⎯
⎯→ MnO2
（碱性介质）(Alkaline medium)
（2）CrO 24− ⎯
⎯→ Cr(OH)3
（碱性介质）(Alkaline medium)
（3）H2O2 ⎯
⎯→ H2O
（4）H3AsO4 ⎯
⎯→ H3AsO3
（5）O2 ⎯
⎯→ H2O2(aq)
（酸性介质）(Acidic medium)
（酸性介质）(Acidic medium)
（酸性介质）(Acidic medium)
3．Having the following substances: KMnO4, K2Cr2O7, CuCl2, FeCl3, I2, Br2, Cl2, F2
Under certain conditions they can be as oxidant. According to electrode potential table
oxidation ability of these substances are arranged according to the order and write the
reduction product of them in an acidic medium
Solution: In an acidic medium , KMnO4, K2Cr2O7, CuCl2, FeCl3, I2, Br2, Cl2, F2 as
oxidants, Its reduction products wereMn2+, Cr3+, Cu, Fe2+, I-, Br-, Cl- ,F-, ϕ θ is larger,
the stronger oxidation ability. According to electrode potential table oxidation
capacity were arranged in descending sequence for F2 >MnO4 - > Cl2 >Cr2O7 2- >Br2
(aq)> Fe3+ >I2 >Cu2+ .)
3．Having the following substances: FeCl2, SnCl2, H2, KI, Li, Mg, Al. They could
serve as reductant. According to the standard electrode potential table reduction
ability of these substances are arranged according to the order. Write the oxidation
products in acid medium of them.)
solution： ϕ θ is smaller, the stronger reduction ability. In an acidic medium Fe2+, Sn2+,
H2, I-, Li, Mg, Al were oxidized to Fe3+, Sn4+, H+, I2, Li+, Mg2+, Al3+ . According to the
electrode potential reduction capacity were arranged in descending sequence for
Li>Mg>Al>H2 >Sn2+ > I-> Fe2+ .)
别
1
1
4．
（1） Cu(s) + Cl2(1.013×105Pa)
2
2
（ 2 ） Cu(s) +2H+( 0.01mol·dm-3)
1
Cu 2+ ( 1mol·dm-3)+Cl − ( 1mol·dm-3)
2
Cu
2+
( 0.1mol·dm-3)+H2
(0.9×1.013×105Pa)
(Battery symbols express following cell reaction.Calculate E and△rG at 298K in
order to show that whether reaction can occur from left to right spontaneously.)
solution
：
The reaction can not occur spontaneously from left to right.
5 ． Given Ag+ +e
3.5×10
−11
−
Ag, ϕ θ =+0.799V, Ag2C2O4 solubility product is
.Calculate standard electrode potential for Ag2C2O4+2e
−
2Ag+C2O 24− .)
solution：
Solution:The standard electrode potential for Ag2C2O4+2e −
the value of ϕ Ag+ / Ag when aC2O42- equals to 1.
6(Consequently;)
(consequently)
2Ag+C2O 24− is
7．Whether MnO 24− disproportionation reaction can occur spontaneously? Known the
θ
θ
= 0.56， ϕ MnO
standard electrode potential of lectrode couple: ϕ MnO
2−
−
/ MnO 2 −
/ MnO
4
4
4
2
=2.26V
Written reaction and the battery symbol.
solution：If it can occur disproportionation reaction and reaction equation is)
8. A mixture contains three ions Cl-,Br-,I- . We wanted I- oxidized to I2 without
oxidating Br-, Cl- between oxidant Fe2 (SO4) 3 and the KMnO4, select which one
can meet the above requirement.)
solution：
(Because the oxidation capacity of MnO −4 is too strong Cl-,Br-,I- were all
oxidized,but the oxidation capacity of Fe3+ is common.It can only oxidate I- and
can not be oxidized,so it should choose Fe2(SO4)3 .)
9． Known H3AsO3 + H2O
3I- = I 3− +2e − , ϕ θ =0.535V.
H3AsO4 + 2H+ + 2e-, ϕ θ = +0.559V;
Calculated the equilibrium constant for following reaction :
H3AsO3 + I 3− + H2O = H3AsO4 + 3I- + 2H+
If the solution of Ph = 7, what direction carried out of reaction?
If the solution of [H+]=6mol·dm-3 , what direction carried out of reaction?)
solution：
10． Known in alkaline media: ϕ Hθ
−
2 PO2
/ P4
= -1.82V；ϕ Hθ
−
2 PO2
/ P4
= -1.18V..Calculate
CLP on the P4-PH3 standard electrode potential, and judge whether P4 can occur
disproportionation reaction.
solution：
(According to the potential diagram of elements we come to a conclusion.)
(Therefore, the disproportionation reaction can occur.)
11． Using redox potential of the table to judge whether the following reactions
occur disproportionation reaction.)
(a)
2Cu
Cu + Cu2+
(b) Hg 22+
Hg + Hg2+
IO- + I- + H2O
(c) 2OH- + I2
(d) H2O + I2
HIO + I- + H+
solution: According to the potential diagram of elements we know that the
disproportionation reaction can occur.
(b) The reaction took place was synproportionation reaction rather than the
disproportionation reaction when the right greater than the left.
(c) According to the potential diagram of elements the disproportionation reaction
can occur.
in alkaline conditions.
(d) The reaction took place was synproportionation reaction rather than the
disproportionation reaction in acidic conditions.)
12． Giving a pressure of 1.013 × 105Pa of the hydrogen electrode and 90% argon gas.
1.013 × 105Pa pressure of the hydrogen electrode placed in hydrochloric acid.
Calculate the battery electromotive force E.
solution：In fact, such cells known as the concentration cell.
From H2(high concentration 1) to H2(low concentration 2)
Note:if E>0, The process is spontaneous.
In fact, H2 from high to lower concentrations diffused,itself is a spontaneous
process. So, it is verify that the calculation result of the above.)
13． Containing copper and nickel in acidic aqueous solution, its concentrations
were [Cu2+]=0.015mol·dm-3，[Ni2+]=0.23mol·dm-3，[H+]=0.72mol·dm-3，What the
kind of material discharge and Precipitation at first ?What the kind of material the
most difficult to precipitate?
solution：
(The higher the electrode potential, oxidation capacity is stronger, the more likely
precipitate.From the calculated results we known Cu precipitated firstly.)
14 ． Calculate the standard molar free energy change △rG θm in the following
reaction.)
MnO2 + 4H+ + 2Br − ⎯
⎯→ Mn2+ + 2H2O + Br2
(a) Br2 + HNO2 + H2O ⎯
⎯→ 2Br − + NO 3− + 3H+
(b) I2 + Sn2+ ⎯
⎯→ 2 I − + Sn4+
(c) NO 3− +3H+ +2Fe2+ ⎯
⎯→ 2Fe3+ +HNO2 + H2O
(d) Cl2 + 2Br − ⎯
⎯→ Br2 +2Cl −
solution： (a) MnO2 + 4H+ + 2e − ⎯
⎯→ Mn2+ + 2H2O
ϕ 正 =1.23V
2Br − —2e − ⎯
⎯→ Br2
ϕ 负 =1.0652V
E= ϕ 正 — ϕ 负 =0.1648V
△rG θm = —nEF= —31.806kJ·mol −1
(b) Br2 + 2e − ⎯
⎯→ 2Br −
ϕ 正 =1.065V
HNO2 + H2O—2e − ⎯
⎯→ NO 3− + 3H+
ϕ 负 =0.94V
E= ϕ 正 — ϕ 负 =0.125V
△rG θm = —nEF= —24.125kJ·mol −1
(c)I2 + 2e − ⎯
⎯→ 2 I − ϕ 正 =0.5355V
Sn2+ —2e − ⎯
⎯→ Sn4+
ϕ 负 =0.15V
E= ϕ 正 — ϕ 负 =0.3855V
△rG θm = —nEF= —74.407kJ·mol −1
(d) 2Fe2++ 2e − ⎯
⎯→ 2Fe3+
ϕ 正 =0.771V
HNO2 + H2O—2e − ⎯
⎯→ NO 3− + 3H+
ϕ 负 =0.94V
E= ϕ 正 — ϕ 负 =—0.169V
△rG θm = —nEF= 32.617kJ·mol −1
(e) Cl2 + 2e − ⎯
⎯→ 2Cl −
2Br − —2e − ⎯
⎯→ Br2
ϕ 正 =1.36V
ϕ 负 =1.065V
E= ϕ 正 — ϕ 负 =0.295V
△rG θm = —nEF= —56.935kJ·mol −1
15．
Known the standard electrode potential in following alkaline medium:
⎯→ Cr(OH)3(s) + 5OH − (aq)
CrO 24− (aq) + 4H2O(l) + 3e − ⎯
[Cu(NH3)]+ (aq) + e − ⎯
⎯→ Cu(s) + 2 NH3(aq)
ϕ θ = -0.11V
ϕ θ = -0.10V
Calculate ϕ θ ,△rG θm and K θ when H2 reduce CrO 24− and [Cu(NH3)]+ . What are the
great disparity
reasons between△rG
θ
m
and K? Although the two series of
ϕ θ approximation.)
solution： CrO 24− ：△ f G θ = +31.8kJ·mol −1 ；K θ =2.7×10 −6
[Cu(NH3)]+：△ f G θ = +11.6kJ·mol −1 ；K θ =9.3×10 −3
16．At 298K Sn2+ and Pb2+ and with their solution of metal powder balance. At
θ
=
low ionic strength solution [Sn2+]/ [Pb2+] equals to 2.98. Giving ϕ Pb
2+
/ Pb
θ
-0.126V and calculating ϕ Sn
.)
2+
/ Sn
solution： Sn + Pb2+
K θ =[Sn2+]/ [Pb2+]=2.98
Sn2+ + Pb
θ
θ
n(ϕ 正
— ϕ负
） 2 ×（ — 0.126 — x)
nE θ
lg K =
=
=
0.0591
0.0591
0.0591
θ
x == —0.14V
θ
ϕ Sn
2+
/ Sn
== —0.14V。
17．At 298K the equilibrium constant for the reaction Fe3+ + Ag
is 0.531.
Fe2+ + Ag+
θ
Giving ϕ Fe
= +0.770V and calculating ϕ θAg + / Ag .)
3+
/ Fe 2 +
θ
θ
n(ϕ 正
— ϕ负
） 1 ×（0.770 — x)
nE θ
=
=
solution： lg K =
0.0591
0.0591
0.0591
θ
x == 0.786V
ϕ θAg
+
/ Ag
== 0.786V
18． Blister copper film often contains impurities of Zn, Pb, Fe, Ag, etc. The blister
copper as anode and pure copper as cathode to carry out electrolytic refining which
can be got 99.99% purity copper. With the electrode potential shows how these four
impurities separated from copper .)
solution：
Electrode potential is relatively large, △rG θm is relatively small,so to make an
exhaustive go ahead.
19． Electrolytic cell of plus external voltage of the two electrode containing
CdSO4 solution and measured the corresponding current. The data is as follows:
E/V 0.5
1.0
1.8
2.0
2.2
2.4
2.6
3.0
I/A 0.002 0.0004 0.007
0.008
0.028
0.069
0.110
0.192
Drawing on the coordinate paper and finding the decomposition voltage.)
solution：
B
3.0
2.5
E/V
2.0
1.5
1.0
0.5
0.00
0.05
0.10
0.15
0.20
I/A
Draw a conclusion: E=3.0V
20． In a copper electrolysis experiment given 5000A current intensity and 94.5%
current efficiency.
How many kilograms electrolytic copper will be got after 3h (hours)?)
q
solution： I=
t
q=I t = 5000×94.5%×3×3600=5.013×10 7 （库仑）
Cu + 2e == Cu
2
64
7
m
5.013×10
m = 1.63×10 7 （千克）
Chapter 12．halogen
1．Which elements is the most lively in halogen? Why has a mutation in lively
changes from fluorine to chlorine？
KEY：The reactivity of elementary substance is sequence for F2>>Cl2>Br2>I2
The reactivity mutation from F2 to Cl2, the reasons attributed to the particularly
small radius of F atoms and F-ionic .
F Cl Br I
F— Cl—
Br— I—
r/pm 64 99 114 133 136 181
195 216
(1) Because the atomic radius of F is very small, the repulsion of F-F between atoms
and non-bonding electron pair are larger, so that F2 dissociation energy
(155KJ/mol) is much smaller than the dissociation of Cl2 (240KJ/mol).
(2) As the F-ionic radiu is very small, and therefore the formation of compounds, the
fluoride ionic bond is stronger and bond energy or lattice energy is larger)
2．The homologous of oxidation of halogen elemental and halide ions X reducing
and explain for the reasons.
KEY：The order of oxidative is F2 > Cl2 >Br2>I2 ; the order of reducing is I>Br->Cl->F-.
Halogen elementary substance is a very strong oxidizing agent ,with the atomic radius
increases, the oxidation ability of halogen followed by diminished. Although the
electron affinity of chlorine in the same family is the highest , but the strongest
oxidizing agent is fluorine. At room temperature a kind of oxidizing agent in water
use its standard electrode potential values indicated that oxidation capacity, , ϕ θ is
related to the following process .(see textbook P524).)
3．Written the reactions about chlorine and titanium, aluminum, hydrogen, water and
potassium carbonate and indicated the necessary reaction conditions.)
：
（1） 2Cl2+Ti =TiCl4
加热，干燥 (heating and drying)
solution：
（2） 3Cl2+2Al =2AlCl3
加热，干燥 (heating and drying)
（3） Cl2+H2 =2HCl
点燃 (light)
（4） 3Cl2+2P(过量)=2PCl3 干燥 (drying)
5Cl2(过量)+2P=2PCl5 干燥（drying）
（5） Cl2+H2O=HClO +HCl
(6) Cl2+2K2CO3+H2O=KCl+KClO+2KHCO3
4．To explain the following phenomena：
（1）I2 dissolved in CCl4 to get purple solution, while I2 dissolved in the ether get
reddish brown.
（2）I2 is difficult to dissolve in water but solube in KI.)
KEY：(1) CCl4 is a non-polar solvents, I2 is still in the molecular state after I2
dissolved in CCl4.
I2 elementary substance shows purple in the steam. But the ether is polar
solvents. I2 and solvents are solvation when I2 dissolved in ethyl ether
solvent adduct formation was no longer its single-vapor color, but was
red-brown.
(2) I2 exists molecular state , in the water disproportionation part is very few,
according to the principle of like dissolves like, non-polar I2 solubility in
water is small.However, in KI solution,I2 interaction with the I-generation
I3-ion, I3-ion solubility in the water is very large, so I2 soluble in KI
solution.)
5．Bromide can replace to the out of iodine from the iodine solution , iodine can
replace the out of bromine from potassium bromate solution , both of which are
contradictory? Why?
solution：
：Eθ（Br2/Br-）> Eθ（I2/I-）
(Therefore, Br2 solution replacement out of I2 from the I- ).
Br2+2I-===2Br-+I2。
θ
θ
E BrO
>
E
−
IO − / I
3 / Br
2
3
2
Therefore, I2 solution replacement out of
Br2 from the KBrO3 .
2BrO3-+I2===Br2+2IO-3
6．Why AlF3 melting point is up to 1563 K while the melting point of AlCl3 was only
463K?
KEY：Look up tables known the electronegativity of elements:Al-1.61 F-3.98
Cl-3.16.
Electronegativity difference is AlF3 2.37；AlC13 1.55.
Generally ,it believed that two elements of the compounds of electro-negativity
difference greater than 1.7 is called ionic compounds or less than 1.7 for the two
elements of the compounds is called covalent compounds. So MF3 is typical ionic
compound, its melting point is high; and AlCl3 is a covalent compound, its melting
point is low.)
7．From the following elements Potential diagram show that the reasons for chlorine
gas pass into the lime get bleaching powder while in the bleach solution by adding
hydrochloric acid can produce chlorine .
ϕ
θ
ϕ
θ
KEY:
HClO
A
ClO −
B
Because
+ 1.63V
Cl2
+ 0.40V
Cl2
pass
+ 1.36V
Cl2
Cl-
+ 1.36V
into
the
Cl −
lime
in
the
alkaline
medium,
and
because ϕ θCl /Cl >θClO /Cl , so Cl2 in alkaline conditions prone to disproportionation
-
2
-
2
reaction.while in the bleach solution after adding hydrochloric acid in the acidic
θ
conditions, ϕ θHClO/Cl > ϕ Cl
/Cl , so the following reaction is going to the right :
−
2
2
HClO + Cl- + H+ = Cl2 + H2O
8．Write the reaction of following preparation and indicate the conditions:
(1) Preparation of chlorine from hydrochloric acid;
(2) Preparation of hypochlorous acid from the hydrochloric acid;
(3) Preparation of potassium perchlorate from chlorate;
(4) The preparation of bromate from the sea.)
solution：
：(1)
(2)
Δ
4HCl(浓)+MnO2 ⎯
MnCl2+Cl2+2H2O
⎯→
Δ
MnCl2+Cl2+2H2O
4HCl(浓)+MnO2 ⎯
⎯→
2Cl2+H2O+CaCO3 → CaCl2+CO2+2HClO
（3）
Δ
KClO3 ⎯
3KClO4+KCl
⎯→
High potassium chlorate solubility
smaller than potassium chloride ,
detachable.
KClO4+H2SO4=KHSO4+HClO4
(1)
Cl2+2Br-=2Cl-+Br2
3Na2CO3+3Br2=5NaBr+NaBrO3+3CO2
5Br-+BrO3-+6H+=3Br2+3H2O
9．Three bottles white solid lose tags, which are KClO, KClO3 and KClO4, what
methods to identify?
KEY:Put a small amount of solid into dry test tube and do the following experiment:
Add dilute hydrochloric acid that is released Cl2 gas is KClO;
KClO+2HCl=KCl+Cl2+H2O
Adding concentrated hydrochloric acid with the release of Cl2 and Yellowing
solution is KClO3; 8KC1O3+24HCl(thick)=9Cl2↑+8KCl+60ClO2(yellow)+12H2O
Another was KClO4 .).
10． which means can get Hydrogen halide? What is the meaning of all methods
in practical applications?)
solution：：CaF2+H2SO4=CaSO4+2HF↑, With fluorspar as raw materials, reaction
carried out in the distillation and platinum kettle.)
NaCl +H2SO4(浓)=NaHSO4+HCl
3NaBr + H3PO4(浓)=Na3PO4+3HBr
3NaBr + H3PO4(浓)=Na3PO4+3HBr。
11．There is a white solid which may be one or both of the mixture:KI, CaI2, KIO3,
BaCl2 ,judge according to the following experiments the composition of a white solid.
(1)A white solid soluble in water are colorless solution; (2)To this solution by adding
a small amount of dilute H2SO4, the solution to turn yellow and get a white precipitate,
meeting starch turns blue immediately; (3)By adding NaOH to alkaline to the blue
solution, the blue disappears and the white has not disappeared.
KEY:The white solid was composed of: CaI2, KIO3 mixture.)
12. what are the rules of two kinds of halogen atoms and oxidation number in the
halogen intermetallic compounds? For examples.
KEY: Two kinds of halogen atoms in the halogen intermetallic compounds is not
arbitrary, the central atom is only one which the electronegativity is small and the
radius is large; ligand number is odd, usually the halogen of large electro-negativity
and small radius, such as ICl, ICl3, IF5, IF7 etc.The number of ligands is related to the
two kinds of halogen-radius ratio, the greater radius ratio ,the more the number of
ligands .However, the number of ligands can not be more than 7, such as iodine and
fluorine can be formed the seven ligand compounds IF7, but the iodine and chlorine
can only form 3 Ligand intermetallic compounds ICl3.The central atom oxidation
number of halogen intermetallic compounds not only related to the radius ratio, but
also related to the two kinds of elements electronegativity difference and
electro-negativity difference.The central atom oxidation number of halogen
intermetallic compounds not only related to the radius ratio, but also related to the two
kinds of elements electronegativity difference and larger electro-negativity
difference,the center can have a higher oxidation number. This is in accord with the
coordination number of the resulted radius ratio.）
13.What thermal Decomposition of Law for polyhalide? Why fluoride is not easy
exist in polyhalide in general?
KEY: The products of multi-halide decomposition generated which can be as large as
possible in the halide lattice. If CsBrCl2 thermal decomposition products as CsCl and
BrCl are not CsBr and Cl2.Fluoride usually difficult to exist in the multi-halide, in
other words, there are poor stability of the multi-halide to participate in the fluoride.
The reason is that F-radius are very small, F electronegativity is large, decomposition
products of multi-halide MF crystal lattice can be particularly large, far more stable
halides.）
14. What is Pseudohalogen? For example several important Pseudohalogen.
KEY: Some one net negative charge anions in the formation of ionic compounds and
covalent compounds, showing similar properties with halogens. In a free state, its
nature is very similar with halogen simple substance, so we call it Pseudohalogen.)
15. By the nature of the comparison of (CN) 2and Cl2 shows that the basic nature of
the halogen.
KEY: ① the free state are all volatile.
② With hydrogen to form acids, in addition to the hydrocyanic acid the majority of
acidity is strong.
③Compounded to metal form the salt
④Reacted with alkali and water and similar with halogen
⑤Formation to complexes similar with the halogens.
⑥Pseudohalogen the same as halide ions are also reductive.)
16. A pure soluble iodide 332mg dissolved in dilute H2SO4, adding accurate weighing
0.002mol KIO3 in solution, the boiling to remove the reaction of iodine, then add
sufficient quantities KI in solution, KI reacted with excessive KIO3 , and then titrated
with sodium thiosulfate ions to form I 3− , spented 0.0096mol sodium thiosulfate by
calculating ,asked what the original compound?
KEY: The original compound is KI.)
17．Make use of thermodynamic data (refer to the books) calculated the approximate
temperature with preparation for HCl reaction proceed to according to the
following reaction equation.)
2Cl(g)+2H20(g)=4HCl（g）+O2（g）
solution：
：△H=4×(-92.31)-2×(-241.82)=114.4KJ/mol
△
S=4×186.8-2×188.7=369.8J/mol.K
T≥△H/△S=114400/369.8=309.4K
18.Using reaction express the following reaction process and indicate the reaction
conditions:
(1) Using excessive HClO3 treat with I2;
(2) chlorine long pass into the KI solution;
(3) chlorinated water dripping KBr and KI mixture.
Answer: (1) purplish black or brown I2 disappeared and a Cl2 gas is generated.
2HCl03 + I2 === 2HI03 + C12
(2) The first,I2 generated and the colorless solution turned to yellow, orange and
brown until the purplish black precipitation precipitated, finally purplish black
precipitate disappeared and got a colorless solution.
Cl2 +2 KI === 2KCl + I2
5Cl2 + I2 +6 H20 === 2HI03 +10 HCl
(3) The solution turn to yellow or orange at first and shallow to nearly colorless, and
finally become yellow or orange.)
