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INORGANIC CHEMISTRY
REASONING QUESTIONS
1. There is a considerable increase in covalent radius
from N to P. However, from As to Bi only small increase
in covalent radius is observed.
Ans. This is due to the presence of
completely filled d and/or f orbital in
heavier members.
2. The ionization enthalpy of the group 15
elements is much greater than that of
group 14 elements in the corresponding
periods.
Ans. Because of the extra stable
half-filled p orbital electronic configuration
and smaller size.
3. R3P=O exist but R3N=O does not.
Ans. Due to the absence of d orbitals in valence shell of
nitrogen and because of inability of Nitrogen to expand its
covalency beyond four, nitrogen cannot form d π–pπ
bond.
4. In solid state PCl5 exists as Ionic
compound.
Ans. Since solid phosphorous pentachloride exists as
[PCl4]+ [PCl6]-and hence exhibit some ionic
character. [PCl4]+ is tetrahedral and the anion [PCl6]– is
octahedral.
5. Tendency to show –2 oxidation state
diminishes from Sulphur to polonium in group
16.
Ans. The outer electronic configuration of
group 16 elements is ns2 np4. These
elements therefore have the tendency to
gain two electrons to complete the octet.
Since elctronegativity and I.E. decrease on
going down the group, tendency to show
–2 oxidation state diminishes.
6. Ozone (O3) act as a powerful oxidising
agent.
Ans. Due to the ease with which it liberates atoms of
nascent oxygen (O3 → O2 + O), it acts as a powerful
oxidizing agent.
7. Halogens are coloured.
Ans. Halogens are coloured. This is due to
absorption of radiations in visible region
which results in the excitation of outer
electrons to higher energy level. By
absorbing different quanta of radiation, they
display different colours. For example, F2
has yellow, Cl2 greenish yellow, Br2 red and
I2 violet colour.
8. Cl2 bleaches a substance permanently
but SO2 does it temporarily.
Ans. Cl2 bleaches a substance permanently because it
is due to oxidation but SO2 does it temporarily because
it is due to reduction.
9. Xe does not forms compounds such as
XeF3 and XeF5.
Ans. By the promotion of one, two or three electrons from
filled p-orbital to the vacant d-orbital in the valence shell,
2,4 or 6 half filled orbitals are formed. Thus Xe can
combine only with even number of fluorine atoms
and not odd.
10. Helium is used for inflating aeroplane tyres &
filling balloons for meteorogical observations.
Ans. Helium is a non-inflammable and light
gas.
Why does nitrogen does not form
pentahalides?
Nitrogen with n=2 has s and p orbitals only.It
does not have d orbitals to expand its coelent
beyond four.Thats why it does not form
pentahalides.
Why does PH3 has lowrer boiling point than
NH3?
Unlike NH3 PH3 molecules are not associated
through hydrogen bonding in liquid
state.That’s why the Boling point of PH3 is
lower than NH3.
Why are Pentahalides more covalent than
Trihalides ?
Higher the positive oxidation state of central
atom ,more will be it’s polarising power
which in turn increases the covalent
character of bond formed between the
central atom and the other atom.
Why is BIH3 the strongest reducing agent
amongst the hydrides of group 15 elements?
BIH3 is the strongest reducing agent amongst
all the hydrides of group 15 element.Because
BIH3 is the least stable amongst the hydrides
of group 15.
Why does NH3 acts as a Lewis Base?
Nitrogen atom in NH3 has one lone pair of
electron which is available for
donation.Therefore it acts as a Lewis base.
Why is the highest oxidation state of metal
exhibited in its oxides or fluorides only?
The highest oxidation state of a metal
exhibited in its oxides or fluorides because of
small size and higher electronegatively oxygen
or fluorine can oxidise the metal to its higher
oxidation state.
Why are Mn2+ compounds more stable than
Fe2+ towards oxidation to their +3 state?
Mn2+ compounds are more stable than Fe2+
towards oxidation to their +3 state because
Mn2+ has 3d configuration which has extra
stabilty.
Why is Actinoids contraction is greater from
elements to elements than lanthanoids
contraction?
The 5F electrons are more effectively shielded
from nuclear charge.In other words the 5F
electrons themseles provide poor shielding
from element to element in the series.
Why Cu+ ions is not stable in aqueous
solution?
Cu+ ions in aqueous solution undergoes
disproportionation ie
2Cu+ -> Cu2+ (aq) + Cu(s)
Why is Cr2+ reducing and Mn3+ oxidising when
both have d4 configuration?
Cr2+ is reducing as its configuration changes
from d4 to d3 later having half filled t2g
level.On the other hand ,the change from
Mn2+ to Mn3+ results in the half filled d5
configuration which has extra stability.
Q.Why is first ionisation energies of boron
and aluminium are lower than that of
beryllium
Ans. In B and Al,the first electron to be removed
lies in p-orbitals while in beryllium and
magnesium the valency electrons of an orbital
are strongly attracted by nucleus and hence,
difficult to remove tan the p-orbitals.
Q. Boric acid can be titrated against sodium
hydroxide using phenolphthalein indicator
only in presence of polyhydroxy compounds
like catechal. Explain .
Ans. Boric acid,B(OH)3 is a very weak lewis acid.
It forms a stable complex with polyhydroxy
compounds. Due to formation of this stable
complex, boric acid acts as a strong acid and
hence can be titrated against NaOH
Q. Why (SiH3)3N is a weaker base than (CH3)3N.
Ans. In (SiH3)3N , lone pair of electrons on
nitrogen is involved in pπ-dπ back bonding,
while in (CH3)3N pπ-dπ back bonding is not
possible because of absence of d-orbitals.
Hence, (CH3)3N is more basic than (SiH3)3N
Q. The experiment determined N-F bond length
in NF3 is greater than the sum of the single
bond covalent radii of N ans F. why
Ans. Nitrogen and flourine both are small and
have high electron density and hence they
repel the bonded pair of electrons leading to
larger bond length than expected.
Q. Why Nitrogen is a gas while other members
of this group are solids.
Ans. Due to smaller size of nitrogen atom,it can
undergo sideways overlap with the result two
π bonds are formed between two nitrogen
atoms (N≡N). The discrete nitrogen molecules
are held together by weak Van Der Waal’s
forces. On the other hand,other atoms are
bigger, so sideways overlap is not strong and
hence multiple bonds are not formed . Thus
more number of atoms are linked together by
single covalent bonds with the result their
molecular weight become high ans hence
these are solids.
Q. Nitrous oxide supports combustion more
vigorously than air.why
Ans. Nitrous oxide decomposes by heat of
burning substances to produce gases having
one third oxygen by volume
2N2O2N2 + O2
while air has one fifth volume of oxygen.