19. Discussion forms on the various oxidation states of chlorine oxygen acids. And
explain acidity, thermal stability and oxidation of homologous and say the reasons.
Answer：HClO,HClO2,HClO3 HClO4
Acid:HClO4>HClO3>HClO2>HClO.
Chapter 13. oxygen group
1. Using molecular orbital theory described the bond, bond orders and magnetic
(paramagnetic and inverse magnetic) and relative stability in the following
species.
(1) O2+ (2 oxy cation)
(2) O2
(3) O2-(superoxide ions)
(4) O22- (peroxide ions)
Solution: in the below table
Species
molecular orbitals
bond mag relative
orde netic stability
rs
2
+
2
2
2
2
1
O2
2.5
forw reduce
KKσ 2 sσ *2 s σ 2 pπ 2 p y π 2 pz π *2 p y
in
ard
2
*2 2
2
2
1
1
O2
2
forw
KK σ 2 sσ 2 s σ 2 pπ 2 p π 2 p π *2 p π *2 p
turn ↓
ard
O21.5
forw
KK σ 22sσ 2*2s σ 22 pπ 22 p y π 22 p z π *22 p y π *12 p z
ard
2
*2
2
2
2
2
2
2O2
1
rever
KK σ 2 sσ 2 s σ 2 pπ 2 p y π 2 p z π *2 p y π *2 p z
se
y
z
y
z
2. what is the differences between heavy water and heavy oxygen water,? Write to
their formula. What use are they? How do I prepare?
A: The heavy water: D2O; heavy oxygen water: H 2 18O ; heavy water is neutron
moderator in nuclear power industry , heavy oxygen water is to study the chemical
reaction ,in particular, the tracer of mechanism of hydrolysis .)
3. Explain why is paramagnetic O2 molecules and anti-magnetic O3?
2
1
Answer: (1) O2 molecular orbital formula is KKσ 22S σ 2*S σ 22 p π 22 p π 22 p π 2* P π 2* P
X
Y
Z
Y
1
Z
This shows O2 molecule has two single-electron which has a paramagnetic. The O3 is
anti-magnetic, because of no single electron in O3 molecule , a large bond π exists
in π 34 in molecular.)
4. How preparation of O3 in the laboratory? What is its significance?
A: Preparation of ozone in the laboratory mainly relies on ultraviolet (<185nm)
exposure to oxygen or to oxygen through the electrostatic discharge device and the
access to a mixture of ozone and oxygen which containing ozone of up to 10%. Ozone
generator schematic shown in Figure 13-10. It is composed of two glass casing, the
intermediate glass walls inlaid with tin-anchor, outer tube outer wall around a copper
wire, When the foil and copper wire connected with high-voltage ,the wall between
the two tube take place silent discharge ( there is no spark discharge), part of 02
transformed into O3 .
5. why changed the painting placed dark and black after a long time? Why H2O2 can
be used to deal with? Write reaction equation.
A: The painting will be dark and hair dark after placed on long time ,the reason is oil
painting of the white pigment containing PbSO4, encountered H2S in the air will
generate PbS
PbSO4 + H2S = PbS (black) + H2SO4
treatment with H2O2 and white again, because the following reaction occurs
PbS + H2O2 = PbSO4 + H2O2 ).
6. Comparison the delivery transgender law of oxygen-group elements and halogen
elements in hydride in acid, reduced thermal stability .
A: The hydrides of oxygen group elements are H20, H2S, H2Se, H2Te
Halogen element hydrides are HF, HCl, HBr, HI
(1) Acid
H20 <H2S<H2Se<H2Te
HF <HCl<HBr<HI
Halogen elements in the acidity of hydride is far stronger than the hydrides of oxygen
group elements of the same cycle.
(2) reductive
H2O <H2S<H2Se<H2Te
HF <HCl<HBr<HI
Halogen elements in the reduction of hydride is weaker than the Chalcogen hydride of
same period .
(3) Thermal Stability
H2O>H2S>H2Se>H2Te
HF>HCl>HBr>HI
Halogen elements in the thermal stability of hydride is higher than the Chalcogen
hydride of same period which can be seen that the nature of homologous of the
hydride of oxygen group elements the same as the halogen element hydride.)
7. Comparison the homologous of the oxygen-containing sulfur and nitrogen acids in
acid, oxidation and thermal stability .
A: H2SO4 no stronger than HClO4 acid, H2SO4 oxidation is strong, HClO4 thermal
stability is strong.)
8. Why do SOCl2 either Lewis alkali or Lewis acid?
A:
MnO2 + 4 H + + 2e = Mn 2+ + H 2 O ϕ ϑ = 1.23V
H 2 O2 + 2 H + + 2e = 2 H 2 O ϕ ϑ = 1.776V
O 2 ( g ) + 4 H + + 2e = H 2 O 2
ϕ ϑ = 0.6824V
By
electrode
the
standard
potential
we
can
see
that
ϕ ϑ ( MnO2 / Mn 2+ ) < ϕ ϑ ( H 2 O2 / H 2 O)
∴ H 2 O2 can oxidated Mn2+ to be MnO2 ϕ ϑ ( MnO2 / Mn 2+ ) > ϕ ϑ (O2 / H 2 O2 )
∴ MnO2 can also be reduced to Mn2+ from H2O2 , H2O2 is oxidized to be O2.
So,to form a continuous cycle of reaction process, Mn2
decomposition of H2O2.)
+
can accelerate the
9. Talking the relationship among SO3, H2SO4 and the fuming sulfuric acid,to write
SO3 structural formula about solid and gaseous.
Answer: SO3 is H2SO4 acid anhydride, SO3 dissolved in water that is H2SO4:
S O3+H2O=H2SO4
Fuming sulfuric acid is after concentrated sulfuric acid absorption in SO3, sulfuric
acid has the color, And there is smoke.Namely, sulfuric acid containing excessive
levels SO3, was fuming sulfuric acid.
Gaseous SO3 molecule shows planar triangular structure, S as the sp2
hybrid ,molecules have a π 46 bond.
Solid SO3 is a multi-polymer, mainly SO4 tetrahedron sharing two vertices , oxygen
form to two kinds of structure both ring and chain.)
10. Write the name and structure of the following anions: S2O32-，S2O42-，S2O62-and
S2O82A; S2O32- thiosulfate;S2O42- dithionous acid radical ;S2O62- dithionic acid radical;
S2O82- peroxydisulfuric acid radical.)
11.Telling S-O bond strength variation rules of OSF2, OSCl2, and OSBr2 molecule,
and to explain the reasons.
A: The S - O bond strength of molecule: SOF2>SOCl2>SOBr2
The three structures are triangular shaped, S as the central atom, and there is a lone
electron pair.
(1) the greater x electronegativity, the stronger the ability to attract the electron, then
the electron density is lower around the S atom , the higher positively charged nature
of sulfur, the stronger S to O polarization, the greater S--O bond covalent composition,
the shorter the bond, so the stronger the S-O bond.
(2) S-O bond exists d-p feedback π bond , the electron density is small around S and
the stronger ability to attract electronics, the stronger bond. Electronegativity of
Elements F>Cl>Br, the electron density of molecules in around
S :SOF2<SOCl2<SOBr2)
12. Using Na2CO3 and sulfur as raw materials, how to get Na2S2O3, write down the
reaction equation.
Solution：S + O2 ==SO2
Na2CO3+ SO2 ==CO2+Na2SO3
Na2S + SO2 + H2O == Na2SO3 + H2S
2H2S + SO2 ==3S + 2H2O
Na2SO3 + S ==Na2S2O3
13. There are four agents: Na2SO4，Na2SO3，Na2S2O3，Na2S2O6 ,their labels have been
lost, design an easy way to identify them.
A: Take a small amount of reagent reacted with dilute hydrochloric acid respectively
(1) without any response to the phenomenon is the Na2SO4
(2)Having irritant gas released, but no precipitation is Na2SO3 or Na2S2O6
SO 32 + 2H + = SO 2 ↑ + H 2 O
−
S 2 O 62 = SO 24 + SO 2 ↑
−
−
And then dropping BaCl2 generated a white precipitate is Na2SO3, the other is
Na2S2O6
(3) irritant gas released at the same time there is precipitation is Na2S2O3 ).
14. By the preparation process of H2S to analyze some of its properties.
A: Preparation of H2S commonly used FeS or Na2S reacted with dilute H2SO4 or
concentration H3PO4, collected H2S gas in the fume hood. This can be judged H2S has
the following properties:
(1) volatile and small solubility, H2S solubility in water is small and volatile, and
therefore can be released H2S gas with dilute sulfuric acid .
(2) Strong reducing; can not be used concentrated sulfuric acid when Preparation H2S,
because of H2S can oxidate concentrated H2SO4
H 2 SO 4 + H 2 S = S ↓ +SO 2 ↑ +2H 2 O
(3) Toxicity:
H2S has toxic,H2S to be prepared within the fume hood, and exhaust to escape
absorption.)
15.A salt A dissolved in water, add dilute HCl, a pungent gas B produced. At the same
time there is a yellow precipitate C precipitates, B gas can bleach KMnO4 solution. If
pass Cl2 in the A solution, Cl2 solution disappearing and get D, D reacted with barium
salts, that is ,produced a white precipitate E which not soluble in dilute nitric acid.
Which substances for A, B, C, D, E exactly? Write each step of reaction equations.
A: A:Na2S2O3; B:SO2 C:S; D:H2SO4 or SO42-; E:BaSO4
Reaction equation for each step is
S2O32- +2H+=SO2+S+H2O
5SO2 + 2MnO4− + 2 H 2 O = 5SO42− + 2Mn 2+ + 4 H +
S 2 O32− + 4Cl 2 + 5H 2 O = 2 SO42− + 8Cl − + 10 H +
Ba 2+ + SO42− = BaSO4
16. Completion and balancing of the following reaction formula)
（1）H2S+H2O2 →
（2）H2S+Br2 →
（3）H2S+I2 →
（4）H2S+O2 →
（5）H2S+ClO −3 +H+ →
（6）Na2S+Na2SO3+H+ →
（7）Na2S2O3+I2 →
（8）Na2S2O3+Cl2 →
（9）SO2+H2O+Cl2 →
（10）H2O2+KMnO4+H+ →
（11）Na2O2+CO2 →
（12）KO2+H2O →
（13）Fe(OH)2+O2+OH- →
（14）K2S2O8+Mn2++H++NO −3 →
（15）H2SeO3+H2O2 →
Solution：
（1）H2S+H2O2=S+2H2O
H2S+4H2O2（过量）=H2SO4+4H2O
（2）H2S+Br2=2HBr+S
H2S+4Br2（过量）+4H2O=8HBr+H2SO4
（1） H2S+I2=2I-+S+2H+
（2） 2H2S+O2=2S+2H2O
（3） 3H2S+ClO3-=3S+Cl-+3H2O
（4） 2S2-+SO32-+6H+=3S+3H2O
（5） 2Na2S2O3+I2=Na2S4O6+2NaI
（6） Na2S2O3+4Cl2+5H2O=Na2SO4+H2SO4+8HCl
（7） SO2+Cl2+2H2O=H2SO4+2HCl
（8） 5H2O2+2MnO4-+6H+=2Mn2++5O2+8H2O
（9） 2Na2O2+2CO2=2Na2CO3+O2
（10）2KO2+2H2O=2KOH+O2+H2O2
（11）4Fe（OH）2+O2+2H2O=4Fe（OH）3
（12）5S2O82-+2Mn2++8H2O=10SO42-+2MnO4-+16H+
（13）H2SeO3+H2O2=H2SeO4+H2O
17. In standard conditions, 50cm3 of oxygen containing O3, if the containing O3
decomposed completely, the volume increased to 52 cm3. If decomposition of
pre-mixed gas pass into the KI solution, how many grams of iodine can precipitate??
What is the volume fraction of O3 in decomposition of pre-mixed gas?
Solution: 5.68mg, 8.1%)
18. 35.00 cm3 Na2S2O3 • 5 H2O solution contains 12.41 grams per liter, so that 50.00
cm3 I −3 solution just fade, calculate the concentration of iodine solution.
Solution: I3-+2S2O32-=S4O62-+3I1.75×10-2mol/L.)
19．In the following reaction △H mθ is 284.5kJ·mol-1 at 298K
3O2
2O3
Known the equilibrium constant of the reaction is 10-54,
Calculate△G mθ and △S mθ in the reaction.)
20. Use of electrode potential to explain the reasons for adding a small amount of Mn2
+
in the H2O2 can promote the decomposition of H2O2.
θ
A: ϕ MnO
2
/ Mn 2 +
= 1.23V
H2O2 as the oxidant
ϕ Hθ O
2
2
/ H 2O
=1.776V
H2O2 can be oxidized Mn2+ Mn2+ + 2H2O2 == MnO2 + 2H2O
H2O2 as a reducing agent ϕ Oθ 2 / H 2O2 = 0.68V〈 1.23V
H2O2 can restore MnO2
MnO2 + 2H2O2 == Mn2+ + O2 + 2H2O)
Chapter 14. nitrogen famliy
(1. Discussed the bonding situation of N2 and NO molecules with MO theoretical and
pointed out how much the bond between them?
KEY: N2 ： KK σ 2 sσ 2 s π 2 p π 2 p σ 2 p
2
*2
2
2
y
2
bond order is 3
z
NO： KK σ 2 sσ 2 s π 2 p π 2 p σ 2 pπ 2 p
2
*2
2
2
y
2
*1
z
bond order is 2.5.)
2. To explain the following issues:
(1) Although the electronegativity of nitrogen higher than phosphorus, but the
chemical properties of phosphorus more lively than nitrogen.
(2) Why Bi (V) oxidation capacity stronger than other elements of the same family?
Answer: (1) electronegativity of nitrogen higher than the P, but N less lively than P,
this is determined by different structures of both elementary substance. Radius of N is
small, but the N triple bond formed between atoms, Triple bond energy is high,
difficult to break, so N2 is not lively.
P has large atomic radius, p orbital overlap is very little among p atoms , do not
create multiple bonds. P-P single-bond energy is very small and is very easy to
disconnect.In particular, bond angle of the white phosphorus P-P-P is small and strain
is arge ,more vivid.
(2) Bi (v) oxidation capacity stronger than the other elements with multi-ethnic, there
has been full of 4f, 5d, and 4f, 5d to the shielding effect of electrons is smaller. 6s
have greater penetrating power, so 6s electron energy level lower significantly which
has the "inert electron pair effect"is not easy to lose, Loss of two 6s electrons Bi
(v )which more likely to be two electrons to form a more stable Bi3 +.
3.Compare acidity and alkaline of NH3、HN3、N2H4 and NH2OH from the molecular
structure.
Answer: NH3 structure is shown in the book 648
HN3 structure is shown in the book 658
N2H4 structure is shown in the book 655
NH2OH structure is shown in the book 658
Results: acidic HN3>NH3> N2H4>NH2OH
On the contrary,Alkaline )
4. Compare the nature of the following compounds:
(1) The oxidation of NO −3 and NO −2 ；
(2)The reaction situation of NO2、NO and N2O in the air；
(3) The reduction of N2H4 and NH2OH.
Answer: (1)The oxidation is NO2->NO3-;
(2) NO2 does not react with air; NO react with air at room temperature, can
produce red-brown smoke. N2O does not react with air.
(3):The reduction is N2H4 >NH2OH.)
5. Ammonium nitrate can have two kinds of thermal decomposition of the following
ways:
NH4NO3(s)=NH3(g)+HNO3(g)
ΔH θ =171kJ·mol-1
NH4NO3(s)= N2O(g)+2 H2O(g) ΔH θ =-23 kJ·mol-1
According to the view of thermodynamic, which way more likely to break down for
ammonium nitrate solids.
Answer: The possibility of decomposition of the latter is large; because ΔG = ΔH-TΔS,
we can see, the direction of the reaction depends mainly onΔH.)
6. How to remove:
(1) nitrogen contained in the trace oxygen;
(2) Thermal Decomposition obtained NO2 by melting NH4NO3 in the system which
was mixed with small amounts of NO;
(3) NO contained in the micro-NO2;
(4) trace amounts of NH +4 ions in solution.
Answer:
(1) to enable the gas through heated copper filings:
O2+2Cu=2CuO
(2) to enable the gas through the FeSO4 solution to remove NO
NO+FeSO4=Fe（NO）SO4
(3) to enable the gas through the water to remove NO2
2NO2+H2O=2HNO3+NO
(4) plus a small amount of NaNO2 solution and heated to remove NH4 +
NH4++NO2-=N2+H2O)
7. Write reaction equation in the following substances when heated:)
（1）NaNO3 （2）NH 4 NO3 （3）NH 4 Cl 和 NaNO 2 的混合物（mixtures）
（4） CuSO4 .5 H 2 O
Solution：
（5） Cu ( NO3 ) 2 .2 H 2 O
（6） NaN 3
Δ
(1)2 NaNO3 ⎯
⎯→
2 NaNO2 + O2 ↑
Δ
N 2 O ↑ +2 H 2 O
(2) NH 4 NO3 ⎯
⎯→
Δ
N 2 ↑ + H 2O
(3) NH 4 NO2 ⎯
⎯→
Δ
CuSO4 + 5 H 2 O
(4)CuSO4 .5 H 2 O ⎯
⎯→
Δ
Cu (OH ) NO3 + HNO3 + 2 H 2 O
(5)Cu ( NO3 ) 2 .3H 2 O ⎯
⎯→
Δ
Cu (OH ) NO3 ⎯
CuO + HNO3
⎯→
Δ
4 HNO3 ⎯
⎯→
4 NO2 ↑ +O2 ↑ +2 H 2 O
Δ
(6)3 NaN 3 ⎯
⎯→
4 N 2 ↑ + Na3 N
Δ
Na3 N ⎯
⎯→
6 Na + N 2 ↑
8．Which are isoelectronicity each other select from the following substances: C 2−2 ，
O2，O −2 ，O 2−2 , N2，NO，NO+，CN-and N2H −3
and to discuss their oxidation capacity strong or weak and acid-base strength.
Answer: Electron donors are C22- ，N2，NO+，CN-；O22-，N2H3- )
9． Complete the following reaction
N 2 H 4 + HNO2 ⎯
⎯→
Δ
NH 4Cl + HNO2 ⎯
⎯→
KI + HNO2 ⎯
⎯→
Δ
KClO3 + HNO2 ⎯
⎯→
Δ
KMnO4 + HNO2 ⎯
⎯→
solution：
Δ
N 2 H 4 + HNO2 ⎯
⎯→
HN 3 + 2 H 2 O
Δ
NH 4 + HNO2 ⎯
⎯→
N 2 ↑ +2 H 2 O + HCl
KI + HNO2 ⎯
⎯→ I 2 + 2 NO ↑ +2 H 2 O
−
Δ
ClO − 3 + 3HNO2 ⎯
⎯→
Cl − + 3 NO3 + 3H +
2 MnO4
−
Δ2
+ 5 HNO2 + H + ⎯⎯→
2 Mn 2+ + 5 NO3− 3H 2 O
10．Starting from sodium nitrate, write the reaction equation which prepared to
nitrite.)
solution：
：2NaNO3=2NaNO2+O2
NaNO2+HCl=HNO2+NaCl
11．Explain the phenomenon of the following reactions:
(1) Why NaNO2 will accelerate the speed of reaction of copper and nitric acid?
(2) Why is phosphorus reacted with KOH solution get PH3 gas which meet air puffs
of white smoke?
(3) Put AgNO3 solution into NaH2PO4 or Na2HPO4 solution would be precipitated
yellow Ag3PO4 sediment?
Answer: (1) NaNO2 reacted with nitric acid generate NO2, NO2 as catalysts play the
role of electron transfer, leaving the reaction rate of nitric acid and copper to speed up.
HNO3 + NO2− + H + = 2 NO2 + H 2 O
NO2 + e − → NO2 −
Can be seen through the NO2 to obtain the reductant electronic Cu, leaving the
reaction rate to accelerate.
2NO2+Cu=2NO2-+Cu2+
(2) Phosphorus reacted with KOH solution get PH3 gas contains a small amount of
P2H4, P2H4 in the air easier to take spontaneous combustion generated P2O3 and
puffed of white smoke.
(3) The solubility product constants of Ag3PO4 are much smaller than the AgH2PO4
and Ag2HPO4 that is, the solubility of Ag3PO4 is smaller, easily precipitate from the
solution..)
12． Complete the following reaction:
(a) P4 + HNO 3 ⎯
⎯→
(b) AsCl 3 + H 2 O ⎯
⎯→
(c) POCl 3 + H 2 O ⎯
⎯→
(d) P4 O10 + H 2 O ⎯
⎯→
(e) P4 O 6 + H 2 O ⎯
⎯→
(f) Zn 3 P2 + HCl(稀 ) ⎯
⎯→
：
solution：
(a)3P4 + 20HNO 3 + 8H 2 O = 12H 3 PO 4 + 20NO
(b)AsCl 3 + 3H 2 O = H 3 AsO 3 + 3HCl
(c)POCl 3 + 3H 2 O = H 3 PO 4 + 3HCl
3Δ
(d)P4 O10 + 6H 2 O ⎯HNO
⎯⎯
→ 4H 3 PO 4
(e)P4 O 6 + 6H 2 O = 4H 3 PO 3
(f)Zn 3 P2 + 6HCl(稀 ) = PH 3 + 3ZnCl 2
13．Try to explain why you can generate diatomic molecule nitrogen N2 and reasons
for the same family can not generate the diatomic molecule.
A:The radius of N is very small, but formed triple bond among the N atoms, Triple
bond energy is high, difficult to break, so N2 is far from lively.
atomic radius of P is very large, p-orbital overlap is very small among P atoms which
do not form multiple bonds.)
14．Point out the number oxidation of P in the following types of phosphate.
H3PO4,
H4P2O7,
H5P3O10,
H6P4O13,
H3P3O9,
H4P4O12,
H4P2O6,
H2PHO3, HPH2O2, H3PO5, H4P2O8,
solution：
：+5，+5，+5，+5，+5，+5，+4，+3，+1，+7，+6。
15．Describe the structure of NO −3 ，PO 3−4 ，Sb(OH) −6 .
A: NO3- : plane triangle; PO 3−4 : regular tetrahedron; Sb(OH) −6 : octahedron.)
16. There are two kinds of the P-O bond which lengths are 139pm and 162pm in
P4O10 respectively, try to explain different reasons.
Answer: There are two P-O bond in P4O10 molecules, one is-P = O, this bond length
which is short is only 139pm, another is P-O-P, the bond length is 162pm.)
17．How to identify the following groups of material?
（1） NH 4 Cl and NH 4 NO3 ；
（2） NH 4 NO3 and NH 4 NO 2 ；
（3） Na 3 PO 4 and Na 2 P 2 O 7 ；
（4） H 3PO 3 and H 3PO 4 ；
（5） H 3AsO 3 and H 3AsO 4 ；
（6） As 3+ 、 Sb 3+ and Bi 3+
Answer: (1) identification with AgNO3.
(2) with potassium iodide starch test paper
(3) identification with AgNO3.
(4) identification with AgNO3
(5) with potassium iodide starch test paper
(6) with Na2S solution.)
18．Write the following thermal decomposition products of oxygen-containing salts;
（1） Na2SO4·10H2O
（2） Ca(ClO4)2·4 H2O
（3） Cu(NO3)2·3 H2O
（4） Al2(SO4)3
（5） NaHSO4
（6） (NH4)2Cr2O7
（7） AgNO2
（8） Na2SO3
（9） KClO3
solution：
：
（1）Na2SO4，H2O；
（2）CaCl2，O2，H2O（3）CuO，NO2，H2O；
（4 ）
（5）
（6）N2，Cr2O3，
（7）Ag，NO2；
（8）Na2S，Na2SO4（9）KCl，O2
19．Compare the nature of As, Sb, Bi sulfides and oxides
A: As, Sb, Bi sulfide:
As2S3
As2S5
Sb2S3
Bi2S3
(1) Acid-base properties :gender Piansuan , acid,
amphoteric
alkali
(2) soluble:
insoluble
soluble
soluble
insoluble
(concentrated hydrochloric acid)
(NaOH)
soluble
soluble
soluble
insoluble
(Sodium sulfide) soluble
soluble
soluble
insoluble
(Na2SX)
soluble,
insoluble ,
soluble,
insoluble
As, Sb, Bi oxide:
As4O6
Sb4O6
Bi2O3
(Water) very slightly soluble insoluble
insoluble
(Acid)
soluble
insoluble
insoluble
( Alkali) soluble
soluble
insoluble
Sb2S5
amphoteric
soluble
soluble
soluble
insoluble ,
20．Write the reaction equation of Cl2 oxidize to Bi(OH)3 in alkaline media and used
ϕ θ to explain the reasons for reactions.
A: Cl 2 + Bi (OH ) 3 + 3 NaOH = NaBiO3 ↓ +2 NaCl + 3H 2 O
θ
Because of φ Cl
2 / Cl − > ϕ BiO
−
3
/ Bi ( OH )
, so the reaction can be carried out to the right.)
3
21．How to interpret causes of the solubility of As2O3 in hydrochloric acid with the
increasing concentration of acid decreases then increases.
A: When the acid concentration is very low, there is the following balance:
As2O3+3H2O=2As（OH）3
Increase the concentration of acid is not conducive to hydrolysis.
Acid concentration increased, the existence of the following reactions:
As2O3+8HCl=2H[AsCl4]+3H2O;
Increasing the concentration of hydrochloric acid is beneficial to the right of the
reaction.)
22．A compound A is a white solid, insoluble in water, heating violent decomposition
to produce a solid B and Gas C. Solid B is not soluble in water or HCl, but dissolved
in hot dilute HNO3, get a solution D and gas E. E is colorless, but turn red in the air.
D solution deal with HCl, may precipitate a white F.
Gas C does not work with ordinary reagents, but its reacts with hot magnesium metal
to get white solid G. Another role of G reacts with water getting a white solid H and a
gas J. Gas J to get moist red litmus paper turn blue, solid H can be dissolved in dilute
H2SO4 solution getting I.
compound A deal with H2S solution, getting a black precipitate K, colorless solution
of L and gas C. After filtration, solid-K was dissolved in concentrated HNO3 to get
gas E, yellow solid N and solution M. M handled with HCl to get precipitation F.
Solution of L in order to deal with NaOH solution is again Gas J.
Please indicate the name of the substance A, B, C, D, E, F, G, H, I, J, K, L, M, N
Answer: AgN3；B：Ag；C：N2；D：AgNO3；E：NO；F：AgCl；G：Mg3N2；
H：Mg（OH）2；J：NH3；I：MgSO4；K：Ag2S；L（NH4）2S；M：AgNO3
Ｎ：S
23．What is an example for inert electron pair effect? What is the reason for produce
this effect?
A: s 2 p 0 — 6 elements have the electron shell configuration , electron pair difficult to
participate in bonding and often form to +（+n-2) oxidation state of compounds and its
oxidation state + n compounds are either difficult to form or it's not stability, a
phenomenon known as inert electron pair effect.
Produce this effect was due to (1) large atomic orbital overlap is not good, (2) the
inner electron repulsion, especially after transition elements Ga and Tl and Pb are in
the most significant.)
24．Known the standard electrode potential of various oxidation states of phosphorus
was as follows in alkaline media:
H3PO4
− 0.23V
H3PO3
− 0.50V
P
+ 0.06V
PH3
Drawing oxidation state - free energy diagram of phosphorus in alkaline media..)
：
solution：
? G/kJ穖 ol
-1
200
150
100
50
0
-2
0
2
? ? ?
4
Chapter 15．carbon famliy
1.What allotropes of carbon elementary substance?How about its structural
characteristics and physical properties?
K: graphite, diamond, fullerenes.
2. How to get carbon dioxide gas in the laboratory? How to get carbon dioxide in
industry?
A: The Laboratory Preparation: CaCO3+2HCl=CaCl2+CO2 ↑ +H2O
⎯ ⎯⎯→ CaO + CO2 ↑ )
Industrial production method: CaCO3 ⎯高温煅烧
3. In Figure 15-6, the three straight lines intersect at one point, it is inevitable or
accidental, to try to discuss why.
A: This is inevitable.)
4. Adding an equal volume of 0.20 mol·dm −3 of the Na2CO3 solution to
0.20mol·dm −3 Mg 2+ and Ca 2+ solution,respectively, How about the situation of
precipitation?try to discuss its regularity.
A: generated Mg（OH）2 and CaCO3 ,respectively.)
5.What is the reason for that CCl4 less prone to hydrolysis and the SiCl4 prone to
hydrolysis?
Answer: C is the second cycle of elements, only the 2s, 2p orbital which the
maximum coordination number is 4 can bond, CCl4 which can accept coordinated the
water is not empty orbit ,therefore it is not hydrolyzed.
Si is the third cycle of elements, after the formation of SiCl4 still has empty 3d orbital,
d orbital to accept the lone electron pair of oxygen atoms of the water molecules to
form a coordination bond and hydrolysis.)
6. By calculating to answer the following questions:
(1) How can occur spontaneously of the reaction for generating water gas at 298K
1.01 × 105Pa? (2) With △rS θ value Judging whether to raise temperature on the
reaction of generation water-gas is beneficial? (3) how high the temperature about the
reaction system of generation water gas to reach equilibrium in 1.01 × 105Pa?
Solution: (1) Tip: judging from ΔGm<0.
(2) because of the reaction C + H 2 O( g ) = CO ( g ) + H 2 ( g )
And ΔS>0,since ΔG=ΔH-TΔS,so raising the reaction temperature is conducive to the
right to respond.)
7. Comparison properties of CO and CO2 .How to remove a small amount of CO2
which be contained in CO? How to remove a small amount of CO which be contained
in CO2?
K: through the red hot toner and the red hot CuO powder.
Tip:
α HCO =
32
α CO =
2−
3
[ H + ]K 1
[ H + ]2
;
;
α
=
HCO
[ H + ] 2 + [ H + ]K 1 + K 1 K 2
[ H + ] 2 + [ H + ]K 1 + K 1 K 2
−
3
K1 K 2
)
[ H + ] 2 + [ H + ]K 1 + K 1 K 2
8．Calculated percentages of H2CO3，HCO −3 ，CO 32− when the solution pH is equal to
4,8,12 .)
：pH=4
solution：
CO2 的饱和状态下，H2CO3 的 pH=3.9，HCO −3 的 pH=8.1，CO 32− 的 pH=12.6
pH=4 时，体系中主要成分为[H2CO3]，[CO 32− ]可忽略
(Solution: pH = 4
CO2 of the saturated, H2CO3 pH = 3.9, HCO −3 pH = 8.1, CO 32− pH = 12.6.
When pH = 4, the system main components is [H2CO3]，[CO 32− ]which can be
ignored.)
H2CO3==H+ + HCO −3
m
10 −4
n
10 −4 xn
=4.2×10 −7
m
(m+n=1)
所以
n
=4.2×10 −3
m
[H2CO3%]=99.4%，[ HCO −3 %]=0.06%
HCO −3 == H+ + CO 32−
0.6%
10 −4
x
10 −4 x
=K a2
0.6%
所以 x=2.4×10 −10 %。
pH=8，12 同理。
9. Olefins can be a stable presence while the silicon olefins H2Si=SiH2 is difficult to
exist. What are the reasons?
K: The radius of carbon atoms is smaller, in addition to the formation of σ bond, p
orbital which can form a stable bond has large overlap by side by side.Thus carbon
atoms can form a single-bond, double bonds or even triple bond, alkene and alkyne
hydrocarbons are all stable.
The radius of silicon atoms is large, p orbital overlap side by side is very little, it is
difficult to form a stable bond, thus the majority of silicon atoms can only form
single-bond, while difficult to form a double bond, silicon olefins is not stable exist.)
10. How to create allochroic silicagel starting from the sodium silicate?
K: Mix up the Na2SiO3 solution with the acid is generated gel.
Na2SiO3+2H+=SiO2.H2O+2Na+
Regulating the amount of acid and Na2SiO3, so that the resulting silica gel containing
8% ~ 10% of SiO2 .The gel put it aside and age a day later and wash soluble salts with
hot water.The gel in 60 ~ 70 ℃ after drying, slowly heated to 300 ℃ activation,
namely, to get porous silica gel. If it activated and dried silica gel prior to use COCl2
soaked, then dried and activated it to get allochroic silicagel.)
11．Chemical reaction equation for the preparation reaction of elemental silicon, the
main chemical property of elemental silicon and preparation and property of silicon
dioxide.)
：
solution：
SiO 2 + 2C + 2Cl 2 → SiCl 4 + 2CO
SiCl 4 + 2Zn → Si + 2ZnCl 2
12. What is zeolite molecular sieve? Discussion on its structural features and
applications?
K: The nature exist in some network-like silicate and aluminate which has a
cage-shaped structures, such uniform cages can adsorb a certain size molecules
selectively, this effect is called molecular sieve effect. Usually the natural silicate and
aluminate zeolites are called zeolite molecular sieve.
In A type zeolite, all the silicon-oxygen tetrahedron and aluminum-oxygen
tetrahedron linked through shared oxygen atoms into a multi-ring.)
13. What measures should be taken for preparation of SnCl2 solution in the laboratory?
What are their purposes?
K: Adding some Sn grain in order to prevent air oxidation to Sn4 +; adding dilute
hydrochloric acid to prevent hydrolysis of Sn2 +.)
14. What allotrope for elementary substance stannum, how about the the differences
of their properties ? What is similarities and differences of the extraction and the
properties of α - stannic acid and β - stannic acid?
A: stannums are: gray stannum, white stannum ,brittle stannum.)
15. Discussion on the basic principle of lead storage battery.
K: discharge: negative
PbSO4 + 2e − == PbSO4 + 2e −
Positive
PbO2 + 4H+ + SO 24− + 2e − == PbSO4 + 2H2O
PbSO4 + 2H2O == PbO2 + 4H+ + SO 24− + 2e −
charging: anode
PbSO4 + 2e − == PbSO4 + 2e − .)
Cathode
16 ． How many oxides of lead ? Talk its properties respectively. How to use
experimental methods proved Pb3O4 have different valence state of lead?
K: The lead oxide are: PbO2，Pb2O3，Pb3O4，PbO.)
：oxides: PbO2，Pb2O3，Pb3O4，PbO
solution：
应用以下反应：
Pb3O4 + 4 HNO3 = 2 Pb（NO3）2 + PbO2 ↓ +2 H 2 O
17. A solution contains Sn(ClO4)2 and Pb(ClO4)2 oscillated with an excess of Sn-Pb
alloy powder in the 298K, measured equilibrium concentration of the solution
[Sn2+]/[Pb2+] is 0.46.
θ
θ
Known ϕ Pb
=- 0.126V, calculate the value of ϕ Sn
.)
2+
2+
/ Pb
/ Sn
： Pb + Sn
solution：
2+
= Sn + Pb 2+
0.05916 [ Sn 2+ ]
(Equilibrium) 达到平衡时,E=0,即 E = E +
lg
= 0;
2
[ Pb 2+ ]
θ
(So) 所以, ϕ θSn
ϕ θ Sn
2+
/ Sn
2+
/ Sn
− (−0.126) +
0.05916 0.46
lg
=0
2
1
= −0.136ev
18. Completed and balanced the following chemical reaction equation:）
(1) Sn+HCl→
(2) Sn+Cl2→
(4) SnCl4+H2O→
(3) SnCl2+FeCl3→
(5) SnS+Na2S2→
(7) Sn+SnCl4→
(9) Pb3O4+HI(过量)→
：
solution：
(6) SnS 32− +H+→
(8) PbS+HNO3→
(10)Pb2++OH-(过量)→
1)Sn + 2HCl = SnCl 2 + H 2 ↑
2)Sn + 2Cl 2 = SnCl 4
3)SnCl 2 + 2FeCl 3 = SnCl 4 + 2FeCl 2
4)3SnCl 4 + 3H 2 O = SnO 2 ⋅ H 2 O + 2H 2 SnCl 6
5)SnS + Na 2 S 2 = Na 2 SnS 3
6)SnS 32− + 2H + = SnS 2 + H 2 S
7)Sn + SnCl 4 = 2SnCl 2
8)3PbS + 8HNO 3 = 3Pb(NO 3 ) 2 + 3S ↓ +2NO ↑ +4H 2 O
9)Pb 3 O 4 + 15I − + 8H + = 3PbI 24− + I 3− + 4H 2 O
2−
10)Pb 2+ + 4OH − (过量）= Pb（OH）4 ；
Pb（OH） 4
2−
+ ClO − = PbO 2 ↓ +Cl − + 2OH − + H 2 O
19. A white solid A which soluble in water produces a white precipitate B, B can be
dissolved in concentrated hydrochloric acid. If solid A dissolved in dilute nitric acid
getting a colorless solution C. Put AgNO3 solution into C, precipitation of a white
precipitate D. D be dissolved in ammonia solution to get E, E produces a white
precipitate D after acidification.
H2S gas will pass into the solution of C to produce brown precipitate F, F soluble in
(NH4)2Sx the formation of solution G. .Acidified solution of G and obtained a yellow
precipitate H.
HgCl2 solution by adding a small amount of solution C was got a white precipitate I,
continued to participate in the solution C, precipitate I gradually dimmed, finally
becomes a black precipitate J.
Key: What is A，B，C，D，E，F，G，H，I，J expressed?
Chapter 16
1. The data of first energy ionization (unit kJ·mol −1 ) about the second and the third
cycle given as the following table, and try to explain why the first energy ionization
on B and Al is lower than its left and right elements?
the second
Li
Be
B
C
N
O
F
Ne
cycle
energy
520
900
801
1086
1402
1314
1681
2081
ionization
the third
Na
Mg
Al
Si
P
S
Cl
Ar
cycle
energy
496
738
578
787
1012
1000
1251
1521
ionization
Solution: Because of the structure of valence band of B，Al element ground state atom
is ns2np1, the outermost layer electron easier be lost, and then form the full-up stable
structure.
2. How to prepare boroethane in experimental laboratory? What is the structure of
boroethane?
Solution: hydride-ion substitution method:
aether
3LiAlH4+4BF3 ⎯⎯⎯
→ 2B2H6+3LiF+3AlF3
aether
3NaBH4+4BF3 ⎯⎯⎯
→ 2B2H6+3NaBF4
the structure of B2H6 follows the textbook P778
3. Explain the boiling point consequence of trihalogen boron and trihalogen aluminum,
and indicate the structure of their steam molecule.
Solution: the boiling point consequence of trihalogen substance follows the textbook
P780
the steam molecule of all trihalogen boron are single molecular, the steam molecule of
AlF3 is single molecular, but the steam molecule of AlCl3 is dimeric molecule, its
structure follows the textbook P781
4. Drawing up the structure of B3N3H6 (inorganic benzene).
Solution:
5. How many kinds of chemical bond in B10H14 structure? How many every kind?
Solution: the structure of B10H14
6. Why is boracic acid a Lewis acid? How do write out the structure of borax? What is
the acid-base property about borax water solution?
Solution: boracic acid is electron-defect compound, there is a empty p orbit in center
atom B, and B can accept electron pair, therefore boracic acid is a Lewis acid;
The structure of borax follows the textbook P787;
It express strong basicity of borax water solution.
7. Try to express the preparing process from borax to the following compound with
chemical equation:
（1）H3BO3 (2) BF3
(3) NaBH4
Solution: after borax concentrated solution and concentrated sulfuric acid have
reacted, chilled it and generated H3BO3
Na 2 [ B4 O5 (OH ) 4 ] + H 2 SO4 + 3H 2 O = 4 H 3 BO3 + Na 2 SO4
8. How to do the preparing process from allume to (1) aluninum hydroxide, (2)
potassium sulfate, (3) potassium aluminate? Write out the chemical equation.
Solution: dissolve allume into water, add proper quantities KOH solution, there is Al
（OH）3 deposition;
Al3++3OH-=Al（OH） ↓
3
after filtering, atmometering, condensing, can get K2SO4
After Al（OH）3 and KOH have reacted, get colorless solution, filtering, atmometering,
get KAlO2
9. Write out the following chemical equation:
(1) Roasting solid mixture of natronite and aluminum oxide, crush up the inker clue
and input the water, generate white creaming;
(2) Liberation of gases when aluminum and hot concentrate NaOH solution reacting;
(3) Adding ammonium chloride to sodium aluminate solution, generating ammonia
gas, at the same time there is milk-white gelatinous precipitate in solution;
(4) Boron fluoride is inlet to natron solution.
Solution: 1）Na2CO3+Al2O3=2NaAlO2+CO2 ↑
NaAlO2+2H2O=Al(OH)3 ↓ +NaOH
（2）2Al+2NaOH+2H2O=2NaAlO2+3H2
（3）Na[Al(OH)4]+NH4Cl=Al(OH)3 ↓ +NH3 ↑ +NaCl+H2O
（4）8BF3+3Na2CO3+3H2O=2Na Al(OH)4+3H2 ↑
10.Why is that corrosion resistance of Al is stronger than Fe, as Al is in front of Fe,
even Cu in metal action sequential table? Why Cu is able to react with cool
concentrate nitric acid, but Al can’t?
Solution: Al with higher stability, can’t react further with oxygen and water, though Al
is active metal, here is single compact oxide overlaid on its surface.
In cool concentrated sulfuric acid or concentrated nitric acid, the surface of Al will be
inactivated which prevent Al from being attacked.
11. Why in elevated temperature sulfur and aluminum react and generate Al2S3, but in
water solution Na2S and aluminum-salt react, can’t generate Al2S3? Try to express
with chemical equation.
Solution: Al2S3+H2O=Al（OH）3 ↓ +H2S ↑
12. Discuss the different and the same points on structure of BN and blacklead, on
property of AlN and carbofrax.
Solution: in structure BN is the same to blacklead.
13. Explain why InCl2 is diamagnetic substance? Why does TlI3 not steady exist?
Solution: InCl2 is diamagnetic substance because that InCl2 is consist of In+ and In3+,
in other word InCl2 can be written In[InCl4], In+ and In3+ are not single electron.
TlI3 doesn’t steady exist because of inert electron pair effect.
14. Known ϕ
θ
Tl + / Tl
= —0.34V， ϕ
θ
Tl 3 + / Tl +
= 1.25V,calculate the value ϕ
298 K the reaction equilibrium constant about 3Tl+(aq)
θ
Tl 3+ / Tl
,and in
2Tl + Tl3+(aq).
Solution: due to 3ϕ θ
= 2ϕ θ
+ ϕθ
3
3
+
+
+
/ TI
/ TI
TI
TI
TI + / TI
solution
ϕθ 3 +
=0.72V
TI
/ TI
15. What similitude are there between compound of Tl（Ⅰ）and Ag(I), and explain the
causation.
Solution: There are similitude between compound of Tl（Ⅰ）and Ag(I), since that the
radial are similarity, electric charge are the same, polarization energy are near to I+
and Ag+;
(1) Slightly solubility of compound
TICl and AgCl are white insoluble compound, TI2S and Ag2S are insoluble compound
(2)Haloid photo-breakdown easier such as TICl, AgCl,etc
and there are some difference on them, for instance:
1) different stability of hydroxid
in room-temperature, AgOH is not stable, easier breakdown to Ag2O, but TlOH is
stable, TlOH solution is strong base, similar to KOH.
2) Ag+ easier generates stable composition, but Tl+ isn’t.
3) the oxidability of Ag+ is stronger, but Tl+ with determinate reducibility.
Tl+ can replace K+ to form alum, but Ag+ can’t.
16. Close:
（1）
1
O2
HNO3
Cl2
Ga
S
3
4
6
Na2CO3
5
Ga(OH) −4
H2O
（2）
1
2
I2
3
CO2
△
OH
Tl
H2SO4
4
−
8
5
△
6
△
HS −
7
Cl2
OH −
9
Solution:（1）1.Ga2O3 2.GaCl3 3.Ga(NO3)3 4. Ga2S3 5.energizing 6.OH − 7.Ga(OH)3
（ 2 ） 1.TlI 2.Tl2CO3 3.TlOH 4.Tl2SO4 5.TlCl 6.Tl2O 7.Tl2S 8.Tl2O3
9.TlCl3
17. What are the rules on exist state of non-metal corpus liberum in normal
temperature and pressure? What are properties on structure of non-metal corpus
liberum?
Solution: the exist states about non-metal corpus liberum in normal temperature and
pressure are basically gassy and liquidness, few are solidity.
Chapter17
1. Try to explain the gradual changing rule on chemical activity according to electric
layer configuration of alkali metals and alkaline-earth metals.
Solution: the chemical activity from strength to strength from the top down since the
number of electronic shell gradually increased, attractive power of atomic nucleus to
outmost electron gradually decreased.
2. Try to distinguish the similarity and distinction on chemical ability about lithium
and kalium; lithium and magnesium.
Solution: caustic potash is dissoluble, lithium hydrate is slightly dissoluble. lithium
and magnesium are similar on diagonal rule.
3. Sodium metallic is strong reducer, try to write out the chemical equation about it
react to the following substance.