Oxygen being a supporter of combustion,
keeps combustion on.
Q. Nitric acid acts only as an oxidising agent
while nitroous acid can act both as an
oxidising and reducing agent.
Ans. In HNO3, N is in its highest oxidation state
,hence, it can only be converted to lower
oxidation state,i.e., it can act as an oxidising
agent. In HNO2, N is in +3 oxidation state
hence, it can be raised to higher oxidation
state or lowered to lower oxidation state .thus
HNO2 can act as an oxidising as well as
reducing agent.
Q. The wooden shelf under reagent bottle
containing conc.H2SO4 blackens after
sometime.
Ans. Conc. H2SO4, being a strong dehydrating
agent ,when trickles down on the bench
removes water from wood (cellulosic material
) leaving behind black carbon.
Q. Chlorine displaces iodine from potassium
salts.
Ans. Due to more electronegativity,chlorine
takes up an electron from the I- ion forming Clion and iodine .
Q. Bleaching of flowers by Cl2 is permanent
while by SO2 it is temporary.
Ans. Cl2 bleaches by oxidation ,while SO2
bleaches by reduction. Hence product
bleached by SO2 is reoxidised by air to its
original form.
Q. Out of cobalt and zinc salts, which is attracted
in a magnetic field . Explain with reason.
Ans. Cobalt has d7 electrons in the outer orbit
I.e., 3 unpaired electrons,hence it will be
attracted in a magnetid field ,while zinc having
d10 electrons will not be attracted.
Q.Why is first ionisation energies of boron
and aluminium are lower than that of
beryllium
Ans. In B and Al,the first electron to be removed
lies in p-orbitals while in beryllium and
magnesium the valency electrons of an orbital
are strongly attracted by nucleus and hence,
difficult to remove tan the p-orbitals.
Q. Boric acid can be titrated against sodium
hydroxide using phenolphthalein indicator
only in presence of polyhydroxy compounds
like catechal. Explain .
Ans. Boric acid,B(OH)3 is a very weak lewis acid.
It forms a stable complex with polyhydroxy
compounds. Due to formation of this stable
complex, boric acid acts as a strong acid and
hence can be titrated against NaOH
Q. Why (SiH3)3N is a weaker base than (CH3)3N.
Ans. In (SiH3)3N , lone pair of electrons on
nitrogen is involved in pπ-dπ back bonding,
while in (CH3)3N pπ-dπ back bonding is not
possible because of absence of d-orbitals.
Hence, (CH3)3N is more basic than (SiH3)3N
Q. The experiment determined N-F bond length
in NF3 is greater than the sum of the single
bond covalent radii of N ans F. why
Ans. Nitrogen and flourine both are small and
have high electron density and hence they
repel the bonded pair of electrons leading to
larger bond length than expected.
Q. Why Nitrogen is a gas while other members
of this group are solids.
Ans. Due to smaller size of nitrogen atom,it can
undergo sideways overlap with the result two
π bonds are formed between two nitrogen
atoms (N≡N). The discrete nitrogen molecules
are held together by weak Van Der Waal’s
forces. On the other hand,other atoms are
bigger, so sideways overlap is not strong and
hence multiple bonds are not formed . Thus
more number of atoms are linked together by
single covalent bonds with the result their
molecular weight become high ans hence
these are solids.
Q. Nitrous oxide supports combustion more
vigorously than air.why
Ans. Nitrous oxide decomposes by heat of
burning substances to produce gases having
one third oxygen by volume
2N2O2N2 + O2
while air has one fifth volume of oxygen.
Oxygen being a supporter of combustion,
keeps combustion on.
Q. Nitric acid acts only as an oxidising agent
while nitroous acid can act both as an
oxidising and reducing agent.
Ans. In HNO3, N is in its highest oxidation state
,hence, it can only be converted to lower
oxidation state,i.e., it can act as an oxidising
agent. In HNO2, N is in +3 oxidation state
hence, it can be raised to higher oxidation
state or lowered to lower oxidation state .thus
HNO2 can act as an oxidising as well as
reducing agent.
Q. The wooden shelf under reagent bottle
containing conc.H2SO4 blackens after
sometime.
Ans. Conc. H2SO4, being a strong dehydrating
agent ,when trickles down on the bench
removes water from wood (cellulosic material
) leaving behind black carbon.
Q. Chlorine displaces iodine from potassium
salts.
Ans. Due to more electronegativity,chlorine
takes up an electron from the I- ion forming Clion and iodine .
Q. Bleaching of flowers by Cl2 is permanent
while by SO2 it is temporary.
Ans. Cl2 bleaches by oxidation ,while SO2
bleaches by reduction. Hence product
bleached by SO2 is reoxidised by air to its
original form.
Q. Out of cobalt and zinc salts, which is attracted
in a magnetic field . Explain with reason.
Ans. Cobalt has d7 electrons in the outer orbit
I.e., 3 unpaired electrons,hence it will be
attracted in a magnetid field ,while zinc having
d10 electrons will not be attracted.
Question:
Out of noble gases, only xenon is known to form chemical compounds.
Answer:
Except radon, which is radioactive, xenon has lowest ionisation enthalpy
among noble gases and hence it readily forms chemical compounds
particularly with O2 and F2
Question:
Fluorine is anomalous in many properties.
Answer:
The anomalous behaviour of fluorine is due to its small size, highest
electronegativity, low F-F bond dissociation enthalpy, and non availability
of d orbitals in valence shell.
Question:
SCl6 is not known but SF6 is known.
Answer:
Because of high electronegativity of fluorine, sulphur exhibits its maximum
oxidation state (+6) in SF6
Question:
N2 is less reactive at room temperature.
Answer:
Because of strong pπ–pπ overlap Nitrogen has triple bond between two
nitrogen atoms N≡N which has high bond dissociation energy. So it is less
reactive
Question:
NO2 is coloured but N2O4 is colourless.
Answer:
NO2 has unpaired electrons therefore it absorbs light from visible and
radiate brown colour whereas N2O4 does not have unpaired electrons so
it does not absorb light from visible region
Question:
Tendency to show –2 oxidation state diminishes from Sulphur to polonium in
group 16.
Answer:
The outer electronic configuration of group 16 elements is ns2 np4. These
elements therefore have the tendency to gain two electrons to complete octet.
Since electronegativity and I.E. decrease on going down the group, tendency
to show –2 oxidation state diminishes.
Question:
There is large difference between the melting point of Oxygen & Sulphur.