H2O, NH3, C2H5OH, Na2O2, NaOH, NaNO2, MgO, TiCl4
Solution: 2Na+2H2O=2NaOH+H2
2Na+2C2H5OH=2C2H5ONa+H2
Na+MgO=Na2O+Mg
4Na+TiCl4=Ti+4NaCl
2Na+2NH3=2NaNH2+H2
2Na+Na2O2=2Na2O
4. Write out the chemical equation about sodium peroxide react to the following
substance:
NaCrO2, CO2, H2O, H2SO4(dilute)
Solution: ：2Na2O2+2CO2=2Na2CO3+O2
3Na2O2+2NaCrO2+2H2O=2Na2CrO4+4NaOH
5. Write out the main chemical ability and application of sodium hydroxide and
calcium hydroxide.
Solution: (1) react to ampholytic metal;
(2) react to nonmetal boron, silicon;
(3) when react to nonmetal as halide etc, nonmetal occur dismutation;
(4) be able to occur neutralization reaction with acid, generate salt and water;
(5) react to acidic oxide to generate salt and water.
the application of sodium hydroxide is as raw mineral materials and breakdown
metasilicate model.
the application of calcium hydroxide is in industrial to use its suspending liquid, i.e.
lime cream.
6. Write out the process of preparing sodium metallic, sodium hydroxide, sodium
peroxide, natronite as table salt is raw, and write out the chemical reaction.
Solution:
electrolyse
2 NaCl ⎯⎯⎯⎯
⎯
→ 2 Na + Cl2 ↑
electrolyse
⎯
→ 2 NaOH + H 2 ↑ +Cl2 ↑
2 NaCl + 2 H 2O ⎯⎯⎯⎯
emblaze
2 Na + O2 ⎯⎯⎯→
Na2O2
2 NaOH + CO2 → Na2CO3 + H 2O
7. The fusing point of alkaline-earth metals is higher than alkali metal, the hardness is
higher too, explain the causation.
Solution: The fusing point and hardness of alkaline-earth metals are higher than alkali
metal, since alkaline-earth metals have two bonding electron, alkali metal only has a
bonding electron, metallic bonding of alkaline-earth metals is stronger than alkali metal.
8. The generation of calcium burned in the air reacts to water, giving out a lot of heat
quantity and odor of ammonia can be smelled. Try to express the action by chemical
equation.
Solution: 3Ca+N2=Ca3N2；Ca3N2+3H2O=3Ca（OH）2+2NH3 ↑
9. Why does the chemical bond have more covalent feature when beryllium bonded
to other nonmetals, and the chemical bond of the other alkali-earth metals and nonmetals
formed has more ion feature?
Solution: In BeCl2 the Be-Cl bond as main the covalent feature, BeCl2 is covalent
compound, since the electronegativity of beryllium(1.57) is bigger, the radial of Be2+
(approx 31 pm)is smaller, its polarization energy is stronger. In MgCl2，CaCl2 the
chemical bond as main the ion feature, compounds are ionic compound, since the
electronegativity of other alkali-earth metals is smaller, the radial is much more bigger
than Be2+.
10. How to distinct and separate Be(OH)2 and Mg(OH)2; BeCO3 and Mg CO3; BeF2
and MgF2 by the property distinction of magnesium and berillium?
Solution: separate Be(OH)2 and Mg(OH)2 by Be(OH)2 is able to dissolve into NaOH,
Mg（OH）2 isn’t able to.
BeCO3 heated is difficult to break up, MgCO3 heated is easier.
separate BeF2 and MgF2 by BeF2 is able to dissolve into water, MgF2 isn’t able to.
11. Write out the preparing process of BaCl2, BaCO3, BaO, BaO2 as barite is raw.
Solution: ：BaSO4 + 4C == BaS + 4CO
BaSO4 + 4CO == BaS + 4CO2
BaS + 4H2O == Ba(HS)2 + 4Ba(OH)2
Ba(HS)2 + CO2 + H2O == BaCO3 ↓ + 2H2S
BaCO3 + 2HCl == BaCl2 + CO2 + H2O
BaCO3
△
BaO + O2
BaO + CO2
点燃
BaO2
12. Write out the chemical equation of added individually in sequence ammonium
carbonate, acetic acid, potassium chromate in BaCl2 water solution and CaCl2 water
solution.
Solution:
BaCl 2 + ( NH 4 ) 2 CO3 = BaCO3 ↓ +2 NH 4 Cl
CaCl 2 + ( NH 4 ) 2 CO3 = CaCO3 ↓ +2 NH 4 Cl
2 BaCO3 + K 2 Cr2 O7 + 2 HAc ⎯
⎯→ 2 BaCrO4 ↓ +2 KAc + H 2 O + 2CO2 ↑
2CaCO3 + K 2 Cr2 O7 + 2 HAc ⎯
⎯→ 2CaCrO4 ↓ +2 KAc + H 2 O + 2CO2 ↑
13. Suppose generate NaCl（s）by two ways, calculate individually △fH mθ NaCl(s) by
Geissler pump and compare the results. (the temperature is 298K)
1
(1) Na(s)+ H2O(l) ⎯
⎯→ NaOH(s)+ H2(g)
2
△rH mθ =-140.89kJ·mol-1
1
1
H2(g)+ Cl2(g) ⎯
⎯→ HCl(g)
2
2
△rH mθ =-92.31kJ·mol-1
HCl(g)+NaOH(s) ⎯
⎯→ NaCl(s)+H2O(l)
△rH mθ =-177.80kJ·mol-1
(2)
1
1
H2(g)+ Cl2(g) ⎯
⎯→ HCl(g)
2
2
△rH mθ =-92.31kJ·mol-1
Na(s)+ HCl(g) ⎯
⎯→ NaCl(s)+
1
H2(g)
2
△rH mθ =-318.69kJ·mol-1
so that both the results of ΔH are equally, they are –411kJ/mol, since the ΔH is
state function.
Chapter 18
1. explain the following appearance with reaction equation:
(1) It will generate a layer of copper green on the surface of copper ware if it is in the
moist air;
(2) Gold dissolves in chloronitrous acid;
(3) The color of CuCl2 concentrated solution will be changed from yellow brown to
green, and then to blue, when it is diluted by water;
(4) It will separate out white deposition when CuSO4 and NaCl concentrated solution
is inletted SO2;
(5) When adding KCN solution, AgNO3 solution will generate white deposition which
then will be disappeared, then adding NaCl solution there is not generating AgCl
deposition, however, if adding a little of Na2S solution, it will separate out Ag2S black
deposition.
Solution:
(1) 2Cu + O2 + H2O + CO2 = Cu2(OH)2CO3
(2) Au + 4HCl + HNO3 = HAuCl4 + NO + 2H2O
(3) [CuCl4]2- + 4H2O = [Cu(H2O)4]2+ + 4Cl[CuCl4]2-:yellow, [Cu(H2O)4]2+:blue ， the solution is green when both
[CuCl4]2- and [Cu(H2O)4]2+ coexist.
(4) 2Cu2+ + 2Cl- + SO2 + 2H2O = 2CuCl + 4H+ + SO42(5) Ag+ + CN- = AgCN
AgCN + CN-=[Ag(CN)2]2[Ag(CN)2]- + S2- = Ag2S + 4CN2. Explain the following experimental facts:
(1) It usually use ZnCl2 concentrated solution first to clean its surface when welding
iron sheet;
(2) Not dissolved in HCl，HNO3 and (NH4)2S, but HgS is dissolved in chloronitrous
acid or Na2S solution;
(3) Slightly dissolved in water, but HgC2O4 can be dissolved in the solution which
includes Cl − ion;
(4) There is not generating anhydrous CuCl2 when heat decompositing CuCl2·2H2O;
(5) Adding ammonia liquor can not generate NH2HgCl white deposition when HgCl2
solution contains NH4Cl.
Solution: (1) ZnCl2 + H2O = H[ZnCl2(OH)]
H[ZnCl2(OH)] can clean out the oxide on metal surface but not damaging its surface,
for its acidity is obvious, for example:
FeO + 2H[ZnCl2(OH)]2 + H2O
(2) HgS can’t dissolve in HCl solution because its solubility product constant is too
small
HgS can’t dissolve in HNO3 solution because HNO3 and it can combine to
indissolvable Hg(NO3)2HgS
HgS can’t dissolve in (NH4)2S solution because (NH4)2S solution hydrolyze to HS-,
and for the concentration of S2- is very low, they can’t generate composition
HgS can dissolve in chloronitrous acid, because chloronitrous acid and it can react to
HgCl42- and S
3 HgS + 8H+ + 2NO3- + 12Cl- = 3HgCl42- + 3S + 2NO + 4H2O
HgS can dissolve in Na2S solution because they react to dissoluble HgS22HgS + S2- = HgS22(3) HgC2O4 can dissolve in HCl, because the acidity of H2C2O4 is much lower than
HCl, and the composition of Cl- and Hg2+ is relatively stable.
(4) The is not CuCl2, because of the polarization ability of copper ion is strong, and
the HCl acid is volatile acid, and CuCl22H2O generates hydrolyzation when it heated
CuCl22H2O ════Cu(OH)Cl + H2O↑ + HCl↑
(5) There is not generate NH2HgCl white deposition because of NH4Cl has depressed
the generation of NH2-, and the solvability of NH2HgCl is high
HgCl2 + 4NH3 = Hg(NH3)42+ + 2Cl3. Selecting matching reagent to individual dissolving the following deposition, and
write out the relative equation.
（1）CuCl （2）Cu(OH)2 （3）AgBr
（4）Zn（OH）2（5）CuS （6）HgS
（7）HgI2
（8）AgI
（9）CuI
（10）NH2HgOH
Solution: (1) CuCl + HCl = HCuCl2
(2) Cu(OH)2 + 2OH- = Cu(OH) 42(3) AgBr + 2S2O32- = Ag(S2O3)23- + Br(4) Zn(OH)2 + 2OH- = Zn(OH)42(5) 2CuS+ 10CN- = 2Cu(CN)43- + 2S2- + (CN)2
(6) 3HgS + 8H+ + 12Cl- + 2NO3- = 3HgCl42- + 3S ↓+ 2NO↑ + 4H2O
(7) HgI2 + 2I- = HgI42(8) AgI + 2CN- = [Ag(CN)2]- + I(9) CuI + 2S2O32- = Cu(S2O3)23- + I(10)NH2HgO
4. Completing the following equations:
(1) Hg22+ + OH- →
(2) Zn2+NaOH（strong） →
(3) Hg2++NaOH →
(4) Cu2++NaOH（strong） →
(5) Cu2++NaOH →
(6) Ag++NaOH →
(7) HgS+Al+OH-（excess） →
(8) Cu2O+NH3+NH4Cl+O2 →
Solution: (1) Hg22+ + 2OH- = HgO↓ + Hg↓ + H2O
(2)Zn2+ + 2OH- = Zn(OH)2
Zn(OH)2 + 2OH- = Zn(OH)42(3)Hg2+ + 2OH- = HgO + H2O
(4)Cu2+ + 4OH- = Cu(OH)42(5)2Cu+ + 2OH- = Cu2O↓ + H2O
(6)2Ag+ + 2OH- = Ag2O↓ + H2O
(7)3HgS + 2Al + 8OH- = 3Hg + 2Al(OH)4- + 3S2(8)2Cu2O + 8NH3 + 8NH4+ + O2 = 4Cu(NH3)42+ + 4H2O
5. Summarizing the following synthetic process:
(1) From CuS to synthesis CuI;
(2) From CuSO4 to synthesis CuBr;
(3) From K[Ag（CN）2] to synthesis Ag2CrO4;
(4) From blister copper ore CuFeS2 to synthesis CuF2;
(5) From ZnS to synthesis ZnCl2 (non-aqueous);
(6) From Hg to prepare K2[HI4];
(7) From ZnCO3 to withdrawal Zn;
(8) From Ag（S2O3） 32+ solution to recover Ag.
Solution: (1) 3CuS + 2HNO3 + 3H2SO4 = 3CuSO4 + 3S↓ + 2NO↑ +4H2O
2CuSO4 + 4KI = 2CuI↓ + I2 + 2K2SO4
(2) CuSO4 + Cu + 2KBr = 2CuBr↓ + K2SO4
(3) 2K[Ag(CN)2] + Zn = Ag↓ + K2[Zn(CN)4]
3Ag + 4HNO3 = 3AgNO3 + NO↑ + 2H2O
2AgNO3 + K2CrO4 = Ag2CrO4↓ + 2KNO3
CuFeS2 + O2 = Cu2S + 2FeS + SO2↑
(4)
Cu2S + 22HNO3 = 6Cu(NO3)2 + 3H2SO4 + 10NO↑ + 8H2O
Cu(NO3)2 + 2NaOH = Cu(OH)2 + 2NaNO3
Cu(OH)2 + 2HF = CuF2 + 2H2O
ZnS + 2HCl + n H2O = ZnCl2nH2O + H2S
(5)
ZnCl2 nH2O = ZnCl2 + n H2O
3Hg + 8HNO3 = 3Hg(NO3)2 + 2NO ↑+ 4H2O
(6)
Hg(NO3)2 + 4KI = K2[HgI4] + 2KNO3
ZnCO3 + H2SO4 = ZnSO4 + CO2 ↑+ H2O
(7)
2ZnSO4 + 2H2O = 2Zn +O2↑ + H2SO4
Ag(S2O3)23- + 8Cl2 + 10H2O = AgCl↓ + 15Cl- + 4SO42- + 2OH-
(8)
AgCl + 2NH3 = [Ag(NH3)2]Cl
[Ag(NH3)2]Cl + HCOOH = Ag↓ + CO2 ↑+ H2O + NH3 ↑+ HCl
2[Ag(NH3)2]+ + Zn = 2Ag↓ + [Zn(NH3)4]22[Ag(NH3)2]+ + CH3CHO + 2OH- = CH3COONH4 + 2Ag↓ + 3NH3 ↑+ H2O
6. Manage to design a proposal on separating the following ions without using H2S:
Ag+， Hg 22+ ， Cu2+， Zn2+， Cd2+， Hg2+ and Al3+
Solution:
mixed solution
HCl
║
AgCl,Hg2Cl2
NH3•H2O
Ag(NH3)2+
CuI
HgNH2I,HgNH2Cl,Al(OH)3
NH3•H2O
Zn(NH3)42+,
HgNH2Cl +Hg
Zn ,Cd2+,HgI42-,Al3+
2+
Cd(NH3)4
2+
Cu2+,Zn2+,Cd2+,Hg2+,Al3+
KI（excess）
NaOH
║Δ
Zn(OH)2 ,Cd(OH)2
HgNH2I,HgNH2Cl
NaOH
Al(OH)4-
Cd(OH)2
2-
Zn(OH)4
7. After have dissolved 1.008g Cu-Au composition metal sample, add to excess iodine
ion, and then generate iodine by 0.1052mol·dm-3Na2S2O3 solution, and totally spend
29.84cm-3Na2S2O3 solution, try to calculate the percentage composition of copper in
composition metal.
Solution: 2Cu2+ + 2I- = CuI + I2
I2 + 2S2O32- = S4O62- + 2I2Cu2+ condign to I2, 2S2O32and so Cu2+equal to S2O32- : 0.1052×29.84×10-3
Cu％ = nCu·MCu/1.008×100％
= 0.1052×29.84×63.546×100％/1.008
=19.8％
8. Calculating the following electrode potential of half-cell reaction:
(1) Hg2SO4+2e-
2Hg+SO 24−
θ
（known ϕ Hg
=0.792V，K θSP ( Hg 2 SO4 ) =6.76×10-7）
2+
/ Hg
2
(2)
ϕ1θ
CuS
ϕ 2θ
Cu2S
ϕ 3θ
Cu
θ
θ
（known ϕ Cu
=0.15V， ϕ Cu
=0.52V，
2+ /
+
/ Cu +
/ Cu
K θsp ( CuS ) =7.94×10-36，K θsp (Cu2 S ) =1.0×10-48）
θ
θ
θ
（ ϕ CuS
/ Cu 2 S = —0.51V， ϕ Cu 2 S / Cu = —0.159V， ϕ CuS / Cu = —0.332V）
（1） Solution： Hg2SO4 + 2e- ＝ 2Hg + SO42Hg22+ + SO42- = Hg2SO4
Kθsp(Hg2SO4) = [Hg22+]·[SO42-]
instar [SO42-] = 1.0mol/dm3
[Hg22+] = Kθsp(Hg2SO4) = 6.76×10-7 mol/dm3
E(Hg22+/Hg) = Eθ(Hg2SO4/Hg)
E(Hg22+/Hg) = Eθ(Hg22+/Hg) + 0.059/2lg[Hg22+]
= 0.792 + 0.059/2lg(6.76×10-7)
= 0.61V
θ
so E (Hg2SO4/Hg) = 0.61V
（2） Solution： 2CuS + 2e- = Cu2S + S2EθCuS/Cu2S = EθCu2+/Cu+ + 0.059/2lg([Cu2+]2/[ Cu+])
= EθCu2+/Cu+ + 0.059/2lg(KθspCuS/ Kθsp Cu2S)
= 0.15 + 0.059/2lg[(7.94×10-36)2/1.0×10-48]
= -0.50V
Cu2S + 2e = 2Cu + S2EθCu2S/ Cu = EθCu+/ Cu + 0.059/2lg[Cu+]2
= EθCu+/ Cu + 0.059/2lg KθspCu2S
= 0.52 + 0.059/2lg1.0×10-48
= -0.90
CuS + 2e = Cu + S2EθCuS/ Cu = EθCu2+/Cu + 0.059/2lg[Cu2+]
= EθCu2+/Cu + 0.059/2lg KθspCuS
=(EθCu2+/Cu + EθCu+/ Cu)/2 + 0.059/2lg KθspCuS
= (0.15 + 0.52) + 0.059/2lg(7.94×10-36)
= -0.70V
9. Dissolved anhydrous copper sulphate into water, make into A solution. Deal A
solution with excess NaCl solution, make into B solution. After dealing B solution
with excess SO2, thinning by water, generate C deposition. After filtering out the
deposition and washing by water, the deposition dissolved into ammonial solution,
make into D colorless solution, when D stewing in air it will change rapidly into
blue-purple E solution. If add mass of copper scale into E solution, and it will make
into colorless F solution.
(a) Explain the color of A solution and B solution, and then write out the main
chemical formula including copper in the solution;
(b) Explain the color of C deposition and its chemical formula;
(c) What is chemical formula including copper in E?
(d) Explain simply the variety when D solution changing into E solution;
(e) Whether there is distinction about compose between D solution and F solution?
Solution: (a) A: blue Cu(H2O)42+
B: yellow CuCl42(b) C: white CuCl
(c) E: Cu(NH3)42+
(d) 4Cu(NH3)2+ + O2 + 2H2O + 8NH3 = 4Cu(NH3)42+ + 4OH(e) D solution and F solution are the same, both including Cu(NH3)2+ ion
10. The two following balance:
Cu2+ + Cu
2Cu+
Hg2+ + Hg
Hg 22+
(1) Why does it apparent this case which is phenomenal repulsion?
(2) In which case the balance will move left? Try to illustrate two instances for each
one.
Solution:
（1）
+0.17V
+0.52V
+0.920
+0.789
Cu+
Cu
Hg2+
Hg22+
Cu2+
Hg
+0.34V
+0.854
move left by adding precipitant of Cu+ ion
(2) 2 Cu+ = Cu2+ + Cu
for instance: Cu2+ + Cu + 2Cl- = 2CuCl
move left by adding precipitant of Hg2+ ion
Hg2+ + Hg = Hg22+
for instance: Hg22+ + H2S = HgS + Hg + 2H+
11. The instability constant of [Ag（CN）2]- is 1.0×10-20, if 1g silver is oxidized and
then dissolved into 1dm-3 solution which including 1.0×10-1mol·dm-3CN-, what is the
density of Ag+ ion in balance?
Solution: [Ag(CN)2]- = Ag+ + 2CNK 不稳 = [Ag+]·[CN-]2/[Ag(CN)2]K 不稳 is very low [Ag(CN)2]- = mAg/(M·V) = 1/108 = 0.009 mol/dm3
[CN-] = C CN- - 2[Ag(CN)2]- = 0.1 - 2×0.009 = 0.081 mol/dm3
[Ag+] = K 不稳×[Ag(CN)2]-/ [CN-]2 = 1.4×10-20mol/dm3
12. Write out the reaction equation about Hg.
Solution:
heat to boil
2Hg + O2 ══════2HgO
3 Hg + 8HNO3════ 3Hg(NO3)2 + 2NO↑ + 4H2O
2HgCl2 + SnCl2 + 2HCl = Hg2Cl2↓ + H2SnCl6
Hg2Cl2 + SnCl2 + 2HCl = 2Hg ↓+ H2SnCl6
NH4Cl + 2K2[HgI4] + 4KOH = Hg2NI·H2O↓ + 7KI + 3H2O
Hg2Cl2 + 2NH3 = HgNH2Cl↓ + NH4Cl + Hg
13. Write out the distinction about Hg2+ and Hg 22+ , the reaction equation of checking
NH4+.
Solution: the distinction about Hg2+ and Hg22+ as the following table:
agent
Hg22+
Hg2+
KOH
generate fuscoue Hg2O deposition
KCl
generate white Hg2Cl2 deposition
KI
generate green Hg2I2 depositionl
generate HgNH2Cl(white) + Hg(black)
hoary
NH3·H20
H2S
generate black HgS + Hg
SnCl2
generate hoary Hg deposition
generate yellow HgO
deposition
generate soluble colorless
HgCl2
generate aurantium
HgI2deposition，if KI excess
will generate colorless
K2[HgI4]
generate white Hg(NH2)Cl
deposition
generate black HgS
deposition
generate white
Hg2Cl2deposition，then
change into hoary Hg
deposition
Hg depositing on the surface of Cu and
the same to the left
forming white Hg
the reaction equation of checking NH4+:
NH4Cl + 2K2[HgI4] + 4KOH = Hg2NI·H2O↓ + 7KI + 3H2O
the mixed solution of K2[HgI4] and KOH is named nessler’s reagent, it will be appear
special red drop some nessler’s reagent, if there is microtier NH4+ ion in solution.
14.What is the distinction on geometric and stability of the composition of Cu（Ⅱ）
and Hg（Ⅱ）?
15. Compare the main chemical property of ⅠB andⅠA，ⅡB andⅡA.
Solution: both the subgroup and the main group element atom have only one electron
in the outermost layer, they both have +1 oxidation state. But the number on
second-outermost layer is different. There are 18 electron on second-outermost layer
of subgroup element atom, the obit has been filled-up, but there are only 8 electron on
second-outermost layer of alkali metal ( Li is 2), non d electron.
Subgroup element is inert heavy metal, after hydrogen in electric series, and the
property of action is taper from Cu to Au. But the main group element is active light
metal, forehand in electric series, and the property of action is gradually reinforce
from Li to Cs.
Subgroup element express multiple oxidation states, but the main group element
express only +1 oxidation state, since the energy gap between (n-1)d and ns electron
is not very big, its second energy ionization is not very big, d electron partly can
participation bonding, forming compound with +2 oxidation state, and even forming
compound with +3oxidation state. In the same time, the energy gap of ns and np
electron of main group metal is very big. General condition, ionizing the second
electron is impossible, express only +1 oxidation state.
The bond types of binary compound of copper group element (for instance CuS)
express considerable covalent property, and copper group element form composition
easily, but alkali metal element form difficult.
Cu
The metallicity of zinc group is thicker than alkaline-earth metals, and follows the
subsequence of Zn-Cd-Hg decline. The metallicity changing direction is contrary to
alkaline-earth metals.
Forming 18 electron configuration M2+ ion after zinc group element atom loss
electron, with stronger polarization energy, easier anamorphosis. And its bond types
of binary compound express considerable covalent property, different from
alkaline-earth metals.
Zinc group element form composition easily, since the energy of bonding electron obit
is near, ion radius is small, d electron screen thick, effective nuclear charge is big in
transition element.
Chapter 19
1. A empirical formula of some substance is PtCl4·2NH3, its water solution is
non-conducting, there is no deposition added in AgNO3, there is no NH3 released
when reacted with strong base, write out its coordination chemical formula.
Solution: [Pt(NH3)2Cl4]
2. In the following compounds which are composition? Which are chelate? Which are
complex salt? Which are simple salt?
（1）CuSO4·5H2O
（2）K2PtCl6
（3）Co（NH3）6Cl3
（4）Ni（en）2Cl2
（5）
（NH4）2SO4·FeSO4·6H2O
（6）Cu（NH2CH2COO）2
（7）Cu（OOCCH3）2
（8）KCl·MgCl2·6H2O
Solution: composition: K2PtCl6, Co(NH3)6Cl3, CuSO4·5H2O
chelate: Ni(en)2Cl2, Cu(NH2CH2COO)2
complex salt: (NH4)2SO4·FeSO4·6H2O KCl·MgCl2·6H2O
simple salt: Cu(OOCH3)2
3. Naming the following compositions and composition ions:
（1）
（NH4）3[SbCl6]
（2）Li[AlH4]
（3）[Co（en）3]Cl3
（4）[Co（H2O）4Cl2]Cl
（5）[Cr（H2O）4Br2]Br·2H2O
（6）[Cr（H2O）
（en）
（C2O4）
（OH）
3（7）Co（NO2）6]
（8）[Co（NH3）4（NO2）C]+
（9）[Cr（Py）2（H2O）Cl3]
（10）[Ni（NH3）2（C2O4）]
Solution: (1) hexachloride ammonium antimonite(III)
(2) lithium aluminate(III) tetrahydride
(3) Trichloride Tris (ethylenediamine) cobalt (III)
(4) 4-Dichloro-hydrated cobalt chloride (III)
(5) Dibromo-2 hydrated bromide • 4 hydrated cobalt (III)
(6) OH • Water • oxalate • ethylenediamine chromium (III)
(7) 6-nitro cobalt (III) anion
(8) Chlorine • Nitro • Four ammonia cobalt (III) cations
(9) Trichloro • Water • 2-pyridine chromium (III)
(10) 2 ammonia • oxalato nickel (II)
4. Point out the space configuration and draw out the possible stereo isomer of the
following composition:
（1）[Pt（NH3）2（NO2）Cl]
（2）Pt（Py）
（NH3）ClBr]
（3）Pt（NH3）2（OH）2Cl2]
（4）NH4[Co（NH3）2（NO2）4]
（5）[Co（NH3）3（OH）3]
（6）[Ni（NH3）2Cl2]
（7）[Cr（en）2（SCN）2]SCN
（8）[Co（en）3]Cl3
（9）[Co（NH3）
（en）Cl3]
（10）[Co（en）2（NO2）2]Cl2
Solution: （1） [Pt(NH3)2(NO2)Cl] plane quadrate
2 stereo isomer
NO2
H3N
NO2
H3N
Pt
Pt
Cl
H3N
Cl
H3N
（2） [Pt(Py)(NH3)ClBr] plane quadrate
3 stereo isomer
Cl
Py
Cl
NH3
Pt
Pt
Br
Br
NH3
Cl
Py
NH3
Pt
Py
Br
（3） [Pt(NH3)2(OH)2Cl2]
NH3
NH3
HO
Cl
octahedron
5 stereo isomer
NH3
Cl
Cl
OH
Cl
NH3
HO
Cl
OH
HO
Cl
NH3
NH3
NH3
OH
Cl
Cl
Cl
NH3
OH
octahedron
（4）[Co(NH3)2(NO2)4]NH3
O2N
O2N
HO
NH3
NO2
HO
H3N
NH3
Cl
2 stereo isomer
NH3
O2N
NO2
NH3
（5）[Co(NH3)3(OH)3]
OH
HN3
NH3
O2N
OH
NH3
（6）[Ni(NH3)2Cl2]
tetrahedroid
（7）[Cr(en)2(SCN)2]SCN
octahedron
[Cr(en)2(SCN)2]SCN
NO2
NO2
2 stereo isomer
OH
H3N
OH
octahedron
OH
H3N
NH3
OH
no stereo isomer
2 stereo isomer
SCN
NCS
en
en
SCN
（8）[Co(en)3]Cl3
[Co(en)3]3+
en
en
octahedron
NH3
2 stereo isomer
（9）[Co(NH3)(en)Cl3]
octahedron
2 stereo isomer
（10）[Co(en)2(NO2)2]Cl2
octahedron
2 stereo isomer
[Co(en)2(NO2)2]2+
5. A certain metal ion its magnetic moment is 4.90B.M in octahedral feeble field, and
its magnetic moment is 0 in octahedral intense field, which the center metal ion may
be?
Solution: the center metal ion may be Fe2+
n(n + 2) = 4.90
n=4
i.e. there are four single electrons in octahedral field
there is no single electron in octahedral intense field
and so Fe2+ match the condition
6. According to experimental determination validate magnetic moment judge the
following coordinate ions which are belonged to high spin? Which are belonged to
low spin? Which are belonged to inner-orbital? Which are belonged to outer-orbital?
（1）Fe（en） 22+
5.5B.M
（2）Mn（SCN） 64−
6.1BM
（3）Mn（CN） 64−
1.8B.M
（4）Co（NO2） 64−
1.8B.M
（5）Co（SCN） 24−
4.3B.M.
（6）Pt（CN） 24−
0B.M.
（7）K3[FeF6]
5.9B.M.
（8）K3[Fe（CN）6]
2.4B.M.
，
（2）
，
（5）
，
（7）
Solution: high spin: （1）
，
（4）
，
（6）
，
（8）
low spin: （3）
，
（4）
，
（6）
，
（8）
inner-orbital: （3）
，
（2）
，
（5）
，
（7）
outer-orbital: （1）
7. Known that [Pd（Cl）2（OH）2] has two different structure, which hybrid orbital do
the bonding electrons occupy?
Solution: dsp2
8. Explaining the causation of generating deposition and dissolving alternately when
adding sequence NaCl，NH3，KBr，Na2S2O3，KI，KCN，Ag2S into dilute AgNO3
solution by soft-hard
acid base theory.
Solution: by soft-hard acid-base rule:’’ hard conversant hard, soft conversant soft, non
business of soft to hard adjacent’’
9. Forecasting the stability of two group coordinate ions formed every following
group, and explaining the causation:
(1) coordinate Al3+ to F- or Cl-;
(2) coordinate Pd2+ to RSH or ROH;
N
;
(3) coordinate Cu2+ to NH3 or
2+
(4) coordinate Cu to NH3 or NH2COOH or CH3COOH.
Solution: (1) the composition consists of Al3+ and F- is more stable, since Al3+ is hard
acid, the hardness of base F- is stronger than Cl(2) the composition consists of Pd2+ and RSH is more stable, since Pd2+ is soft acid,
RSH is soft base, ROH is hard base.
(3) the composition consists of Cu2+ and NH3 is more stable.
(4) the composition consists of Cu2+ and NH2CH2COOH is more stable, because of
the former ligand includes N, the later ligand includes only O, the ligand ability N is
stronger than O.
10. Adding solid KCl avouch the density of Cl- is 0.10mol·dm-3 into 0.1mol·dm-3
K[Ag（CN）2] solution, what will be happen?
Solution: suppose [Ag+] is x mol•dm-3
Ag(CN)2Ag+ + 2CNx
2x
0.1-x
0.1 − x
= 1.25 × 10 21
x(2 x) 2
result: x = 2.7×10-8mol•dm-3
Q = [Ag+][Cl-] = 2.7×10-8×0.10 = 2.7×10-9
Q > KspΗ（AgCl）
，hence there will generate AgCl deposition
11. Adding 0.01mol solid CuSO4 into 1dm-36mol·dm-3 ammonia solution, and then
add 0.01mol solid NaOH, whether the cuprammonium composition will be destroyed
or not?
Solution: suppose after generating Cu(NH3)42+ the density of Cu2+ is xmol•L-1
Cu 2+ + 4 NH 3 ⇔ Cu ( NH 3 ) 4
x
2+
6-4(0.01-x)
0.01-x
0.01 − x
= 2.09 × 1013
4
x[6 − 4(0.01 − x)]
result： x = 3.79 × 10-19 mol•dm-3
Q = [Cu2+] – [OH-]2 = 3.79 × 10-19×0.012
= 3.79 × 10-23
Q less than KspΗ, hence the cuprammonium composition will not be destroyed.
12. When NH4SCN and a little Fe3+ coexist in the solution and have reached balance
state, added NH4F and made [F − ]=[SCN − ]=1 mol·dm-3, what is concentration ratio of
[FeF6] 3− and
[Fe（SCN）3] in the solution?
（Ksta，Fe（SCN）3=2.0×103，Ksta，[FeF6]3-=1×1016）
Solution: Fe3+ +6F −
a
Fe3+ + 3SCN −
[FeF6] 3−
1
x
Ksta，[FeF6]3- =
a
x
= 1×1016
1 a
6
Ksta，Fe（SCN）3 =
[Fe（SCN）3]
1
y
y
=2.0×103
1 a
3
x
= 5×1012
y
so that concentration ratio of [FeF6] 3− and [Fe（SCN）3] is 5×1012
13. Planning to dissolve 1×10-5mol solid AgI into 1cm-3 ammonia liquor, try to
calculate out what is the lowest density of ammonia liquor on theory?
（Ksta，Ag（NH3） +2 =1.12×107；K θsp , AgI =9.3×10-17）
Solution: KspΗAgI = [Ag+][I-]
（NH 3）2+ =
KstaAg
[ Ag
（NH 3）2+ ]
[ Ag + ][ NH 3 ]2
result：Kspθ AgI / [ I −] =
[ Ag ( NH 3 ) 2+ ]
Ksta Ag ( NH 3 ) 2+ • [ NH 3 ]2
[ I − ] = 1× 10−5 mol • cm −3
[ Ag ( NH 3 ) 2+ ] = 1× 10−5 mol • cm −3
[ NH 3 ] =
1× 10−5 × 1× 10−5
= 0.31mol • cm −3 = 310mol • cm −3
1.12 ×107 × 9.3 × 10−17
14. Known that Au+ + e—=Au ϕ θ =1.691V, request the result of ϕ θ on Au（CN）−2 +
e—
Au+2CN—. （Ksta，Au（CN） −2 =2×1038）
Solution:
Au (CN ) 2− + e− ⇔ Au + 2CN −
[ Au
in standard condition：
（CN）2− ] = [CN − ] = 1.0mol • dm −3
Kstaθ [ Au
（CN）2− ] =
[ Au
1
（CN）2− ]
=
+
− 2
[ Au ][CN ] [ Au + ]
θ
E θ [ Au
Au + / Au）+ 0.059lg[ Au + ]
（CN）2 − / Au ] = E（
=1.691 + 0.059lg1/KstaΗ
=1.691+0.059lg1/2×1038
=-0.57V
15. A copper electric pole dipped into a solution which including 1.00 mol·dm-3
ammonia and 1.00Cmol·dm-3Cu（NH3） 24+ coordinate ion, if the standard hydrogen
electrode as positive pole, the difference of potential between it and copper electric
pole is 0.0300V through the experimental determination. Try to calculate the stablility
θ
=0.34V）.
constant of Cu（NH3） 24+ （known ϕ Cu
2+
/ Cu
Solution: EΗ = EΗ（H+/H2）- EΗ[Cu（NH3）42+/Cu] = 0.0300V
i.e. EΗ[Cu（NH3）42+/Cu] = -0.0300V
reason EΗ[Cu（NH3）42+/Cu] = EΗ（Cu2+/Cu） + 0.059/2lg[Cu2+]
Cu 2+ + 4 NH 3 ⇔ Cu
（NH 3）4 2+
result：Kstaθ =
1
i.e.[Cu 2+ ] = 1/ Kstaθ
[Cu 2+ ]
E θ [Cu
（NH 3）4 2+ / Cu ] = 0.34 + 0.059 / 2lg1/ Ksta
0.059
lg Kstaθ
2
θ
result：Ksta = 3.49 ×1012
16, Why is Co3+ able to oxidate water, and [Co（NH3）6]3+ isn’t able to in water
solution?
−0.0300 = 0.34 −
Ksta，Co（NH3） 62+ =1.38×105；KstaCo（NH3） 36+ =1.58×1035；Kb，NH3=1.8×10-5
θ
ϕ Co
3+
/ Co 2 +
=1.808V； ϕ Oθ 2 / H 2O =1.229V； ϕ Oθ 2 / OH − =0.401V
Solution: in water solution,
E θ (Co3 + / Co 2+) > E θ (O2 / H 2 O)
so that Co3+ is able to oxidate water
4Co3++2H2O=4Co2+O2↑+4H+
E θ [Co( NH 3 )63+ / Co( NH 3 )6 2+ ] = Eθ (Co3+ / Co 2+ ) + 0.059lg
= Eθ (Co3 + / Co2+ ) = 0.059lg
[Co3+ ]
[Co 2+ ]
Kstaθ [Co( NH 3 )6 2+ ]
Kstaθ [Co( NH 3 )6 2+ ]
1.38 ×105
= 0.04V
1.58 × 1035
suppose[ NH 3 ] = 1.0mol • dm −3then：
= 1.81 + 0.059lg
[OH − ] = 1.0 ×1.8 ×10−5 = 4.24 × 10−3 mol • dm −3
O2 + 4 H 2O + 4e − = 4OH −
E (O2 / OH − ) = Eθ (O2 / OH − ) +
3+
0.059 p(O2 ) / pθ
0.059
1
lg
= 0.401 +
lg
= 0.45V
− 4
4
[OH ]
4
(4.24 ×10−3 ) 4
2+
E θ [Co( NH 3 ) 6 / Co( NH 3 ) 6 ] < E (O2 / OH − )
hence
Co（NH3）63+ is not able to oxidate water.
17. After adding 0.010mol·dm-3 Fe（NO3）3 into 1.0mol·dm-3 HCl acid solution, which
is the biggest coordinate ion density in the solution? (known that step stability
constant in this system is k1=4.2，k2=1.3，k3=0.040，k4=0.012)
Solution: by reason that stability constant of all levels is small, Cl- excess much more,
hence believe that
[Cl-] = 1.0 mol•L-1
[Cl-] + [FeCl2+] + 2[FeCl2+] + 3[FeCl3] + 4[FeCl4-] =1.0 mol•L-1
[Fe3+] + [FeCl2+] + [FeCl2+] + [FeCl3] + [FeCl4-]=0.010 mol•L-1
[ FeCl 2+ ]
= 4.2
k1 =
[FeCl2+] = 4.2[Fe3+]
[ Fe 3+ ][Cl − ]
+
k2 =
[ FeCl 2 ]
= 1.3
[ FeCl 2+ ][Cl − ]
[FeCl2+] = 1.3×4.2[Fe3+]
=5.5[Fe3+]
k3 =
[ FeCl3 ]
+
−
[ FeCl 2 ][Cl ]
= 0.040
[FeCl3] = 0.040[FeCl2+]
= 0.04×1.3×4.2[Fe3+]
=0.22[Fe3+]
−
k4 =
[ FeCl 4 ]
= 0.012
[ FeCl3 ][Cl − ]
[FeCl4-]=0.012[FeCl3]
=0.012×0.04×1.3×4.2[Fe3+]
=0.0026[Fe3+]
then：[Fe3+] + 4.2[Fe3+] + 5.5[Fe3+] +0.22[Fe3+] + 0.0026[Fe3+]=0.010mol•L-1
result：[Fe3+]=9.2×10-4 mol•L-1
thereout result：the density of FeCl2+ is the biggest.
20th Chapter
1. What is the main minerals of titanium? Please briefly describe reaction principle of
preparation of titanium dioxide from ilmenite.
Solution: The main minerals are ilmenite titanium FeTiO2
Reaction principle:
FeTiO3 + 2H2SO4 = TiOSO4 + FeSO4 + 2H2O
TiOSO4 + 2H2O = TiO2 • H2O ↓ + H2SO4
TiO2 • H2O = TiO2 + H2O
2. Explain why the compound TiCl3 and [Ti (O2) OH (H2O) 4]+ are colorful.
Solution: TiCl3 has color because the generated electricity of nuclear transitions, [Ti
(O2) OH (H2O) 4] + is colorful because ion O22- has strong deformation, d-d
transition.
3. Complete and equilibrate the following reaction equations.
(1) Ti + HF
(2) TiO2 + H2SO4
(3) TiCl4 + H2O
(4) FeTiO3 + H2SO4
(5) TiO2 + BaCO3
(6) TiO2 + C + Cl2
Solution: (1) Ti + 5HF = H2TiF6 + 2H2 ↑
(2) TiO2 + H2SO4 = TiOSO4 + H2O
(3) TiCl4 + 2H2O = TiO2 + 4HCl
(4) FeTiO3 + 2H2SO4 = TiOSO4 + FeSO4 + 2H2O
(5) TiO2 + BaCO3 = BaTiO3 + CO2 ↑
(6) TiO2 + 2C + 2Cl2 = TiCl4 + 2CO ↑
4. To accomplish the following reaction
(1) TiI4 was intensely heated in a vacuum;
(2) Mixture FeTiO3 and carbon were heated in the atmosphere of chlorine gas.
(3) Into the aqueous solution containing TiCl62- was added excessive ammonia;
(4) Into VCl3 aqueous solution was added excess of ammonia;
(5) Add VCl2 solid to the aqueous HgCl2.
Solution: (1) TiI4 = Ti + 2I2
(2) 2FeTiO3 + 6C + 5Cl2 = 2FeCl3 + 2TiCl4 + 6CO
(3) TiCl62-+ NH3 = [Ti (NH3) 6] 4 + + 6Cl (4) VCl3 + Na2SO3 =
(5) VCl2 + HgCl2 =
5. Based on the following experiment, write down the reaction equation: To open the
cork a bottle of TiCl4 immediately produce the white smoke. To the bottle by adding
concentrated HCl solution and the zinc is generated when the purple solution, slowly
adding NaOH solution until the solution is alkaline, so there purple precipitate.
Sedimentation After filtration, the first dealing with HNO3, and then dilute alkali
solution treatment to produce a white precipitate.
Solution: TiCl4 + 3H2O = H2TiO3 + 4HCl ↑
2TiCl4 + Zn = 2TiCl3 ↓ + ZnCl2
TiCl3 + 3NaOH = Ti (OH) 3 + 3NaCl
3Ti (OH) 3 + 7HNO3 = 3TiO (NO3) + 2 NO ↑ + 8H2O
TiO2 + + 2OH-+ H2O = Ti (OH) 4 ↓
6. The use of standard electrode potential data to determine H2S, SO2, SnCl2, and
metal Al can put TiO2 + ion reduced to Ti3 + ions?
Solution: from look-up table may:
Because EθAl3 + / Al <EθTiO2 + / Ti3 +, so that only Al can restore the TiO2 + ions
into the Ti3 + ion
7. To the steps outlined in the following reaction
(1) from MnSO4 prepared K5Mn (CN) 6;
(2) from MnO2 prepared K3Mn (CN) 6;
(3) from MnS prepared by KMnO4;
(4) from MnCO3 preparation K2MnO4;
(5) The preparation of manganese by the BaMnO4.
Solution: (1) MnSO4 + 6KCN = K4 [Mn (CN) 6] + K2SO4
(2) 4MnO2 +6 H2SO4 = 2Mn2 (SO4) 3 +6 H2O + O2
Mn2 (SO4) 3 +12 KCN = 2K3Mn (CN) 6 +3 K2SO4
(3) MnS + 2O2 = MnO2 + SO2
2MnO2 + 4KOH + O2 = 2K2MnO4 + 2H2O
2K2MnO4 + 2H2O = 2KMnO4 + 2KOH + H2
(4) MnCO3 + 2HNO3 = Mn (NO3) 2 + H2O + CO2
Mn (NO3) 2 = MnO2 + 2NO2
2MnO2 + 4KOH + O2 = 2K2MnO4 + 2H2O
(5) BaMnO4 + K2SO4 = K2MnO4 + BaSO4
K2MnO4 + CO2 = KMnO4 + MnO2 +2 K2CO3
KMnO4 + SO32-+ H2O = 2MnO2 + SO42-+2 OH-(neutral conditions)
3MnO2 = Mn3O4 + O2 (heating)
Mn3O4 +8 Al = 4Al2O3 +9 Mn
8. Potassium chromate in aqueous solution to pass into the CO2, it will change?
Solution: solution from yellow to orange
9. Completed and the following reaction equation balancing
(1) V2O5 + NaOH
(2) V2O5 + HCl
(3) VO43-+ H + (excess)
(4) VO2 + + Fe2 + + H +
(5) VO2 + + H2C2O4 + H +
Solution: (1) V2O5 + 2NaOH = 2NaCO3 + H2O
V2O5 + 6NaOH = 2Na3VO4 + 3H2O
(2) V2O5 + 6HCl = 2VOCl2 + Cl2 + 3H2O
(3) VO43-+ 4H + (excess) = VO2 + + 2H2O
(4) VO2 + + Fe2 + + 2H + = VO2 + + Fe3 + + H2O
(5) 2VO2 + + H2C2O4 + 2H + = 2VO2 + + 2CO2 ↑ + 2H2O
10. Vanadium ion composition in aqueous solution depends on which factors? How
do these factors affect the balance VO V2O (V3O, V10O, etc.).
Solution: Depending on (1) solution pH, with the pH value decreased degree of
polymerization increased, but by adding an adequate amount of acid, the solution
existence and stability of VO2 + ions.
(2) the size of the concentration of vanadium acid radical concentration of less
than 10-4mol • dm-3, the solution exists only monomer and acid-type vanadium
vanadium carboxylate ion, if the concentration is greater than 10-4mol • dm-3, then
when the acidity is too high, pH value of about 2, it will precipitate a red-brown
vanadium pentoxide hydrate.
11. Why does the new generation of hydroxide precipitation of the following changes
will occur
(1) Mn (OH) 2 is almost white, in the air becomes dark brown?