Answer:
The large difference between the melting and boiling points of oxygen and
sulphur may be explained on the basis of their atomicity; oxygen exists as
diatomic molecule (O2) whereas
sulphur exists as polyatomic molecule (S8)
Question:
All the bonds in the molecules of PCl5 are not equal
Answer:
PCl5 has a trigonal bipyramidal shape in the gas space. A trigonal
bipyramidal is an irregular structure in which some bond angles are 90
degree and others of 120 degree resulting in unequal P-Cl bond lengths
Question:
In solid state PCl5 exists as Ionic compound.
Answer:
Since solid phosphorous pentachloride exists as [PCl4]+ [PCl6]- and hence
exhibit some ionic character. [PCl4]+ is tetrahedral and the anion, [PCl6]–
octahedral.
Question:
There is a considerable increase in covalent radius from N to P. However, from
As to Bi only small increase in covalent radius is observed.
Answer:
This is due to the presence of completely filled d and/or f orbital in heavier
members
Question:
Nitrogen does not form pentahalides.
Answer:
Nitrogen with n = 2, has s and p orbitals only. It does not have d orbitals to
expand its covalence beyond four. That is why it does not form pentahalides.
Question:
H3PO2 and H3PO3 act as as good reducing agents while H3PO4 does not.
Answer:
In H3PO2, two H atoms are bonded directly to P atom & in H3PO3 one H atom is
bonded directly to P atom which imparts reducing character to the acid,
whereas in H3PO4 there is no H atom bonded directly to P atom
Question:
Dioxygen is a gas but Sulphur is a solid.
Answer:
Because O2 is Diatomic molecules, hence weak Vander Waal’s force of
attraction thus is a gas whereas S8 is octaatomic hence Stronger Vander
Waal’s force of attraction thus it is solid.
Question:
OF2 should be called oxygen fluoride and not fluorine oxide.
Answer:
OF2 should be called oxygen fluoride and not fluorine oxide. Since oxygen
is less electronegative than fluorine, OF2 should be called oxygen
diflouride.
Question:
Xe does not forms compounds such as Xe F3 and XeF5.
Answer:
By the promotion of one, two or three electrons from filled p-orbital to the
vacant d-orbital in the valence shell, 2,4 or 6 half filled orbitals are formed.
Thus Xe can combine only with even number of fluorine and not odd.
• Q:Boron forms no compound in unipositive state
while thallium is quite stable in unipositive state.
Explain.
• A:m+ ionic state of grp 13 elements exists due to
inert pair effect when ns2 electrons penetrate in
(n-1)d subshell to become inert . the inert pair
effect begins from n>4 and increases with
increase in value of n. The n n i.E., Outermost
shell for boron is 2 and thus it does not form B+
ion.
• Q:Although the ionization energy of boron
(8.30ev) is less than gold(9.22ev), yet former is
semi-metal while later is metal.
• A: Here metallic and non-metallic characters
are explained in terms of the lattice structure of
the solid. Boron has 6 or less atoms as nearest
neighbors in solid state while gold has 12 atoms.
In general, good metals have a large number of
neighbors atoms, while non-metals have relatively
less atoms.
• Q:Pure BBr3 and BI3 are colorless, but they
become colored on exposure to light, why?
• A: Due to liberation of free halogen by
photolysis
• Q: Why is aluminium used in in making
electrical cables despite its low conductivity as
compared to copper.
• A: it is cheaply available. However, Since it has
lesser conducting nature than copper, thicker
aluminium wires are used.
• Q: Why are cryolites used in the electrolytic
manufacturers of Al from alumina?
• A: Alumina has very high m.pt. and fuses at
about 2000 degree Celsius. Addition of cryolite
lowers the fusion temperature of mixture as well
as increases the conductance of fusion mixture.
• Q: Alum is often used to stop bleeding from
cuts. why?
• A: Al3+ ions liberate from alum are highly
effective in coagulating negatively charged
colloids like blood.
• Q: Aluminium is a very reactive metal, but is
still used for making pans. Explain.
• A: Due to formation of a protective oxide film
on the surface. It does not corrode easily.
• Q: Aluminium containers can be used to store
conc. HNO3. Explain.
• A: Al on contact with conc. HNO3 becomes
passive due to the coating of aluminium oxide on
its surface and thus it can store the acid.
• Q: Molten Albr3 is a poor conductor of current.
Explain.
• A: AlBr3 is covalent in nature and thus
during fusion does not produce Al3+ and Brions. Hence it is poor conductor of current.
• Q: It is difficult to obtain pure crystalline boron,
why?
• A: Its melting point is very high and liquid is
corrosive
WHY IS N2 LESS
REACTIVE AT ROOM
TEMPERATURE ?
IT IS BECAUSE OF THE STRONG
pл-pл OVERLAP RESULTING
INTO THE TRIPLE BOND, (N≡N),
CONSEQUENTLY HIGH BOND
ENTHALPY.
NOBLE GASES HAVE
VERY LOW BOILING
POINT. GIVE REASON ?
NOBLE GASES ARE MONOATOMIC. THEIR
ATOMS ARE HELD TOGETHER BY WEAK
DISPERSION FORCES AND HENCE CAN BE
LIQUIFIED AT VERY LOW TEMPERATURES.
SO, THEY HAVE LOW BOILING POINTS.
WHY DO THE TRANSITION
ELEMENTS EXHIBIT HIGHER
ENTHALPIES OF ATOMISATION
?
BECAUSE OF LARGE NUMBER OF
UNPAIRED ELECTRONS IN THEIR ATOMS
THEY HAVE STRONGER INTERATOMIC
INTERACTION AND HENCE STRONGER
BONDING BETWEEN ATOMS RESULTING
IN HIGHER ENTHALPIES OF
ATOMISATION.
PH3 HAS LOWER
BOILING POINT THAN
NH3. WHY ?
UNLIKE NH3, PH3 MOLECULES ARE
NOT ASSOCIATED THROUGH
HYDROGEN BONDING IN LIQUID
STATE. THAT IS WHY BOILING
POINT OF PH3 IS LOWER THAN
NH3.
WHY DOES V2O5 ACT
AS CATALYST ?
V2O5 ACT AS CATALIST BECAUSE IT
HAS A LARGE SURFACE AREA. IT CAN
FORM UNSTABLE INTERMEDIATES
WHICH READILY CHANGE INTO
PRODUCTS.
TETRAHEDRAL COMPLEXES
DO NOT SHOW
GEOMETRICAL ISOMERISM,
WHY ?
IT IS BECAUSE THE RELATIVE
POSITION OF THE LIGANDS
ATTACHED TO THE CENTRAL
METAL ATOM ARE THE SAME
WITH RESPECT TO EACH OTHER.