(2) white Hg (OH) 2 immediately into a yellow?
(3) The blue Cu (OH) 2, when heated Why black?
Solution: (1) 2Mn (OH) 2 + O2 = 2MnO (OH) 2 MnO (OH) 2 as a dark brown
(2) Hg (OH) 2 = HgO + H2O HgO yellow
(3) Cu (OH) 2 = CuO + H2O CuO is black
12. Accordance with the following various experimental phenomena, write the
corresponding chemical reaction equation
(1) to Cr2 (SO4) 3 solution was added dropwise NaOH, the first green flocculent
precipitate onion precipitation, then dissolved, this time by adding bromine water, the
solution changed from green to yellow. Instead of bromine water with H2O2, but also
get the same result.
(2) When the yellow BaCrO4 precipitate was dissolved in concentrated HCl solution
when you get a green solution.
(3) in an acidic medium, using zinc reduction Cr2O, the solution color from orange
through green become blue. Shiyou placed back into the green.
(4) to H2S leads to acidification K2Cr2O7 have spent H2SO4 solution, the solution
color from orange to green, while precipitation of milky white precipitate.
Solution: (1) Cr3 + + 3OH-= Cr (OH) 3 ↓
Cr (OH) 3 + OH-= Cr (OH) 4 -Cr (OH) 4 - + Br2 = Br-+ CrO42-+ H2O
2 Cr (OH) 4 - + 3H2O2 + 2OH-= 2CrO42-+ 8H2O
(2) 2BaCrO4 + 2H + = Cr2O72-+ 2Ba2 + + H2O
Cr2O72-+ 6Cl-+ 14H + = 2Cr3 + + 3Cl2 ↑ + 7H2O
(3) Cr2O72-+ 3Zn + 14H + = 2Cr3 + + 3Zn2 + + 7H2O
2Cr3 + + Zn = 2Cr2 + + Zn2 +
4Cr2 + + O2 + 4H + = 4Cr3 + + 2H2O
6H2S + Cr2O72-+ 8H + = 2Cr3 + + 6S ↓ + 7H2O
13. If the Cr ions and Al coexist, how to separate them? When Zn ions are also
co-exist, how to separate?
Solution: Cr3 +, Al3 +, Zn2 +
(1) H2O2, OH-(excess), Δ
(2) adjust pH = 4-5, BaCl2
BaCrO4 Al3 +, Zn2 +
NH3.H2O
Al (OH) 3 Zn (NH3) 42 +
14. Cl, and CrO-ion containing a mixed solution by adding drop-by AgNO3 solution,
if [Cl] and [CrO] are 10 mol • dm, then who should precipitate? Can the basic
separation of the two.
Solution: KspAgCl = 1.8 × 10-10 KspAg2CrO4 = 2.0 × 10-12
= [Ag +] [Cl-] = [Ag +] 2 [CrO42-]
Therefore obtain an AgCl precipitate
Subject the AgCl precipitation [Ag +] =
Make Ag2CrO4 precipitation [Ag +] = it is when Ag2CrO4 precipitated when the
[Cl-] =
100% = 99.6%
Therefore, the basic separation can be
15. With H2S or sulfide to the following ions from their mixture to separate.
Hg, Al3 +, Cu2 +, Ag +, Cd2 +, Ba2 +, Zn2 +, Pb2 +, Cr3 +
Solution: Al (OH) 3, Cr (OH) 3 (plus hydrochloric acid) Al3 +, Cr3 +, Zn2 +
Hg2S, CuS, Ag2S, CdS, PbS (plus concentrated hydrochloric acid)
Cd2 +, Pb2 + (plus ammonia)
HgS + Hg, CuS, Ag2S (plus nitric acid) Ag +, Cu2 + (Cl)
Ba2 +
16. Pointed out that the color of the following compounds and indicate which are dd
transitions, which are caused by charge transfer?
Cu (NH3) 4SO4 (dark blue), [Mn (CN) 6] (SO4) 2 (dark purple),
Cis - [Co (NH3) 4Cl2] Cl (purple), trans - [Co (NH3) 4Cl2] Cl (green),
K2 [MnBr4] (yellow-green), Mn (H2O) 6 (SO4) (light red),
KFe [Fe (CN) 6] (dark blue), Ag2CrO4 (orange red)
BaFeO4 (dark red), Na2WS4 (orange)
Au [AuCl4] (dark red), Cu [CuCl3] (brown color)
Solution: (1) dark blue (2) deep purple (3) Purple (4) Green
(5), yellow-green (6) pink (7) Deep Blue (8) orange-red
(9) Magenta (10) Orange (11) dark red (12) tan
(9), (10), (11) for the charge transfer is caused due to the remaining Jie Wei dd
transitions.
17. According to 2CrO +2 H + Cr2O + H2O K = 10 tests aim at 1 mol • dm potassium
chromate solution, pH value is the number of hours
(1) chromium ion brown dichromate ion concentrations are equal;
(2) The concentration of chromium ion accounted for 99%;
(3) The concentration of dichromate ion accounted for 99%.
Solution: (1) Solution: Let Cr2O72-concentration of x mol • dm-3
2 CrO42-+ 2H + Cr2O72-+ H2O
2x x
21
[CrO42-] = [Cr2O72-] is: 1 - 2x = xx = 1 / 3 mol • dm-3
Was: Ph =-lg [H +] = 6.76
18. The potassium dichromate solution of silver nitrate solution mixed with the brown,
the precipitation of what is precipitation?
Solution: Build Ag2CrO4 (brick red) precipitation
Cr2O72-+ 4Ag + + H2O = 2 Ag2CrO4 ↓ + 2H +
19. In order to dissolve a 20g 6 hydrated chloride Chromium (Ⅲ) to precipitate out of
solution quickly re-chlorine needed 75ml 2 mol • dm silver nitrate solution, please
write the basis of these data chromium chloride 6 hydrate (Ⅲ)-type of coordination
(structure type).
Solution: Let structure of [Cr (H2O) 6Cl (3-x)] Clx is 1mol of the material can be
ionized out of x mol of Cl Cl-+ Ag + = AgCl
0.15mol 75 × 2 × 10-3mol
11
6 The amount of hydrated chromium chloride: 20g/266.45g/mol = 0.075mol
Then: x / 1 = 0.15/0.075 = 2
The structural formula: [Cr (H2O) 6Cl] Cl2
20. Why are alkali metal dichromate in water of pH <7?
Solution: Because there is in the solution 2CrO42-+2 H + = Cr2O72-+ H2O alkali
metal dichromate so acidic.
21. How the impact of solution pH, chromium ion, molybdenum and tungsten ion ion
composition? Can exist in any medium, CO, MO, MO-ion?
Solution: pH down to weak acidic when CrO42-, MoO42-, WO42-will gradually feed
into the multi-ion. In acidic media.
22. According to memory, write from the manganese ore of tungsten metal tungsten
powder prepared by the whole reaction process.
Solution: 6MnWO4 + 6Na2CO3 + O2 = 2Mn3O4 + 6Na2WO4 + 6CO2
Na2WO4 + 2HCl = H2WO4 + 2NaCl
H2WO4 + 2NH3 • H2O = (NH4) 2WO4 + 2H2O
(NH4) 2WO4 = 2NH3 ↑ + WO3 + H2O
WO3 +3 H2 = W + 3H2O
23. Using standard electrode potential, determine the direction of the following
reactions:
6MNO + 10Cr + 11H2O 5Cr2O + 6Mn2 + + 22H +
Solution: EθMnO4-, Mn2 + = 1.51V
EθCr2O72-, Cr3 + = 1.33V
EθMnO4-, Mn2 +> EθCr2O72-, Cr3 +
Therefore, the right response to
24. Try seeking the following reaction equilibrium constants and estimate the reaction
is reversible
MnO + 5Fe2 + + 8H + Mn2 + + 5Fe3 + + 4H2O
Solution:
EθMnO4-, Mn2 + = 1.51V
EθFe3 +, Fe2 + = 0.771V
LgK =
Were: K = 4.24 × 1062 it is an irreversible reaction
25. Using manganese dioxide as raw material, how to prepare: (1), manganese sulfate;
(2), manganese potassium; (3) potassium permanganate.
Solution: MnO2 + H2SO4 + H2O2 = MnSO4 + O2 ↑ + 2H2O
2MnO2 + 4KOH + O2 = 2K2MnO4 + 2H2O
2K2MnO4 + Cl2 = 2KMnO4 + 2KCl
26. Based on the following potential diagram
MnO MnO2 Mn
IO I2 I
Written when the solution PH = 0 when the following conditions, potassium
permanganate and potassium iodide reaction equation:
(1) excessive potassium iodide;
(2) The excess of potassium permanganate.
Solution: (1) 2MnO4-+ 10I-+ 16H + = 2Mn2 + + 5I2 + 8H2O
(2) 6MnO4-+ 5I-+ 18H + = 6Mn2 + + 5IO3-+ 9H2O
27. What is with the multi-acid and heteropoly acid? Ammonium write Chung, 12 tungsten sodium phosphate formula. Drawing said Chung-Mo carboxylate, 12 tungsten phosphate structure.
Solution: Same as acid: A number of water molecules and two or more similar the
composition of the acid anhydride
Heteropoly acids: A number of water molecules and the composition of two or more
different types of acid anhydride
Zhong ammonium molybdate: (NH4) 6 (Mo7O24)
12 - W sodium phosphate Na3 [P (W12O40)]
28. Weigh 10.00g chromium and manganese have just kind of, after appropriate
treatment, chromium and manganese are oxidized to Cr2O and MnO solution, a total
of 250.0cm3. Exact amount to take the above-mentioned solution 10.00cm3, adding
BaCl2 solution and adjust the acidity of solution, precipitation of chromium all down,
get 0.0549gBaCrO4. A copy of the solution, an alternative 10.00cm3, using Fe2 + in
acidic medium solution 0.075mol • dm-3) titration, spent 15.95cm3. Calculation just
like the mass fraction of chromium and manganese.
Solution: Cr% =%
Mn% =
= 1.50%
29. A solution 1dm3, which contains KHC2O4 • H2C2O4 • 2H2O50.00g. There is a
certain volume of KMnO4 solution 40.00cm3 oxidized the solution, while the same
volume of solution of these just 30.00cm3 of 0.5mol • dm-3NaOH solution. Test
seeking:
(1) the solution of the C2O ions and H + concentration;
(2) KMnO4 solution concentration (mol • dm-3).
Solution: (1) nH + = nOH = 0.5 × 0.03 = 0.015mol
[H +] = 0.015mol / L
(2) The 30Ml0.5N NaOH solution and the quality of the KHC2O4 • 2H2O:
W = n × M = 30 × 0.5 × 10-3 × 254.2 / 3
= 1.271g
Was 40mL xmol • dm-3 KMnO4 oxidation KHC2O4 • 2H2O solution quality as
follows:
W = 40 • x • 254.2 / 4 × 10-3 = 2.542x g
Both are equal:
1.271g = 2.542x g
Was: [MnO4-] = 0.50mol • dm-3
30. According to the standard heat of formation of material, seeking balance the
thermal effect of the following reaction
(NH4) 2Cr2O7 (s) N2 (g) + C2O3 (g) + H2O (g)
Known:
△ H =-1807.49K • moi-1
△ H = -1129.68 K • moi-1
△ H = -97.49 K • moi-1
(NH4)Η = - ΔH ΗSolution: ΔH Cr2O3Η H2O + ΔH Η2Cr2O7 + 4 ΔH
= - (-1807.49) + 4 × (-97.49) + (-1129.68)
= 287.85KJ • mol-1
Chapter 21
1. To explain the following issues:
(1) In the Fe3 + ions in solution by adding KSCN solution occurs when the blood red,
but adding a little iron, the blood red and disappear, what kind of logic is that?
(2) Why Fe3 + salt is stable, while the Ni3 + salt, has not yet obtained?
(3) Why can not by Fe3 + in aqueous solution of salt and KI obtained FeI3?
(4) When acting on the FeCl3 solution, Na2CO3 solution, why get too is Fe (OH) 3
and the step is Fe2 (CO3) 3?
(5) What color gel that contains ingredients? Dry Why was blue, pink after absorbing
water change?
Solution: (1) Fe3 + + nSCN-= Fe (SCN) n3-n
n = 1-6 blood-red by adding iron, the iron to Fe (SCN) n3-n reduction, the
resulting Fe2 + is not with the SCN-generated colored complex, and therefore red
blood disappearance
2Fe (SCN) n3-n + Fe = 3Fe2 + + nSCN (2) iron outer electron configuration of 3d64s2, Fe3 + half-full of the 3d5 electron
configuration to make it more than Fe2 + electronic configuration of 3d6 stable outer
electron configuration of nickel as 3d84s2, of which more than half-full 3d electronic
state of more difficult to lose, so under normal circumstances, the performance of the
+2 oxidation state of nickel, under special conditions can also be unstable Ni (III)
compounds.
(3) Eθ (Fe3 +, Fe2 +)> Eθ (I2/I-), in aqueous solution, Fe3 + will be I-oxidation is
not FeI3
2 Fe3 + + 2 I-= 2Fe2 + + I2
(4) Na2CO3 in water hydrolysis, CO32-+ H2O HCO3-+ OH Solution [CO32-] and [OH-] or less, while the Fe (OH) 3 is much smaller than the
solubility of Fe2 (CO3) 3, so the event of FeCl3 with Na2CO3 solution to produce Fe
(OH) 3 precipitation, rather than generating Fe2 (CO3) 3 precipitation.
2Fe3 + + 3CO32-+ 6H2O = 2 Fe (OH) 3 ↓ + 3H2CO3
(5) color gel in the shouting moisture indicator CoCl2, old crystal water contained in
the number of different colors are different. CoCl2 • 6H2O CoCl2 • 2 H2O CoCl2 •
H2O CoCl2
Pink purple blue-violet blue
Therefore, the blue anhydrous CoCl2 significantly, CoCl2 crystal water were more
noticeable pink.
2. Write with the following experiments were reactive phenomenon:
To the solution containing Fe2 + ions were generated after adding NaOH, white-green
sediment, graded brown. After filtration, dissolve brown precipitate with hydrochloric
acid, the solution yellow. Add a few drops of KSCN solution, an immediate change of
blood red, which leads to SO2 when the red disappears. Dropping KMnO4 solution,
the purple will legs to go. Finally, when adding yellow blood salt solution to generate
a blue precipitate.
Solution: Fe2 + + 2OH-= Fe (OH) 2
4Fe (OH) 2 + O2 + 2H2O = 4Fe (OH) 3
Fe (OH) 3 + 3HCl = Fe3 + + 3H2O +3 Cl Fe3 + + SCN-= FeSCN2 +
2FeSCN2 + + SO2 + 2H2O = 2Fe2 + + SO42-+ 4H + + 2SCN MnO4-+ 5Fe2 + + 8H + = Mn2 + + 5Fe3 + + 4H2O
Fe3 + + Fe (CN) 64 - + K + = KFe [Fe (CN) 6]
3. That the following experiment was experimental phenomena, and write a reaction
formula:
(1) treatment with concentrated hydrochloric acid Fe (OH) 3, Co (OH) 3, Ni (OH) 3
precipitation phenomenon when what?
(2) FeSO4, CoSO4, NiSO4 by adding ammonia solution, and what phenomenon?
Solution: (1) with concentrated HCl treatment Fe (OH) 3, the brown precipitate
dissolution, was yellow solution
Fe (OH) 3 + HCl = FeCl3 + 3H2O
With concentrated HCl treatment Co (OH) 3, brown Co (OH) 3 dissolved in solution
was pink, with chlorine gas release
2 Co (OH) 3 + 2Cl-+ 6H + + 6H2O = 2Co (H2O) 62 + (pink) + Cl2 ↑
Concentrated HCl solution when dealing with excessive blue
Co (H2O) 62 + + 4Cl-= CoCl42-(Blue) + 6H2O
Concentrated hydrochloric acid treatment Ni (OH) 3, the black Ni (OH) 3 was
dissolved and a green solution of chlorine released 2 Ni (OH) 3 + 2Cl-+ 6H + + 6H2O
= 2Ni (H2O) 62 + + Cl2 ↑
(2) In FeSO4 solution by adding ammonia, because of the solution dissolved O2, not
white, but blue-green precipitate precipitation, precipitation in the air, slowly turned
brown.
Fe2 + + 2NH3 + 2H2O = Fe (OH) 2 ↓ + 2NH4 +
4 Fe (OH) 2 + O2 + 2H2O = 4Fe (OH) 3
CuSO4 was added dropwise to the solution of ammonia, the first, blue-green
precipitate generated excess ammonia, the precipitate was dissolved brown solution
placed in the air, solutions become darker in color slightly.
2Co2 + + 2 NH3 + 2H2O = Co (OH) 2 • CoSO4 ↓ + 2 NH4 +
Co (OH) 2 • CoSO4 + 2 NH4 + +10 NH3 = Co (NH3) 62 + + SO42-+ 2H2O
Co (NH3) 62 + + O2 + 2H2O = 4 Co (NH3) 63 + + 4OH The NiSO4 solution by adding ammonia, the first green precipitate generated when
the green precipitate dissolved in excess ammonia to produce a blue Ni (NH3) 63 +
2Ni2 + + 2NH3 + 2H2O + SO42-= Ni (OH) 2 • NiSO4 + 2NH4 +
Ni (OH) 2 • NiSO4 + 2 NH4 + + 10 NH3 = 2 Ni (NH3) 63 + + + 2H2O + SO42 4. Comparison of Fe (OH) 3, Al (OH) 3, Cr (OH) 3 in nature.
Solution: (1) Color Fe (OH) 3 Brown Al (OH) 3 White Cr (OH) 3 gray blue
(2) and excessive NaOH reaction
Al (OH) 3 and Cr (OH) 3 were significant gender, soluble in excess NaOH, and Fe
(OH) 3 is alkaline, insoluble in excess NaOH solution.
(3) The redox
Fe (OH) 3 to restore the capacity of poor, but when you have some acid oxidation
capacity
2 Fe (OH) 3 + 6H + + 2I-= 2Fe2 + + I2 + 3H2O
Al (OH) 3 is neither oxidizing nor reducing, Cr (OH) 3 without significant oxidation,
reduction under weak acidic conditions, but under alkaline conditions to restore the
ability.
2Cr (OH) 3 + 3H2O = 4OH-= 2CrO42-+8 H2O
(4) to generate complex and the dissolution of
Fe (OH) 3, Cr (OH) 3 complexes and the dissolution of generating capacity to
stronger, while the Al (OH) 3 can only be generated with a small number of ligand
complexes and the dissolution of
Fe (OH) 3 + 6CN-= Fe (CN) 63 - + 3OH Al (OH) 3 + OH-= Al (OH) 4 -Cr (OH) 3 + 6CN-= Cr (CN) 63 - + 3OH Cr (OH) 3 + OH-= Cr (OH) 4 -5. Metal M dissolved in dilute hydrochloric acid is generated when MCl2, the
magnetic moment of 5.0B.M.. In the oxygen-free operating conditions MCl2 solution
to produce a white Shen A. A green contact with air gradually, and eventually brown
precipitate B. When burning, B generates a brown-red powder C, C is generated by
the incomplete reduction of the black magnetic iron material D.
D dissolved in dilute hydrochloric acid solution generated E, it allows oxidation of KI
solution, I2, but by adding KI before adding NaF, the KI will not be E by the
oxidation.
If the concentrated NaOH to B, which leads to suspension of a purple-red solution of
Cl2 gas available to F, by adding BaCl2 will precipitate out when the red-brown
solid-G. G is a strong oxidant.
Test to determine the compounds represented by the letter symbols, and write the
reaction equations.
Solution: A: Fe (OH) 2, B: Fe (OH) 3, C: Fe2O3, D: K3 [Fe (C2O4) 3] • 3H2O
E: FeCl3, F: Na2FeO4, G: BaFeO4
Fe + 2HCl = FeCl2 + H2 ↑
FeCl2 + 2NaOH = Fe (OH) 2 + 2NaCl
4Fe (OH) 2 + O2 + 2H2O = 4Fe (OH) 3
2Fe (OH) 3 = Fe2O3 + 3H2O
Fe (OH) 3 + 3KHC2O4 = K3 [Fe (C2O4) 3] • 3H2O
Fe (OH) 3 + 3HCl = FeCl3 + 3H2O
2FeCl3 +2 KI = FeCl2 + I2 + 2KCl
FeCl2 + I2 +12 NaF = 2Na3 [FeI6] + 2NaI + 4NaCl
2Fe (OH) 3 + 3Cl2 + 10NaOH = 2 Na2FeO4 + 6NaCl + 8H2O
Na2FeO4 + BaCl2 = BaFeO4 ↓ + 2NaCl
6. To accomplish the following reaction equation:
(1) FeSO4 + Br2 + H2SO4 →
(2) Co (OH) 2 + H2O2 →
(3) Ni (OH) 2 + Br2 + OH-→
(4) FeCl3 + NaF →
(5) FeCl3 + Ca →
(6) FeCl3 + H2S →
(7) FeCl3 + KI →
(8) Co2O3 + HCl →
(9) Fe (OH) 3 + KCrO3 + KOH →
(10) K4Co (CN) 6 + O2 + H2O →
Solution: (1) 2FeSO4 + Br2 + H2SO4 = Fe2 (SO4) 3 + 2HBr
(2) Co (OH) 2 + H2O2 = 2Co (OH) 3
(3) 2Ni (OH) 2 + Br2 + 2OH-= 2Ni (OH) 3 + 2Br (4) FeCl3 + NaF = Na3 [FeF6] + 3NaCl
(5) 2FeCl3 + 3Ca = 2Fe + 3CaCl2
(6) 2FeCl3 + H2S = 2FeCl2 + S + 2HCl
(7) 2FeCl3 +2 KI = 2FeCl2 + I2 + 2KCl
(8) Co2O3 + 6HCl = 2CoCl + Cl2 + 3H2O
(9) 2Fe (OH) 3 + KCrO3 + 4KOH = 2K2FeO4 + KCl + 5H2O
(10) 4K4Co (CN) 6 + O2 + 2H2O = 4K3 [Co (CN) 6] + 4KOH
7. Known = +0.771 V, =-0.44V, trial calculations.
Solution: Fe3 + + e Fe2 + 0.771V
Fe2 + + 2e Fe-0.44V
Fe3 + + 3e Fe
8. Cited the identification of Fe3 +, Fe2 +, Co2 + and Ni2 + ions commonly used
method.