THE TRANS ISOMERS OF
COMPLEX CoCl2 (en)2 IS
OPTICALLY INACTIVE , WHY
?
IT IS BECAUSE THE TRANS
ISOMER HAS A PLANE OF
SYMMETRY AND CAN BE
DIVIDED IN TWO EQUAL
HALVES.
WHY CHELATE COMPLEXES
ARE MORE STABLE THAN
UNCHELATE COMPLEXES?
WHEN A CHELATING LIGAND
ATTACHES TO THE CENTRAL METAL
ATOM THE PROCESS IS
ACCOMPANIED BY THE INCREASE IN
ENTROPY RESULTING IN THE
FORMATION OF THE STABLE
COMPLEX.
EXPLAIN WHY K3[Fe(CN)6]
IS MORE STABLE THAN
K4[Fe(CN)6]?
IT IS BECAUSE THE STABILITY OF
COMPLEX DEPENDS UPON THE
CHARGE DENSITY ON CENTRAL ION.
MORE IS THE CHARGE DENSITY
GREATER IS THE STABILITY.
THE SPECIES [CuCl4]³ˉ
EXISTS WHILE [CuI4]³ˉ DOES
NOT WHY?
THIS IS BECAUSE IODINE ATOM IS
MUCH LARGER IN SIZE AND
COPPER ATOM IS NOT ABLE TO
ACCOMMODATE FOUR SUCH BIG
ATOMS AROUND IT.
WHY DOES NH3 FORM
HYDROGEN BOND BUT
PH3 DOES NOT?
• Because of high electro-negativity
and small size of nitrogen, ammonia
forms hydrogen bonds. On the other
hand phosphorous has low electronegativity and large size and hence
cannot form hydrogen bonds.
WHY IS NH3 BASIC WHILE BiH3
FEEBLY BASIC?
• Both N and Bi have a lone pair of electron in NH3
and BiH3 respectively. They can donate the
electron pair and therefore behave as lewis base.
• In NH3, nitrogen has small size and the lone pair is
concentrated on a small region and the electron
density on it is maximum.
• Consequently it has greater electron releasing
tendency. But the size of Bi is large and the
electron density of the lone pair is less. As a
result it has lesser tendency to donate electron
pair. Hence, NH3 is basic while BiH3 is feebly basic.
WHY DOES NITROGEN SHOW
CATENATON PROPERTIES LESS
THAN PHOSPHOROUS?
• The single N-N bond is weaker than
the single P-P bond because of high
inter-electronic repulsion of the nonbonding electrons, owing to the small
bond length. As a result the catenation
tendency is weaker in nitrogen.
WHY DOES R3P=O EXIST BUT
R3N=O DOES NOT?
• R3N=O does not exist because N
cannot have covalency more than 4.
Moreover, R3P=O exists because P can
extend its covalency more than 4 as
well as can form dπ-pπ bond whereas
N cannot.
WHY ARE HALOGENS STRONG
OXIDISING AGENTS?
• Halogens have a strong tendency
to accept electrons and
therefore act as strong oxidizing
agents.
WHY DOES FLUORINE FORM ONLY
1 OXOACID, HOF?
• Due to small size and high electronegativity fluorine cannot act as
central atom, it cannot form a
higher oxo-acid.
WHY IS H2O A LIQUID WHILE
H2S A GAS?
• Because of small size and electronegativity of oxygen, molecules of water
are highly associated through hydrogen
bonding resulting in its liquid state.
WHY DOES O3 ACT AS A POWEFUL
OXIDIZING AGENT?
• Ozone is thermodynamically unstable with
respect to oxygen since its decomposition into
oxygen results in the liberation of heat and an
increase in large negative Gibbs energy
change for its conversion into oxygen.
• Therefore it easily liberates atoms of nascent
oxygen and acts as a powerful oxidizing agent.
H2S IS LESS ACIDIC THAN H2Te.
WHY?
• Due to the decrease in bond dissociation
enthalpy down the group, acidic character
increases down the group.
• Hence, H2Te is more acidic than H2S.
FLUORINE EXHIBITS ONLY -1
OXIDATION STATE WHEREAS OTHER
HALOGENS EXHIBIT +1, +3, +5 AND +7
OXIDATION STATES ALSO. EXPLAIN.
• Fluorine is the most electro-negative element
and cannot exhibit any positive oxidation
state. Other halogens have d orbitals and
therefore, can expand their octets and show
+1 , +3, +5 and +7 oxidation states also.
Q .INCREASING ORDER OF
ELECTRONEGATIVITY 0F
+
O <O<O
THE POSITIVE CHARGE ON ATOM
INCREASES ITS ELECTONEGATIVITY
WHILE NEGATIVE CHARGE
DECREASES ITS
ELECTRONEGATIVITY
Q. INCREASING BOND LENGTH OF
N2 < O2 < F2 < Cl2
NITROGEN CONTAINS TRIPLE BOND ,
OXYGEN CONTAINS DOUBLE BOND
AND FLUORINE AND CHLORINE
CONTAIN A SINGLE BOND EACH .
CHLORINE INVOLVES BONDING OF 3p
ORBITALS WHILE FLUORINE INVOLVES
2p ORBITALS.
INCREASING BASICITY OF
I-< Br- < Cl- < F-
STRONGER THE ACID , WEAKER ITS
CONJUGATE BASE
INCREASING ORDER OF BOILING
POINT OF PH3 < AsH3 < NH3 < SbH3
BOILING POINT INCREASES WITH
INCREAES IN MOLECULAR MASS
WITH EXCEPTION IN NH3 BECAUSE
OF HRDROGEN BONDING
WHY IN MOIST AIR COPPER
CORRODES TO PRODUCE A GREEN
LAYER ON THE SURFACE
IN THE PRESENCE OF MOIST AIR A
THIN FILM OF GREEN BASIC COPPER
CARBONATE IS FORMED ON ITS
SURFACE AND HENCE COPPER
CORRODES AND FORM
CuCO3.Cu(OH)2 WHICH IS GREEN
WHY DOES AgNO3 PRODUCE A
BLACK STAIN ON THE SKIN
IN THE PRENSENCE OF ORGANIC
MATTER (SKIN) AND LIGHT , AgNO3
DECOMPOSES TO PRODUCES A
BLACK STAIN OF METALLIC SILVER
INCREASING IONIC CHARACTER OF
LiBr < NaBr < KBr < RbBr < CsBr
THE LARGER THE DIFFERENCE
BETWEEN THE
ELECTRONEGATIVITY ,
GREATER THE IONIC CHARACTER
INCREASING ORDER OF
HYDRATION ENERGY
+2
+2
+2
+2
+2
Ba < Sr < Ca < Mg < Be
THE SMALLER THE SIZE , MORE THE
HYDRATION ENERGY
INCREASING BOND ANGLE
NF3 < NH3
THERE IS LESSER REPULSION IN
BONDING PAIRS IN NH3
INCREASING BOILING POINT IN
HCl < HBr < HI < HF
THE LARGER THE DIFFERENCE
BETWEEN THE
ELECTRONEGATIVITY ,
GREATER THE IONIC CHARACTER
AND HENCE GREATER THE BOILING
POINT. ANOMALOUS BEHAVIOUR
OF HF IS DUE TO HYDROGEN
BONDING
1. WHAT IS THE COVALENCE OF NITROGEN IN
N2O5 ?
AnsCOVALENCY DEPENDS UPON THE NUMBER OF
SHARED PAIR OF ELECTRONS. NOW IN N2O5,
NITROGEN ATOM HAS FOUR SHARED PAIR OF
ELECTRONS.
THERFORE THE VALENCY OF N IN N2O5 IS 4.
2. WHY ARE PENTAHALIDES MORE COVALENT
THAN TRIHALIDES?
AnsAS WE KNOW, HIGHER THE POSITIVE
OXIDATION STATE OF CENTRAL ATOM, MORE
WILL BE ITS POLARISING POWER WHICH, IN
TURN, INCREASES THE COVALENT CHARATER OF
BOND FORMED BETWEEN THE CENTRAL ATOM
AND THE OTHER ATOM.
+
3.BOND ANGLE IN PH4 IS HIGHER IN PH3.
WHY?
AnsBOTH ARE sp3 HYBRIDISED. IN PH4+ ALL THE FOUR
ORBITALS ARE BONDED WHEREAS IN PH3 THERE
IS A LONE PAIR OF ELECTRONS ON P, WHICH IS
RESPONSIBLE FOR LONE PAIR-BOND PAIR
REPULSION IN PH3 REDUCING THE BOND ANGLE
TO LESS THAN 109°28’.
4.WHAT HAPPENS WHEN PCl5 IS HEATED?
AnsPCl5 HAS THREE EQUATORIAL (202 pm) AND TWO
AXIAL (240 pm) BONDS. SINCE AXIAL BONDS ARE
WEAKER THAN EQUATORIAL BONDS, THEREFORE
WHEN PCl5 IS HEATED, THE LESS STABLE AXIAL
BONDS BREAK TO FORM PCl3.
5.WHY DOES OZONE ACT AS A POWERFUL
OXIDISING AGENT?
AnsDUE TO THE EASE WITH WHICH IT LIBERATES
ATOMS OF NASCENT OXYGEN, IT ACTS AS A
POWERFUL OXIDISING AGENT.
O3 → O2 + O (NASCENT OXYGEN)
6.WHY DOES THE REACTIVITY OF NITROGEN
DIFFER FROM PHOSPHORUS?
AnsNITROGEN EXISTS AS DIATOMIC MOLECULE. DUE
TO THE PRESENCE OF A TRIPLE BOND BETWEEN
THE TWO N ATOMS, THE BOND DISSOCIATION
ENERGY IS LARGE (941.4 KJ/ mol). THUS
NITROGEN IS INERT AND UNREACTIVE IN ITS
ELEMENTAL STATE. PHOSPHORUS EXISTS AS A
TETRATOMIC MOLECULE(P4). SINCE THE P—P
SINGLE BOND IS MUCH WEAKER (213
KJ/mol),THERFORE, PHOSPHORUS IS MUCH
MORE REACTIVE THAN NITROGEN.
7.WHY DOES NH3 FORM HYDROGEN BONDS BUT
PH3 DOES NOT?
AnsTHE ELECTRONEGATIVITY OF N(3.0) IS MUCH
HIGHER THAN THAT OF H(2.1). HENCE,N—H
BOND IS QUITE POLAR AND HENCE NH3
UNDERGOES INTERMOLECULAR H—BONDING.
ON THE OTHER HAND, BOTH P AND H HAVE AN
ELECTRONEGATIVITY OF 2.1 THEREFORE, P—H
BOND IS NOT POLAR AND HENCE PH3 DOES NOT
UNDERGO H—BONDING.
8.WHY DOES R3P=0 EXIST BUT R3N=O DOES NOT
(R IS ALKYL GROUP)?
AnsN DUE TO THE ABSENCE OF d-orbitals, CANNOT
FORM p -d MULTIPLE BONDS. THUS, N CAN’T
EXPAND ITS COVALENCY BEYOND FOUR BUT IN
R3N=O, N HAS A COVALENCY OF 5. SO THE
COMPOUND R3N=O DOES NOT EXIST. P DUE TO
THE PRESENCE OF d-orbitals FORMS p -d
MULTIPLE BONDS AND HENCE CAN EXPAND ITS
COVALENCY BEYOND 4. THEREFORE, P FORMS
R3P=O IN WHICH THE COVALENCY OF P IS 5.
9.WHY DOES NITROGEN SHOW CATENATION
PROPERTIES LESS THAN PHOSPHORUS?
AnsTHE PROPERTY OF CATENATION DEPENDS UPON
THE BOND STRENGTH OF THE ELEMENT. SINCE
THE N—N (159 KJ/mol) BOND STRENGTH IS
MUCH WEAKER THAN P—P (213 KJ/mol) BOND
STRENGTH, HENCE, NITROGEN SHOWS LESS
CATENATION PROPERTIES THAN PHOSPHORUS.
10.EXPLAIN WHY INSPITE OF NEARLY THE SAME
ELECTRONEGATIVITY, OXYGEN FORMS
HYDROGEN BONDING WHILE CHLORINE DOES
NOT?
AnsALTHOUGH O AND Cl HAVE THE SAME
ELECTRONEGATIVITY, YET THEIR ATOMIC SIZE ARE
MUCH DIFFERENT i.e. O=66pm AND Cl=99pm.
THUS, ELECTRON DENSITY PER UNIT VOLUME ON
OXYGEN ATOM IS MUCH HIGHER THAN THAT OF
ON CHLORINE ATOM. HENCE, OXYGEN FORMS
HYDROGEN BONDS WHILE CHLORINE DOES NOT,
EVEN THOUGH BOTH HAVE APPROX THE SAME
ELECTRONEGATIVITY.
Why is nitrous acid oxidant as well as reductant?
• The oxidation state of N in nitrous acid is +3
which lies in between its lowest oxidation
state of -3 and highest oxidation state of +5.
Since its oxidation state +3 to any lower value
to -3,therefore it acts as oxidizing agent .