Solution: (1) identification of Fe3 +
Fe3 + case of KSCN to immediately change the blood red Fe3 + + SCN-=
FeSCN2 +
Fe3 + case of K4 [Fe (CN) 6] Blue Fe3 + + Fe (CN) 64 - + K + = KFe [Fe (CN) 6]
(2) Identification of Fe2 +
Fe2 + case of K3 [Fe (CN) 6] Blue Fe2 + + Fe (CN) 63 - + K + = KFe [Fe (CN) 6]
Fe2 + allows acidic potassium permanganate fading
5Fe2 + + MnO4-+ 8H + = 5Fe3 + + Mn2 + + 4H2O
(3) Identification of Co2 +
After the effects of Co2 + with KSCN extracted with ether, ethyl ether layer is
blue
Co2 + + 4SCN-= Co (SCN) 42 -Co2 + the first case of NaOH generated a blue precipitate, the precipitation is
gradually turning into pink
Co2 + + 2OH-= Co (OH) 2 ↓
(4) Identification of Ni2 +
Ni2 + with a small amount of ammonia, the reaction of excess ammonia, the
precipitation of the green was blue solution
Ni2 + + 6NH3 = Ni (NH3) 62 +
Ni2 + with dimethylglyoxime in ammonia solution, the reaction of bright red
precipitate
9. How do I separate Fe3 +, Al3 +, Cr3 + and Ni2 + ions?
Solution: Fe3 +, Al3 +, Cr3 +, Ni2 +
Excess NaOH
Fe (OH) 3, Ni (OH) 2 Al (OH) 4 -, Cr (OH) 4 -NH3 • H2O (1) H2O2, Δ;
(2) PH = 4 ∽ 5, BaCl2
Fe (OH) 3 Ni (NH3) 42 +
BaCrO4 Al3 +
10. For preparation of pure ZnSO4, known crude solution containing Fe (Fe3 +, Fe2 +)
0.56g • dm-3, Cu2 +0.63 g • dm-3, without the introduction of impurities (including
Na +) cases, how to design This process? Write reaction equation.
Solution: Flow Chart is as follows:
ZnSO4, Cu2 +, Fe2 +, Fe3 +
Filter plus Zn powder
Sulfuric acid
Cu, Zn ZnSO4, Fe2 + (a little Fe3 +)
H2SO4 acidified filter pass into the air
Filter pH = 3 ∽ 4
Cu ↓ ZnSO4 ZnSO4 Fe (OH) 3 ↓
Concentrated crystal filter
Mother liquor ZnSO4 • 7H2O
In addition to Cu: Zn + Cu2 + = Zn2 + + Cu ↓
pH 3-4
In addition to Fe: Fe2 + + H2O Fe (OH) 3 ↓
O2
pH 3-4
Fe3 + + H2O Fe (OH) 3 ↓
11. There is a coordination compound is determined by Co3 + ions, NH3 molecules
and ions formed by Cl. From 11.67g of the coordination compounds precipitate out of
Cl ions, the need 8.5gAgNO3, then the same amount of decomposition of the
coordination compounds obtained 4.48dm3 ammonia (standard mode). Is known that
the coordination compounds of molecular weight of 233.3, and seeking its chemical
type, and pointed out that its internal community, the composition of the outside
world.
Solution: Solution: 8.5g AgNO3 is equivalent to nCl-=
nNH3 =
11.6g complexes, with a median of 6 it is chemical formula as follows:
[CoШ (NH3) 4Cl2] Cl (calculated molecular weight of 233.4)
AgNO3 precipitation by external Cl-ion is Cl-ions
12. 521mg of pure iron wire in an inert atmosphere, dissolved in excess HCl solution
added to the above hot solution 253mgKNO3. After reaction, the remaining solution
with the concentration of Fe2 + ions 16.7mol • dm Cr2O solution for titration, a total
dichromate solution 18.00cm3. Try to export the reaction between Fe and NO
stoichiometric relationship.
Solution: 14H + + 6Fe2 + + Cr2O == 6Fe3 + + 2Cr3 + + 7H2O
61
6 × 0.3006 16.7 × 0.018 = 0.3006
= 1.8036
The original n Fe = = 0.009mol
n KNO3 = = 0.0025mol
2H + + 3Fe2 + + NO == 3Fe3 + + NO + 2H2O
13. Pointed out that the following magnetic properties of coordination compounds:
(1) K4 [Fe (CN) 6]
(2) K3 [Fe (CN) 6]
(3) Ni (CO) 4
(4) [Co (NH3) 6] Cl3
(5) K4 [Co (CN) 6]
(6) Fe (CO) 5
Solution: (1) K4 [Fe (CN) 6]
Fe (И) 3d64s04p0, d2sp3, anti-magnetic
(2) K3 [Fe (CN) 6]
Fe (Ш) 3d54s04p0, d2sp3, paramagnetic
(3) Ni (CO) 4 excited
Ni (0) 3d84s24p0 3d104s04p0, sp3 anti-magnetic
(4) [Co (NH3) 6] Cl3
Cl (Ш) 3d64s04p0, d2sp3, anti-magnetic
(5) K4 [Co (CN) 6] excitation
Co (И) 3d74s04p0 3d64s04p05s1, d2sp3, paramagnetic
(6) Fe (CO) 5 excitation
Fe (0) 3d64s24p0 3d84s04p0, dsp3 anti-magnetic
14. Platinum group elements in the main minerals that? How to extract the metal
platinum?
Solution: Answer: The main minerals in the original platinum mines, due to the nature
of platinum group elements are similar, almost six elements co-exist, such as:
Osmium Iridium in the form of the alloy exists in the original platinum mines, due to
the stability of platinum group elements so that they ore a single qualitative state of
existence.
Wang first used to extract metals platinum ore treatment, so that platinum and
osmium, iridium basic separation of platinum into the solution, and then restore
glucose or other reducing agents, under certain conditions, make Palladium and other
impurities are all separated out, while the platinum to remain , the third step from the
solution with (NH4) 2PtCl6 the form of precipitation of platinum, burning
decomposition, melting. Δ
(NH4) 2PtCl6 ═ ═ ═ ═ Pt + 2NH4Cl + Cl2
15. Links the chemical properties of platinum in the platinum system that is able to
engage in the following containers of various reagents involved in chemical reactions:
(1) HF (2) aqua regia (3) HCl + H2O2 (4) NaOH + Na2O2
(5) Na2CO3 (6) NaHSO4 (7) Na2CO3 + S
Solution: A: The Platinum system, the container can be carried out with (10), (5), (6)
in the reagent involved in the reaction, because HF, Na2CO3, NaHSO4 not alone with
the Pt reaction, and (2), (3) , (4), (7), the reagent can not be involved in the reaction
system, platinum container, Pt with aqua regia, and H2O2 + HCl, Na2O2 + NaOH,
Na2CO3 + S can react with
3Pt + 4HNO3 + 18HCl = 3H2PtCl6 + 4NO + 8H2O
Pt + 2H2O2 + 6HCl = H2PtCl6 + 4H2O
Pt + 2Na2O2 = PtO2 + 2Na2O
Pt + 2S = PtS2
16. That the following two kinds of complexes the number of geometric isomers, and
draw their structural style to:
(1) [Pt (NO2) Cl2 (NH3) (en)] Cl (2) [Pt (NO2) BrCl (NH3) (en)] Cl
Solution: (1) four kinds of
(2) six
17. Fill in the blanks:
(1) Cr2O (color) (2) CrO (color) (3) CrO5 (color)
(4) MnO (color) (5) MnO (color) (6) Cu2 + (color)
(7) Cu (NH3) (color) (8) CuCl (color) (9) Cu (CN) (color)
(10) Fe (OH) 3 (color) (11) Fe (OH) 2 (color) (12) Co (OH) 2 (color)
(13) Zn (OH) 2 (color) (14) Cu (OH) 2 (color)
Solution: (1) Orange (2) yellow (3) dark blue (4) Purple (5) Green
(6) Blue (7) Deep Blue (8) Yellow (9) Blue (10) red-brown
(11) white case of O2 into a blue-green (12) pink (13) White (14) Blue
Chapter 21
1. To explain the following issues:
(1) In the Fe3 + ions in solution by adding KSCN solution occurs when the blood red,
but adding a little iron, the blood red and disappear, what kind of logic is that?
(2) Why Fe3 + salt is stable, while the Ni3 + salt, has not yet obtained?
(3) Why can not by Fe3 + in aqueous solution of salt and KI obtained FeI3?
(4) When acting on the FeCl3 solution, Na2CO3 solution, why get too is Fe (OH) 3
and the step is Fe2 (CO3) 3?
(5) What color gel that contains ingredients? Dry Why was blue, pink after absorbing
water change?
Solution: (1) Fe3 + + nSCN-= Fe (SCN) n3-n
n = 1-6 blood-red by adding iron, the iron to Fe (SCN) n3-n reduction, the
resulting Fe2 + is not with the SCN-generated colored complex, and therefore red
blood disappearance
2Fe (SCN) n3-n + Fe = 3Fe2 + + nSCN (2) iron outer electron configuration of 3d64s2, Fe3 + half-full of the 3d5 electron
configuration to make it more than Fe2 + electronic configuration of 3d6 stable outer
electron configuration of nickel as 3d84s2, of which more than half-full 3d electronic
state of more difficult to lose, so under normal circumstances, the performance of the
+2 oxidation state of nickel, under special conditions can also be unstable Ni (III)
compounds.
(3) Eθ (Fe3 +, Fe2 +)> Eθ (I2/I-), in aqueous solution, Fe3 + will be I-oxidation is
not FeI3
2 Fe3 + + 2 I-= 2Fe2 + + I2
(4) Na2CO3 in water hydrolysis, CO32-+ H2O HCO3-+ OH Solution [CO32-] and [OH-] or less, while the Fe (OH) 3 is much smaller than the
solubility of Fe2 (CO3) 3, so the event of FeCl3 with Na2CO3 solution to produce Fe
(OH) 3 precipitation, rather than generating Fe2 (CO3) 3 precipitation.
2Fe3 + + 3CO32-+ 6H2O = 2 Fe (OH) 3 ↓ + 3H2CO3
(5) color gel in the shouting moisture indicator CoCl2, old crystal water contained in
the number of different colors are different. CoCl2 • 6H2O CoCl2 • 2 H2O CoCl2 •
H2O CoCl2
Pink purple blue-violet blue
Therefore, the blue anhydrous CoCl2 significantly, CoCl2 crystal water were more
noticeable pink.
2. Write with the following experiments were reactive phenomenon:
To the solution containing Fe2 + ions were generated after adding NaOH, white-green
sediment, graded brown. After filtration, dissolve brown precipitate with hydrochloric
acid, the solution yellow. Add a few drops of KSCN solution, an immediate change of
blood red, which leads to SO2 when the red disappears. Dropping KMnO4 solution,
the purple will legs to go. Finally, when adding yellow blood salt solution to generate
a blue precipitate.
Solution: Fe2 + + 2OH-= Fe (OH) 2
4Fe (OH) 2 + O2 + 2H2O = 4Fe (OH) 3
Fe (OH) 3 + 3HCl = Fe3 + + 3H2O +3 Cl Fe3 + + SCN-= FeSCN2 +
2FeSCN2 + + SO2 + 2H2O = 2Fe2 + + SO42-+ 4H + + 2SCN MnO4-+ 5Fe2 + + 8H + = Mn2 + + 5Fe3 + + 4H2O
Fe3 + + Fe (CN) 64 - + K + = KFe [Fe (CN) 6]
3. That the following experiment was experimental phenomena, and write a reaction
formula:
(1) treatment with concentrated hydrochloric acid Fe (OH) 3, Co (OH) 3, Ni (OH) 3
precipitation phenomenon when what?
(2) FeSO4, CoSO4, NiSO4 by adding ammonia solution, and what phenomenon?
Solution: (1) with concentrated HCl treatment Fe (OH) 3, the brown precipitate
dissolution, was yellow solution
Fe (OH) 3 + HCl = FeCl3 + 3H2O
With concentrated HCl treatment Co (OH) 3, brown Co (OH) 3 dissolved in solution
was pink, with chlorine gas release
2 Co (OH) 3 + 2Cl-+ 6H + + 6H2O = 2Co (H2O) 62 + (pink) + Cl2 ↑
Concentrated HCl solution when dealing with excessive blue
Co (H2O) 62 + + 4Cl-= CoCl42-(Blue) + 6H2O
Concentrated hydrochloric acid treatment Ni (OH) 3, the black Ni (OH) 3 was
dissolved and a green solution of chlorine released 2 Ni (OH) 3 + 2Cl-+ 6H + + 6H2O
= 2Ni (H2O) 62 + + Cl2 ↑
(2) In FeSO4 solution by adding ammonia, because of the solution dissolved O2, not
white, but blue-green precipitate precipitation, precipitation in the air, slowly turned
brown.
Fe2 + + 2NH3 + 2H2O = Fe (OH) 2 ↓ + 2NH4 +
4 Fe (OH) 2 + O2 + 2H2O = 4Fe (OH) 3
CuSO4 was added dropwise to the solution of ammonia, the first, blue-green
precipitate generated excess ammonia, the precipitate was dissolved brown solution
placed in the air, solutions become darker in color slightly.
2Co2 + + 2 NH3 + 2H2O = Co (OH) 2 • CoSO4 ↓ + 2 NH4 +
Co (OH) 2 • CoSO4 + 2 NH4 + +10 NH3 = Co (NH3) 62 + + SO42-+ 2H2O
Co (NH3) 62 + + O2 + 2H2O = 4 Co (NH3) 63 + + 4OH The NiSO4 solution by adding ammonia, the first green precipitate generated when
the green precipitate dissolved in excess ammonia to produce a blue Ni (NH3) 63 +
2Ni2 + + 2NH3 + 2H2O + SO42-= Ni (OH) 2 • NiSO4 + 2NH4 +
Ni (OH) 2 • NiSO4 + 2 NH4 + + 10 NH3 = 2 Ni (NH3) 63 + + + 2H2O + SO42 4. Comparison of Fe (OH) 3, Al (OH) 3, Cr (OH) 3 in nature.
Solution: (1) Color Fe (OH) 3 Brown Al (OH) 3 White Cr (OH) 3 gray blue
(2) and excessive NaOH reaction
Al (OH) 3 and Cr (OH) 3 were significant gender, soluble in excess NaOH, and Fe
(OH) 3 is alkaline, insoluble in excess NaOH solution.
(3) The redox
Fe (OH) 3 to restore the capacity of poor, but when you have some acid oxidation
capacity
2 Fe (OH) 3 + 6H + + 2I-= 2Fe2 + + I2 + 3H2O
Al (OH) 3 is neither oxidizing nor reducing, Cr (OH) 3 without significant oxidation,
reduction under weak acidic conditions, but under alkaline conditions to restore the
ability.
2Cr (OH) 3 + 3H2O = 4OH-= 2CrO42-+8 H2O
(4) to generate complex and the dissolution of
Fe (OH) 3, Cr (OH) 3 complexes and the dissolution of generating capacity to
stronger, while the Al (OH) 3 can only be generated with a small number of ligand
complexes and the dissolution of
Fe (OH) 3 + 6CN-= Fe (CN) 63 - + 3OH Al (OH) 3 + OH-= Al (OH) 4 -Cr (OH) 3 + 6CN-= Cr (CN) 63 - + 3OH Cr (OH) 3 + OH-= Cr (OH) 4 -5. Metal M dissolved in dilute hydrochloric acid is generated when MCl2, the
magnetic moment of 5.0B.M.. In the oxygen-free operating conditions MCl2 solution
to produce a white Shen A. A green contact with air gradually, and eventually brown
precipitate B. When burning, B generates a brown-red powder C, C is generated by
the incomplete reduction of the black magnetic iron material D.
D dissolved in dilute hydrochloric acid solution generated E, it allows oxidation of KI
solution, I2, but by adding KI before adding NaF, the KI will not be E by the
oxidation.
If the concentrated NaOH to B, which leads to suspension of a purple-red solution of
Cl2 gas available to F, by adding BaCl2 will precipitate out when the red-brown
solid-G. G is a strong oxidant.
Test to determine the compounds represented by the letter symbols, and write the
reaction equations.
Solution: A: Fe (OH) 2, B: Fe (OH) 3, C: Fe2O3, D: K3 [Fe (C2O4) 3] • 3H2O
E: FeCl3, F: Na2FeO4, G: BaFeO4
Fe + 2HCl = FeCl2 + H2 ↑
FeCl2 + 2NaOH = Fe (OH) 2 + 2NaCl
4Fe (OH) 2 + O2 + 2H2O = 4Fe (OH) 3
2Fe (OH) 3 = Fe2O3 + 3H2O
Fe (OH) 3 + 3KHC2O4 = K3 [Fe (C2O4) 3] • 3H2O
Fe (OH) 3 + 3HCl = FeCl3 + 3H2O
2FeCl3 +2 KI = FeCl2 + I2 + 2KCl
FeCl2 + I2 +12 NaF = 2Na3 [FeI6] + 2NaI + 4NaCl
2Fe (OH) 3 + 3Cl2 + 10NaOH = 2 Na2FeO4 + 6NaCl + 8H2O
Na2FeO4 + BaCl2 = BaFeO4 ↓ + 2NaCl
6. To accomplish the following reaction equation:
(1) FeSO4 + Br2 + H2SO4 →
(2) Co (OH) 2 + H2O2 →
(3) Ni (OH) 2 + Br2 + OH-→
(4) FeCl3 + NaF →
(5) FeCl3 + Ca →
(6) FeCl3 + H2S →
(7) FeCl3 + KI →
(8) Co2O3 + HCl →
(9) Fe (OH) 3 + KCrO3 + KOH →
(10) K4Co (CN) 6 + O2 + H2O →
Solution: (1) 2FeSO4 + Br2 + H2SO4 = Fe2 (SO4) 3 + 2HBr
(2) Co (OH) 2 + H2O2 = 2Co (OH) 3
(3) 2Ni (OH) 2 + Br2 + 2OH-= 2Ni (OH) 3 + 2Br (4) FeCl3 + NaF = Na3 [FeF6] + 3NaCl
(5) 2FeCl3 + 3Ca = 2Fe + 3CaCl2
(6) 2FeCl3 + H2S = 2FeCl2 + S + 2HCl
(7) 2FeCl3 +2 KI = 2FeCl2 + I2 + 2KCl
(8) Co2O3 + 6HCl = 2CoCl + Cl2 + 3H2O
(9) 2Fe (OH) 3 + KCrO3 + 4KOH = 2K2FeO4 + KCl + 5H2O
(10) 4K4Co (CN) 6 + O2 + 2H2O = 4K3 [Co (CN) 6] + 4KOH
7. Known = +0.771 V, =-0.44V, trial calculations.
Solution: Fe3 + + e Fe2 + 0.771V
Fe2 + + 2e Fe-0.44V
Fe3 + + 3e Fe
8. Cited the identification of Fe3 +, Fe2 +, Co2 + and Ni2 + ions commonly used
method.
Solution: (1) identification of Fe3 +
Fe3 + case of KSCN to immediately change the blood red Fe3 + + SCN-=
FeSCN2 +
Fe3 + case of K4 [Fe (CN) 6] Blue Fe3 + + Fe (CN) 64 - + K + = KFe [Fe (CN) 6]
(2) Identification of Fe2 +
Fe2 + case of K3 [Fe (CN) 6] Blue Fe2 + + Fe (CN) 63 - + K + = KFe [Fe (CN) 6]
Fe2 + allows acidic potassium permanganate fading
5Fe2 + + MnO4-+ 8H + = 5Fe3 + + Mn2 + + 4H2O
(3) Identification of Co2 +
After the effects of Co2 + with KSCN extracted with ether, ethyl ether layer is
blue
Co2 + + 4SCN-= Co (SCN) 42 -Co2 + the first case of NaOH generated a blue precipitate, the precipitation is
gradually turning into pink
Co2 + + 2OH-= Co (OH) 2 ↓
(4) Identification of Ni2 +
Ni2 + with a small amount of ammonia, the reaction of excess ammonia, the
precipitation of the green was blue solution
Ni2 + + 6NH3 = Ni (NH3) 62 +
Ni2 + with dimethylglyoxime in ammonia solution, the reaction of bright red
precipitate
9. How do I separate Fe3 +, Al3 +, Cr3 + and Ni2 + ions?
Solution: Fe3 +, Al3 +, Cr3 +, Ni2 +
Excess NaOH
Fe (OH) 3, Ni (OH) 2 Al (OH) 4 -, Cr (OH) 4 -NH3 • H2O (1) H2O2, Δ;
(2) PH = 4 ∽ 5, BaCl2
Fe (OH) 3 Ni (NH3) 42 +
BaCrO4 Al3 +
10. For preparation of pure ZnSO4, known crude solution containing Fe (Fe3 +, Fe2 +)
0.56g • dm-3, Cu2 +0.63 g • dm-3, without the introduction of impurities (including
Na +) cases, how to design This process? Write reaction equation.
Solution: Flow Chart is as follows:
ZnSO4, Cu2 +, Fe2 +, Fe3 +
Filter plus Zn powder
Sulfuric acid
Cu, Zn ZnSO4, Fe2 + (a little Fe3 +)
H2SO4 acidified filter pass into the air
Filter pH = 3 ∽ 4
Cu ↓ ZnSO4 ZnSO4 Fe (OH) 3 ↓
Concentrated crystal filter
Mother liquor ZnSO4 • 7H2O
In addition to Cu: Zn + Cu2 + = Zn2 + + Cu ↓
pH 3-4
In addition to Fe: Fe2 + + H2O Fe (OH) 3 ↓
O2
pH 3-4
Fe3 + + H2O Fe (OH) 3 ↓
11. There is a coordination compound is determined by Co3 + ions, NH3 molecules
and ions formed by Cl. From 11.67g of the coordination compounds precipitate out of
Cl ions, the need 8.5gAgNO3, then the same amount of decomposition of the
coordination compounds obtained 4.48dm3 ammonia (standard mode). Is known that
the coordination compounds of molecular weight of 233.3, and seeking its chemical
type, and pointed out that its internal community, the composition of the outside
world.
Solution: Solution: 8.5g AgNO3 is equivalent to nCl-=
nNH3 =
11.6g complexes, with a median of 6 it is chemical formula as follows:
[CoШ (NH3) 4Cl2] Cl (calculated molecular weight of 233.4)
AgNO3 precipitation by external Cl-ion is Cl-ions
12. 521mg of pure iron wire in an inert atmosphere, dissolved in excess HCl solution
added to the above hot solution 253mgKNO3. After reaction, the remaining solution
with the concentration of Fe2 + ions 16.7mol • dm Cr2O solution for titration, a total
dichromate solution 18.00cm3. Try to export the reaction between Fe and NO
stoichiometric relationship.