• Further since the oxidation state of N in HNO2
can be increased from +3 to +5 , therefore it
acts as a reducing agent.
All bonds in PCl5 are equivalent. Justify.
• PCl5 has trigonal bipyrimidal structure. It has
3 equivalent equatorial and two equivalent
axial P-Cl bonds. However due to greater bond
pair – bond pair repulsions , the two axial P-Cl
bonds are longer and hence different from the
three equatorial bonds
• Why is OF6 not known?
• Due to absense of d-orbitals in its valence
shell, oxygen shows a maximum oxidation
state of +2
• Why does Mn(||) shows maximum
paramagnetic character amongst the bivalent
ions of the first transition state?
• Mn2+ has maximum no. of unpaired electrons
.i.e. 3d5
• Give reasons indicate which of the following
would be colored?
• Cu+, VO2+, Sc3+, Ni2+
• (at. No. Cu=29
V=23
Sc=21
Ni =28 )
• Ni2+ due to incompletely filled d-orbitals.
• Zinc, Cadmium and Mercury are generally not
considered as transition metals. Why?
• These metals in their most common oxidation
state of +2 have completely filled d orbitals.
• Why KMnO4 is used in cleaning surgical
instruments in hospitals?
• This is because KMnO4 has a germicidal effect.
• How would you account for the increasing
oxidising power in the series
VO2+ < Cr2O7 2- <MnO4- ?
• This is due to the increasing stability of the
lower species to which they are reduced.
• Why Cr, Mn and Fe has nearly same atomic
radii?
• As effective nuclear charge is nearly same in
all 3 cases.
• Why in permanganate ion, there is a
covalency between Mn and Oxygen?
• In MnO4- ion Mn is in highest oxidation state.
(+7). In higher oxidation state transition
metals forms covalent bonds.
Q1 On what ground you can say that scandium (z=21) is a transition element
but zinc (z=30) is not?
•
Ans : On the basis of incompletely filled 3d orbitals in case of scandium atom in its
1
ground state (3d) , it is regarded as a transition element. On the other hand, zinc
10
atom has completely filled d orbitals (3d ) in its ground state as well as in its
oxidised sate , hence it is not regarded as a transition element.
Q2 Silver atom has completely filled d orbitals (4d ) in its 10ground state .
How can you say that it is a transition element ?
•
Ans: Silver (z=47) can exhibit +2 oxidation state wherein will have incompletely
filled d-orbitals (4d) , hence a transition element.
Q3 In the series Sc (z=21) to Zn (z=30), the enthalpy of atomisation of zinc is
-1
the lowest , i.e., 126 kJ mol . Why?
•
Ans : In the formation of metallic bonds , no electrons from 3d-orbitals are
involved in case of zinc, while in all other metals of 3d series , electrons from the
d-orbitals are always involved in the formation of metallic bonds.
Q4
Which of the 3d series elements of the transition metals exhibits the
largest number of oxidation states and why?
•
Manganese (z=25) , as its atom has the maximum number of unpaired electrons.
2+
3+
4
Q5 Why is Cr reducing and Mn oxidising when both have d configuration ?
•
Cr 2+ is reducing as its configuration changes from d to d 4 , the3 latter having a
2+
half-filled t level . On
to Mn results
in
2g the other hand , the change from Mn
3+
5
the half-filled (d ) configuration which has extra stability.
Q6 How would you for the irregular variation of ionisastion enthalpies (first
and second) in the first series of the of the transition elements ?
•
Ans : Irregular variation of ionisation enthalpies is mainly attributed to varying
10
0
5
degree of stability of different 3d-configurations ( e.g. d , d , d ,are exceptionally
stable)
Q7 Why is the oxidation state of a metal exhibited in its oxide or fluoride ?
•
Ans: Because of small size and high electronegativity oxygen or fluorine can
oxisidise the metal to its highest oxidation state.
2+agent Cr or Fe
Q8 Which is a strong oxidising
and why ?
2+
•
•
•
•
Ans : Cr is stronger reducing than Fe .
4
3
2+
3+
Reason : d
d occurs in case of Cr to Cr
6
5
3+
But d
d occurs in case of Fe to 2+
Fe
In a medium (like water) d is more stable as compared to d
+
Q9 Explain why Cu ion is not stable in aqueous solutions ?
•
•
•
Ans : Cu +in aqueous solution undergoes disproportionation, i.e. ,
2+
+
2Cu (aq)
Cu (aq) + Cu(s)
The E value for this is favourable.
Q10 Actinoid contraction is greater from element to element than
lanthanoid contraction. Why?
•
The 5f electrons are effectively shielded from nuclear charge . In other words the
5f electrons themselves provide poor shielding from element to element to
element in the series.
Why are pentahalides more covalent than trihalides?
Higher the positive oxidation state of the
central atom, more will be its polarising
power which, in turn , increases the
covalent character of bond formed between
the central atom and the other atom. Thus
the pentahalides, which have a more positive central atom than
trihalides are
more covalent than trihalides.
Though nitrogen exhibits +5 oxidation state, it does not
form a pentahalide. Give reason.
Nitrogen with principle quantum no. n=2, has s and p
orbitals only. It does not have d orbital to expand its
covalence beyond four. That is why it dos not form
pentahalides.
PH3 has lower boiling point than NH3. Why?
Unlike NH3, PH3 molecules are not associated through
hydrogen bonding in liquid state. That is why the boiling
point of PH3 is lower than that of NH3.
Why is BiH3 the strongest reducing agent amongst all the
hydrides of Group 15 elements?
The stability of group 15 hydrides decreases from NH3 to
BiH3. Consequently the reducing character increases from
NH3 to BiH3.These makes BiH3 the strongest reducing
agent of all the hydrides of group 15 elements.
Why is N2 less reactive at room temperature?
N2 is inert at room temperature because of the high bond
enthalpy N N triple bond but its reactivity increases rapidly
with temperature.
Why does NH3 act as a Lewis base?
Nitrogen atom in NH3 has one lone pair of electrons which
is available for donation. Therefore, it acts as a Lewis base.
Lone pair
..
N
H
H
H
PH3 is basic in nature. Give reason.
The lone pair on phosphorus of PH3 makes it a Lewis base.
Further the following reaction proves that PH3 is a base
because it reacts with an acid.
PH3 + HI
PH4I
+
Bond angle in PH4 is higher than that in PH3. Why?
3hybridised.
+
Both are sp
In PH4 all the four orbitals are
bonded but in PH3 only three orbitals are bonded and the
fourth one carries a lone pair which brings the bonded
orbitals closer due to bond pair-lone pair repulsion, this
3
results in a smaller angle than expected for sp
hybridisation.
Why does PCl3 fume in moisture?
PCl3 reacts readily with moisture to form orthophosphorous
acid and gives out HCL fumes.
PCl3 + 3H2O
H3PO3 + 3HCl
Why is H2O a liquid and H2S a gas?
Due to small size and high electronegativity of oxygen H2O
molecules are highly associated with inter-molecular
hydrogen-bonding which results in it’s liquid state. There is
no such bonding in H2S and so it’s a gas.
• Q. Nitrogen shows anomalous
behaviour ?
• Ans : Nitrogen differs from the
rest of the members of this
group due to its smaller size,
high electro negativity, high
ionization enthalpy and nonavailability of d orbitals.
• Q. Pentahalides of group 15 are
more covalent than trihalides ?
• Ans: Higher the positive
oxidation state of central atom,
more will be its polarizing
power which, in turn, increases
the covalent character of bond
formed between the central
atom and the other atom.
• Q. Bond angle in PH4+ is higher
than that in PH3 ?
• Ans: Both are SP3 hybridised.
In PH4+ all the four orbitals are
bonded whereas in PH3 there is
a lone pair of electrons on P,
which is responsible for lone
pair-bond pair repulsion in
PH3 reducing the bond angle to
less than 109° 28′.
• Q.NO2 dimerises to form N2O4 ?
• Ans:NO2 contains odd number
of valence electrons. It behaves
as a typical odd molecule. On
dimerisation, it is converted to
stable N2O4 molecule with even
number of electrons.
• Q. Nitrogen is fairly inert gas ?
• Ans: Nitrogen exists as triply
bonded diatomic non polar
molecule. Due to short
internuclear distance between
two nitrogen atoms the bond
strength is very high. It is,
therefore, very difficult to break
the bond.
• Q. Nitrogen exists as diatomic
molecule & phosphorus as P4 ?
• Ans: Because N2 is Diatomic
molecules, hence weak Vander
Waal’s force of attraction thus
is a gas whereas
P4 is tetraatomic hence
Stronger Vander Waal’s force of
attraction thus it is solid.
• Q. NO2 is coloured but N2O4 is
colourless ?
• Ans. NO2 has unpaired
electrons therefore it absorbs
light from visible and radiate
brown colour whereas
N2O4 doesnot have unpaired
electrons so it does notabsorb
light from visible region.
• Q. Tendency to show –2
oxidation state diminishes from
sulphur to polonium in group 16
?
• Ans: The outer electronic
configuration of group 16
elements is ns2 np4. These
elements therefore have the
tendency to gain two electrons
to complete octet. Since
elctronegativity and I.E.
decrease on going down the
group, tendency to show –2
oxidation state diminishes.
• Q. Oxygen generally exhibit
oxidation state of –2 only
whereas other members of the
family exhibit +2, +4, +6
oxidation states also ?
• Ans: Oxygen is a
electronegative element thus
exhibit oxidation state of –2.
.Other members of the family
have d orbitals and therefore,
can expand their octets and
show + 2, + 4, + 6 oxidation
states also.
• Q. Sulphur vapours exhibits
paramagnetism ?
• Ans :In vapour state sulphur
partly exists as S2 molecule
which has two unpaired
electrons in the antibonding π*
orbitals like O2 and, hence,
exhibits paramagnetism.
Ques1.
The solubility of calcium acetate
decreases while that of the lead nitrate
increases with increase in
temperature?
Ans1.
Because dissolution of calcium acetate
is a exothermic process while that of
lead nitrate is a endothermic process
Ques2.
H3PO2 and H3PO3 act as as good
reducing agents while
H3PO4 does not?
Ans2:
In H3PO2, two H atoms are bonded
directly to P atom & in H3PO3 one H
atom is bonded directly to P atom
which imparts reducing character to
the acid, whereas in H3PO4 there is no
H atom bonded directly to P atom
Ques3.
NO2 is coloured but N2O4 is colourless?
Ans3.
NO2 has unpaired electrons therefore it
absorbs light from visible and radiate
brown colour whereas N2O4 doesnot
have unpaired electrons so
it does notabsorb light from visible
region.
Ques4.
Sulphur vapours exhibits
paramagnetism
Ans4.
In vapour state sulphur partly exists as
S2 molecule which has two unpaired
electrons in the antibonding π* orbitals
like O2 and, hence, exhibits
paramagnetism.
Ques5.
NOis paramagnetic in the gaseous
state but dimagnetic in liquid and
solid state ?
Ans5.
NO = 5 + 6 = 11 e–, it has odd pair of e–
and hence paramagnetic in gaseous
state, but in liquid and solid state, it
exists as dimer
Ques6.
Why is Cr2+ reducing and Mn3+
oxidising when both have d4
configuration ?
Ans6.
Cr2+ is reducing as its configuration changes from d4
to d3, the d3 has half-filled t2g level.
n the other hand, the change from Mn2+ to Mn3+
results in the half filled (dS)
configuration which has extra stability.
Cr2+ = 3 d4 4 s0
Mn3+ = 3 d4 4 s0
Cr3+ = 3 d3 4 s0
Cr3+ = 3 d3 4 s0
have half-filled t2g level.
Mn2+ = 3 d5 4 s0
half-filled extra stable.
Ques7.
Although Cu+ has configuration 3 d10 4 s0
(stable) and Cu2+ has
configuration 3 d9 (unstable configuration) still
Cu2+ compounds are more
stable than Cu+?
Ans7.
It is due to much more (–) D Hydration
H– of Cu2+ (aq) than Cu+, which is
more than compensates for the II
ionization enthalpy of Cu.
Ques8.
When mercuric iodide is added to an
aqueous solution of KI, the freezing
point is raised?
Ans8.
Due to the fomation of complex K2 (Hg
I4), number of particles in the solution
decreases and hence the
freezing point is raised
Ques9.
redox couple has less
positive electrode potential than
couple?
Ans9.
In Mn2+ d5 configuration leads to
extrastability of half filled configuration, so
Mn3+ / Mn2+ (d4) tends to get
converted to stable d5, configuration of
Mn2+, by accepting an electron so
Mn3+/Mn2+ redox couple has more
positive potential than
couple.
Ques10.
ClF3 exists but FCl3 does not. Why?
Ans10.
Due to unavailability of d-orbitals in
fluorine atom it cannot expand its valence
shell. Therefore it is unable to form
FCl3 whereas Chlorine has vacant dorbitals. Hence it can promote one of the 3
p electrons to the 3-d sub shell and shows
+3 oxidation state and forms ClF3
• Give reason: The maximum no. of
covalent bonds formed by nitrogen is 4.