Solution: 14H + + 6Fe2 + + Cr2O == 6Fe3 + + 2Cr3 + + 7H2O
61
6 × 0.3006 16.7 × 0.018 = 0.3006
= 1.8036
The original n Fe = = 0.009mol
n KNO3 = = 0.0025mol
2H + + 3Fe2 + + NO == 3Fe3 + + NO + 2H2O
13. Pointed out that the following magnetic properties of coordination compounds:
(1) K4 [Fe (CN) 6]
(2) K3 [Fe (CN) 6]
(3) Ni (CO) 4
(4) [Co (NH3) 6] Cl3
(5) K4 [Co (CN) 6]
(6) Fe (CO) 5
Solution: (1) K4 [Fe (CN) 6]
Fe (И) 3d64s04p0, d2sp3, anti-magnetic
(2) K3 [Fe (CN) 6]
Fe (Ш) 3d54s04p0, d2sp3, paramagnetic
(3) Ni (CO) 4 excited
Ni (0) 3d84s24p0 3d104s04p0, sp3 anti-magnetic
(4) [Co (NH3) 6] Cl3
Cl (Ш) 3d64s04p0, d2sp3, anti-magnetic
(5) K4 [Co (CN) 6] excitation
Co (И) 3d74s04p0 3d64s04p05s1, d2sp3, paramagnetic
(6) Fe (CO) 5 excitation
Fe (0) 3d64s24p0 3d84s04p0, dsp3 anti-magnetic
14. Platinum group elements in the main minerals that? How to extract the metal
platinum?
Solution: Answer: The main minerals in the original platinum mines, due to the nature
of platinum group elements are similar, almost six elements co-exist, such as:
Osmium Iridium in the form of the alloy exists in the original platinum mines, due to
the stability of platinum group elements so that they ore a single qualitative state of
existence.
Wang first used to extract metals platinum ore treatment, so that platinum and
osmium, iridium basic separation of platinum into the solution, and then restore
glucose or other reducing agents, under certain conditions, make Palladium and other
impurities are all separated out, while the platinum to remain , the third step from the
solution with (NH4) 2PtCl6 the form of precipitation of platinum, burning
decomposition, melting. Δ
(NH4) 2PtCl6 ═ ═ ═ ═ Pt + 2NH4Cl + Cl2
15. Links the chemical properties of platinum in the platinum system that is able to
engage in the following containers of various reagents involved in chemical reactions:
(1) HF (2) aqua regia (3) HCl + H2O2 (4) NaOH + Na2O2
(5) Na2CO3 (6) NaHSO4 (7) Na2CO3 + S
Solution: A: The Platinum system, the container can be carried out with (10), (5), (6)
in the reagent involved in the reaction, because HF, Na2CO3, NaHSO4 not alone with
the Pt reaction, and (2), (3) , (4), (7), the reagent can not be involved in the reaction
system, platinum container, Pt with aqua regia, and H2O2 + HCl, Na2O2 + NaOH,
Na2CO3 + S can react with
3Pt + 4HNO3 + 18HCl = 3H2PtCl6 + 4NO + 8H2O
Pt + 2H2O2 + 6HCl = H2PtCl6 + 4H2O
Pt + 2Na2O2 = PtO2 + 2Na2O
Pt + 2S = PtS2
16. That the following two kinds of complexes the number of geometric isomers, and
draw their structural style to:
(1) [Pt (NO2) Cl2 (NH3) (en)] Cl (2) [Pt (NO2) BrCl (NH3) (en)] Cl
Solution: (1) four kinds of
(2) six
17. Fill in the blanks:
(1) Cr2O (color) (2) CrO (color) (3) CrO5 (color)
(4) MnO (color) (5) MnO (color) (6) Cu2 + (color)
(7) Cu (NH3) (color) (8) CuCl (color) (9) Cu (CN) (color)
(10) Fe (OH) 3 (color) (11) Fe (OH) 2 (color) (12) Co (OH) 2 (color)
(13) Zn (OH) 2 (color) (14) Cu (OH) 2 (color)
Solution: (1) Orange (2) yellow (3) dark blue (4) Purple (5) Green
(6) Blue (7) Deep Blue (8) Yellow (9) Blue (10) red-brown
(11) white case of O2 into a blue-green (12) pink (13) White (14) Blue
Chapter 22
1. To write in the correct order of lanthanide and actinide elements names and
symbols, along with the column order of their atoms.
Solution: Lanthanide:
Element Name
La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
La Ce Pr Nb Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
57 58 59 60 61 62 63 6,465,666,768,697,071
Actinide
Actinium thorium protactinium uranium neptunium plutonium americium curium
Berkelium Californium Einsteinium Fermium Mendelevium Nobelium rhodium
Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
89 9,091,929,394,959,697 9,899,100,101,102,103
2. The characteristics of lanthanide oxidation state of +3, why Ce, Pr, Tb, Dy tends to
+4 oxidation state, while the Sm, Eu, Tm, Yb +2 oxidation state was able to show?
A: The characteristics of lanthanide oxidation states are + Ш, this is because the
lanthanide atoms first, second, third, and the ionization potentials is not large, the
energy released when the bond sufficient to cover the atoms in the ionization time
energy consumption, and therefore their + Ш oxidation state are stable.
+ Ш characteristics and differences in oxidation states of various elements in place
of regular distribution of the La (+ Ш),
Gd (+ Ш), Lu (+ Ш) in the vicinity. This is demonstrated by changes in the laws of
atomic structure be explained. La (+ Ш),
Gd (+ Ш), Lu (+ Ш) with the 4f orbital were all empty, half full, all over the stability
of the electron shell structure.
Lanthanide oxidation state depends on its outer 6s electron, 4f and 5d electrons, when
they lost two 6s electron, which can then lose a 4f 1 5d electronic or e-when the screw
is not required for ionization is too high, Therefore, to demonstrate the characteristics
of ethnic ШB oxidation state (+3).
Ce, Pr, Tb, Dy addition to generating +3 oxidation state, but also can be to lose an
electron, 4f orbit at or near full-full or half-full, into a stable structure, so they often
appear +4 oxidation state.
The Sm, Eu, Tm and Yb, loss of two electronic, you can make 4f orbit at or near the
half-full, or full-full stability, and Changcheng emerged in +2 oxidation state.
3. Interpretation of lanthanide elements in the chemical nature of the similarities.
A: The price of lanthanide electron shell structure of 4f1-145d0-16s2 view of the
energy of 4f comparison with the 5d
Close, so to lose three electrons +3 oxidation state characteristics of the energy
required when more or less, even in aqueous solution, because of their ionic radius
close to making ion hydration can also more or less, so whether in the ionization
energy, or electro-negativity and the standard electrode potential, the lanthanide
elements are closer, indicating that their simple substance in any state, demonstrated
by the chemical activity is similar, in the +3 oxidation state, the ionic configuration
and ion radius is also similar, so the nature of Ln3 + ions are very similar.
4. Why is called "lanthanide contraction"? Discuss the reasons for this phenomenon
and its 6th cycle of lanthanide behind the nature of the various elements of the impact
of what happened.
A: The lanthanide contraction, namely: lanthanide atomic radius and ionic radius turn
slowly narrowing phenomenon.
Lanthanide hand reduction is because the lanthanide atom in turn increase the
number of e-filling the outer layer on the third floor of the 4f electrons in the filled 4f
at the same time, the nucleus of nuclear charge in the gradual increase in the
attractiveness of the electron shell is gradually increased, the results of so that the
entire electronic shell of a gradual contraction. However, away from the 4f core layer
closer, shielding the role of large, partially offset by matching the outer electrons
gravity, nuclear effective charge although increased, but not by much larger sequence,
so although the atomic or ionic radius of the contraction However, the value of a
small decrease, in addition to half-filled 4f orbital and full-filled with the Er and Yb,
other general trends of atomic radius is shrinking, while the characteristics of ionic
radius of +3 prices showing a very regular decrease in turn.
Lanthanide contraction results to make the back of the transition elements of
atomic radius with the corresponding element of the same family above a radius of
almost equal, such as IVB of Zr and Hf; VB of Nb and Ta; VIB of the Mo and W,
atomic, ionic radius very close to , similar in nature is difficult to separate.
5. Why lanthanides between differences in the chemical nature of the actinides than in
smaller multi-between?
A: The lanthanides show features more than +3 oxidation state, and the structure and
the radius are also similar, therefore, regardless of their quality or chemical
combination of single-state differences are relatively small. As for the actinide
elements, due to valence electrons filling in the 7s, 6d, 5f orbit, while the light
actinide elements 6d, 5f than the energy difference between the lanthanide in the 5d,
4f energy difference between the even smaller, so , light actinides than lanthanides
easy to show a high oxidation state, such as Ac, Th, Pa, U, Np, Pu followed by the
stability of the oxidation state of +3, +4, +5, +6, +5, + 4. The remaining elements are
characterized by +3, and the actinides are more lanthanides show more kinds of
oxidation states, such as uranium are +3, +4, +5, +6 and other features with multiple
oxidation states, so that actinide elements, whether the differences between the quality
or chemical combination in a single state was relatively large.
6. Why is the formation of lanthanide complexes are mostly ionic? Try to discuss the
stability of lanthanide coordination compounds the law and why.
Answer: Ln3 + ions are outer structure of an 8-electron shell Xe, which is not easy to
participate in the inner 4f electron bond, with higher energy orbital (5d, 6s, 6p)
although may participate in bonding, but the ligand field stabilization energy is small
(only 1 kcal), which shows Ln3 + ion and ligand orbital interaction between the very
weak and therefore difficult to form a covalent complex, but can only rely on Ln3 +
ions on the ligand the formation of electrostatic force ionic complex, but because of
Ln3 + ionic radius greater gravity of the ligand is weak, so the formation of
complexes in general is not stable enough, but due to lanthanide contraction in the
lanthanide ions in coordination with the Center for gradually increase the role of body
gravity, the stability of the complexes also turn into a increments.
7. There is a sample containing uranium re-1.6000g, can be extracted 0.4000g of
U3O8 (relative molecular mass of 842.2), the samples of uranium (relative atomic
mass 238.1) is the mass fraction of what is?
Solution: U% =%
8. Hydrated rare earth chloride Why should it be carried out under a certain vacuum
degree of dehydration? What this and other common water chloride dehydration
similar?
A: They wanted dehydration of hydrated rare earth chloride, often used heating
method heating has promoted the hydrolysis of rare-earth ions and thus are not
anhydrous chloride, in order to prevent hydrolysis, can be used control method of
reading a certain vacuum, so that on the one hand the water vapor out, inhibit
hydrolysis, on the other hand, may also play a role and reduce the dehydration
temperature BeCl2 • 4H2O, ZnCl2 • H2O, MgCl2 • 6H2O, etc. similar to the
dehydration of hydrated chloride
9. Completion and balancing of the following reaction equation:
(1) EuCl2 + FeCl3
(2) CeO2 + HCl EuCl2 + FeCl3
(3) UO2 (NO3) 2
(4) UO3 UO2 (NO3) 2
(5) UO3 +2 HNO3 EuCl2 + FeCl3
(6) UO3 + HF EuCl2 + FeCl3
(7) UO3 + NaOH EuCl2 + FeCl3
(8) UO3 + SF4 EuCl2 + FeCl3
Solution: (1) EuCl2 + FeCl3 = EuCl3 + FeCl2
(2) 2CeO2 + 8HCl = 2CeCl3 + Cl2 + 4H2O
(3) 2UO2 (NO3) 2 = 2UO3 + 4NO2 + O2
(4) 6UO3 = 2U3O8 + O2
(5) UO3 +2 HNO3 = UO (NO3) 2 + H2O
(6) 6UO3 + HF = UO2F2 + H2O
(7) 2UO3 + 2NaOH = Na2U2O7 + H2O
(8) UO3 + 3SF4 = UF6 + 3SOF2
10. Nitrate leaching hydroxide Ce (Ⅳ) and the hydroxide of rare earth (Ⅲ) in mixture,
allowing for control of solution acidity pH = 2.5, rare earth hydroxide into the
solution, but remain in the cerium hydroxide precipitation, through the calculation
instructions.
Solution: KspCe (OH) 4 = 1.5 × 10-51
Therefore, when p H = 2.5, ie [OH-] = 3.162 × 10-12 mol • dm-3 Shi
[Ce4 +] = mol • dm-3
Shows the basic precipitation Ce4 + ions completely,
And Ln (OH) 3 of the Ksp ranged between 1.0 × 10-19-2.5 × 10-4
When the [Ln3 +] = 0.1 mol • dm-3, has begun to precipitate Ln3 + ions, compared
with pH values of 7.82-6.30, so that when pH = 2.5 or even larger when the re-Ce4 +
to Ce (OH) 4 precipitation in the form of removal, while the rare-earth Ln3 + ions,
none precipitation, and so can be separated Ln3 + and Ce4 +.
11. Brief extract from the Monazite principle of mixed rare earth chloride.
A: The monazite is rare earth, thorium, uranium, phosphate ore, available acid
(concentrated sulfuric acid) and soda (concentrated NaOH) to rare earth and thorium,
uranium, phosphorus and other elements of separation, with mixed rare earth chloride,
alkaline more superior. When the monazite concentrate with 50% of the concentrated
NaOH solution, when heated, phosphate into a rare earth chloride and hydrated
sodium hydroxide and generate salt, and washing with hot flooding taking separate
soluble phosphate, the acid precipitate solvent, according to rare-earth than the
thorium, uranium and other acid soluble in nature, when dissolved with hydrochloric
acid to control pH, a certain priority can enable the majority of rare earth leaching,
while uranium, thorium and other rare dissolved, so that can get mixed rare earth
chloride, and with the radioactive elements uranium and thorium separation.
Main reaction:
REPO4 + 3NaOH = RE (OH) 3 + Na3PO4
RE (OH) 3 + 3HCl = RECl3 + 3H2O
Th3 (PO4) 4 + 12NaOH = 3Th (OH) 4 + 4Na3PO4
2U3O8 + O2 + 6NaOH = 3Na2U2O7 + 3H2O
12. Preparation of rare earth metal solution Why when electricity can not be carried
out in aqueous solution?
A: Because the rare earth elements to restore the strong reaction with water,
replacement of hydrogen out of water, so the electrolysis reaction in aqueous solution
and can not get their simple substance.
13. Mainly rare earth (Ⅲ) oxidation state exists in the form of compounds, after
enrichment, it is often prepared by melting electrolytic pure metal or mixed metal,
what compounds commonly used as an electrolyte?
A: Generally with KCl as the electrolyte by adding a certain amount of KCl in order
to reduce the melting point of mixed rare earth, so that electrolysis at a lower
temperature.
14. Lanthanides and actinides with the Department, a Deputy tribe (Ⅲ B), why the
oxidation state of actinides than lanthanides many species?
A: This is from the actinides electronic shell structure of the decision, the actinide
elements in the first half of the 5f electronic and nuclear activity than lanthanides 4f
electron is weak, and therefore can not only track the 6d and 7s e as the valence
electrons are given, but can also track e-5f valence electrons involved in bonding as
the formation of high steady-state. The lanthanide 4f orbit because of its role in the
nucleus of a strong, 4f orbit of the electron can not be the valence electrons involved
in bonding, showing a type of low oxidation states.
15. What are the characteristics of thorium halide?
Solution: thorium halides are the high melting point of white crystals, are vulnerable
to hydrolysis.
Chapter 23
1. Why is nuclear fusion only in very high temperatures to occur?
A: polymer for light nuclei encounter heavier nuclei, and the huge energy release
process is called nuclear fusion, due to the nuclear have brought positron, and mutual
exclusion by static electricity, so in the general conditions, the occurrence of fusion
probability is very small, to make light nuclei must have met a high enough
temperature to make it automatically continue.
2. Why is the need to accelerate α-particles can be caused by nuclear reactions, while
the neutrons can cause nuclear reactions do not need to accelerate it?
A: Because the natural radioactive substances in a-ray energy is not high. Use it to
bombard atomic number greater nuclear, and nuclear reactions can not occur. Must
therefore make a particle acceleration, so that a particle has enough kinetic energy to
overcome their mutual repulsion, mutual collisions, there were a nuclear reaction,
because the neutron is not charged, and the role of the nuclear non-exclusive, so they
do not need to accelerate, they cause nuclear reactions .
3. Explain why lithium and hydrogen nuclei between the fusion temperature than the
fusion between the H2 and H2-core high temperature?
A: The nuclear charge of lithium than the large hydrogen, lithium between the nuclear
and nuclear exclusion of hydrogen than the hydrogen nucleus repulsion between the
large lithium-nuclear and hydrogen nuclei to overcome the exclusion between the
energy required for larger, the required temperature more high.
4. Explain why uranium can not occur in an explosive chain of incidents?
A: Because the U-235 uranium and U-238 mixed together, so that most of the
neutrons will be U-238 "eat", leaving fewer and fewer neutrons, so that can not satisfy
the chain reaction of the conditions (number of neutrons is to maintain at least the
same or slightly increased.
5. Td-232 into a Pb-208 to be emitted by the number of α particles and the number of
β particles?
A: The six a particles and 4 ß particles.
6. To determine all cases arising from the following core?
(a) As (α, n); (b) Li (p, n); (c) P (H, p)
Solution: (a) (b) (c)
7. 1mol of CH4 combustion calculations for the loss of quality? In this process, the
system released 890Kj of energy.
Solution: E = mc2
890 × 103 = m × (3 × 108) 2
m = 9.89 × 10-12kg
8. 1molCo-60 by the decay, it loses or how much energy? (Co → e + Ni, Co core
quality is 59.9381u, Ni core quality is 59.9344u; e quality is 0.000549u.
Solution: ΔE = Δmc2
= (59.9344 + 0.000549 - 59.9381) × 10-6 × (3.0 × 108) 2
=- 2.8 × 108kJ
9. Co-60 half-life was 5.3 years, 1mg of the Co-6015.9 years later, how much is left?
Solution: k = 2.303lg
= 0.693/5.3 = 0.13 / year
Get x = 0.125mg
10. Of a uranium-containing 4.64mg and 1.22mg of U-238 lead -206, estimate the age
of this mineral? U-238 half-life of T1 / 2 = 4.51 × 109 years.
Solution: k = 0.693/t1/2 = 0.693 / (4.51 × 109)
= 1.54 × 10-10 / year
x0 = 4.64 + 1.22 × 238/206 = 6.45mg
Was: t = 1.72 × 109 years
11. With the helium core of nuclear bombardment of Al-27, I get the P-30 and a
neutron, write balanced nuclear reactions in this relationship.
Solution:
12. To write the following transformation process of the nuclear balance equation:
(A) Pu-241 by the decay; (b) Th-232 decay into Ra-228;
(c) Y-84 release of a positron; (d) Ti-44 capture an electron;
(E) Am-241 by the decay; (f) Th-234 decay into Pr-234;
(g) Cl-34 decay into S-34.
Solution: (a)
(B)
(C)
(D)
(E)
(F)
(G) → +
13. Completed and equipped with flat-type the following nuclear reaction
(1) Kr → e +?
(2) Fe + e →?
(3) Cr + He → n +?
(4)? → Mg + e
(5) U → He +?
(6) Cu → e +?
(7) Mg + n → H +?
(8) Be + H → Li +?
(9) U + n → Sr + Te +?
(10)? → e + B
Solution (1) Kr → e + Rb
(2) Fe + e → Mn
(3) Cr + He → n + Fe
(4) Na → Mg + e
(5) U → He + Th
(6) Cu → e + Ni
(7) Mg + n → H + Na
(8) Be + H → Li + He
(9) U + n → Sr + Te + 2 n
(10) Be → e + B
14. To balance the equation written in the following expression:
(A) N (n, p) C; (b) N (p,) C; (c) Cl (n, p) S
Solution (a) N + n → C + H
(b) N + H → C + He
(c) Cl + n → S + H
15. A radioactive isotope decays at 5.0 years, 75%, the number of half-life of the
element?
Xie N = N ()
m = m ()
m = m ()
= () So T = 2.5 years
Chapter 24
1. Examples of what life's elements? What kinds of sub-elements of life?
Answer to the elements needed to sustain life as an essential element for organisms,
called life's elements.
Is divided into (1) a large number of elements and trace elements;
(2) the necessary elements and harmful elements.
2. Exist in the body's Chromium, What is the impact on the body?
A chromium is a human body in trace elements, trace elements in the biological
content of the body, though small, but they are in life activities, role in the process are
very important. Chromium is an indispensable work of insulin substances, direct
intervention in the body of sugar and fat metabolism.
3. Iron is an essential element? What does it do? What is the form of iron in the
presence of harmful to human health?
A. Iron is an important trace elements in the human body, normal body iron has been
almost entirely limited to the specific structure of biological macromolecules of the
closed state of siege, so that it can play a normal physiological role. Once iron ion to
escape or escape closed, free iron ions catalyzed oxidation reactions, so that the body
produces hydrogen peroxide and a number of free radicals, these products would lose
cell metabolism and division, affecting growth and development, and even cause
cancer or death.
4. Ca2 + ions in the organism during growth and development do?
Yes (1) is a major component of bone;
(2) as the enzyme activator;
(3) Messenger role;
(4) and nucleic acid in the phosphate ester oxygen combine to make the stabilization
of the double helix structure.
5. Why does the same amount of pollution, pesticide residues than the metal ions is
more harmful to the body?
Answer as many of the degradation of organic matter can be to reduce toxicity, while
the contamination of metal ions in the body is not biodegradable, they
Often accumulate in specific organs and tissues, constitutes a long-term in vivo role
of pathogenic factors. Heavy metal accumulation and body are closely related to
aging and cell aging. Contamination of metal ions in biological tissue the effective
concentration, the fundamental reason is that these metals pose a biological
macromolecules with amino acids or bases and other active components to form
complexes.
6. A wide range of food intake of food and picky people, health What's the difference?
A wide range of intake of food, people will need the full complement of the various
elements of the human body.
The critical food is very difficult to get full from food elements, so he's health
because of the lack of certain elements of the difference.
7. Usually in two or more of the metal chelate atoms are in close proximity in order to
form a stable chelate, why Cu-Zn superoxide dismutase in the two atoms separated 57
amino acid residues, but the can form stable chelating macrocyclic.
A because it formed a large chelation, the size of its cavity in the ligand close.
8. CO2 gas emitted from the body if it is possible that the physiological role of
non-CO2 contributions to the human body?
A. Although CO2 gas emitted from the body, but not that the physiological role of
non-CO2 contributions to the human body, because Zn2 + with CO2 to form
complexes Zn (OH)-CO2, because CO2 in the cavity of the hydrogen bond to other
groups under the action of, CO2 carbon atoms are bonding with the Zn2 + the
nucleophilic attack of the OH, while the formation of HCO.
9. The human body mainly hydrolase enzyme which type of metal?
Yes (1) iron enzyme; (2) zinc-containing enzyme.
10. The human body the lack of Na +, K +, Ca2 + ions may give rise to what the
disease?
A human body, the lack of Na +, K + cell swelling;
The lack of Ca2 + ions in the human body, PDF is not energy, protein, calcium media,
the lack of Ca2 + there is no activity.