• Ans. Nitrogen has 3 unpaired electrons &
1 lone pair of electrons , therefore it can
form 3 covalent bonds & 1 coordinate
bond therefore its covalency is 4 e.g. in
NH4 .
• Explain why oxygen is a gas while sulphur is a solid.
• Oxygen is diatomic molecule & has double
bond between 2 oxygen atoms, therefore
there is weak van der Waals force of attraction
between molecules ,therefore it is a gas.
Sulphur is octaatomic molecule (S8 ) & has
more intermolecular force of attraction , that
is why it exists as solid.
• Explain why fluorine always exhibit an
oxidation state of -1 only.
• Ans. Fluorine shows only -1 oxidation state
because it is most electronegative element
7 does not have d- orbitals.
• Assign reasons for the following observations.
• (1) Hydrogen iodide is a stronger acid than
hydrogen fluoride in aq. solution.
• (2) The basic character among the hydrides of
group 15 elements decreases with increasing
atomic no.’s.
• Ans. (1) HI has lower bond dissociation energy
than HF.
• (2) It is due to increase in atomic size of group
15 elements , lone pair of electrons is less
available for bond formation.
The d and f – Block Elements
• Why do transition metals have high enthalpy
of hydration?
• Ans. Transition metal ions are smaller in size &
have higher charge, therefore , have higher
enthalpy of hydration.
• Explain why do transition elements show
variable oxidation states. Write all the possible
oxidation states of an element having atomic
no. 25.
• Ans. They show variable oxidation state
because electrons from ‘s’ as well as d –
orbitals take part in bond formation. Mn (25)
can show +2, +3, +4, +6, & +7 oxidation states.
• Give reasons for each of the following :
• (1)Size of trivalent lanthanoid cations
decreases with increase in the atomic no.
• (2) chemistry of all the lanthanoids is quite
similar.
• Ans. (1) It is due to poor shielding effect of f –
electrons , effective nuclear charge increases ,
ionic size decreases.
• (2) It is due to similar ionic size which is due to
lanthanoid contraction , they resemble in their
properties.
• K2PtCl6 is well known compound
whereas corresponding compound of Ni
is not known, Why?
• Ans. K2PtCl6 is known because energy
required to remove 4 electrons in Pt is
less as compared to Ni, therefore ,
corresponding compound of Ni is not
known.
• Explain the following :
• (1) transition elements tend to be unreactive
with increasing atomic no. in the series.
• (2) d – Block elements exhibit more oxidation
states than f – block elements.
• Ans. (1) Transition metals form layer of oxides
on their surface due to which they become
unreactive.Secondly, reactivity decreases with
increase in atomic no. due to decrease in size
& increase in ionisation energy.
• (2)In d- block elements, e- of s- orbital & dorbital both take part in bond formation. In fblock elements due to poor shielding effect of
f-electrons effective nuclear charge increases,
therefore lesser no. of oxidation states are
shown.
• Give reasons :
• (1) Cr2+ is a strong reducing agent whereas
Mn2+ is not.
• (2) The transition metal ions such as Cu+, Ag+,
& Sc3+ are colourless.
• Ans. (1) Cr2+ is less stable than r3+ , therefore
it is good reducing agent whereas Mn2+ is
stable due to half filled d –orbitals therefore it
is not reducing agent.
• (2) Cu+, Ag+, & Sc3+ are colourless because
they do not have unpaired electrons.
• 1) Bond angle PH4 + is higher than in PH3.
Why?
• Ans: Both PH4 + and PH3 involve sp3
hybridisation of P atom. In PH4 + all the four
orbitals are bonded, whereas in PH3 there is a
lone pair of electron on P. In PH4 +, the HPH
bond angle is tetrahedral angle of 109.5°. But
in PH3, lone pair-bond pair repulsion is more
than bond pair-bond pair repulsion so that
bond angles become less than normal
tetrahedral angle of 109.5°. The bond angle in
PH3 has been found to be about 93.6°.
• 2) Out of noble gases, only Xenon is known to
form chemical compounds. Explain.
• Ans: Except Radon, which is radioactive,
Xenon has lowest ionisation enthalpy among
noble gases and hence it readily forms
chemical compounds particularly, with O2 and
F2.
• 3) HCl when reacts with finely divided iron
forms ferrous chloride and not ferric chloride.
Why?
• Ans: HCl reacts with finely divided iron and
produces H2 gas.
•
2Fe + 6HCl → 2FeCl3 + 3H2
• Liberation of hydrogen prevents the formation
of ferric chloride.
• 4) Halogen have maximum negative gain
enthalpy in the respective periods of the
periodic table. Why?
• Ans: . The halogens have the smallest size in their
respective periods and therefore, high effective
nuclear charge. Moreover, they have only one
electron less than the stable noble gas
configuration (ns²np6).
• Therefore, they have strong tendency to accept
one electron to acquire noble gas electronic
configurations and hence have maximum
negative electron gain enthalpy in their
respective periods.
• 5) Explain why inspite of nearly same
electronegativity, Oxygen forms hydrogen
bonding while chlorin does not.
• Ans: Oxygen has smaller size than chlorine.
The smaller size of oxygen favours hydrogen
bonding. In other words, though
electronegativity of Cl is same as that of O, it
does not form hydrogen bonding because of
its larger size.
• 6) Why is europium (II) more stable than
cerium (II)?
• Ans: Europium (II) has electronic
configuration [Xe] 4f7 5d0 while cerium (II)
has electronic configuration [Xe] 4f1 5d1. In
Eu2+, 4f subshell is half filled and 5d-subshell
is empty. Since half filled and completely
filled electronic configurations are more
stable, Eu2+ ion is more stable than Ce2+ in
which neither 4f subshell nor 5d subshell is
half filled or completely filled.
• 7) Why are Mn2+ compounds more stable
than Fe2+ towards oxidation to their +3 state?
• Ans: Mn2+ compounds are more stable than
Mn3+ due to stable 3d5 (half filled)
configuration. On the other hand, Fa3+ is
more stable than Fe2+ because of half filled
configuration of Fe3+. Therefore, Mn2+
compounds are more stable than Fe2+
towards oxidation to their +3 state.
• 8) K2PtCl6 is known but Ni compound is not
known. State reason for it?
• Ans: This is because Pt4+ is more stable than
Ni4+ as a sum of four ionisation energies of Pt
is less than that of Ni.
• 9) Why do transition elements exhibit higher
enthalpies of atomization?
• Ans: The high enthalpies of atomization are
due to large number of unpaired electrons in
their atoms. Therefore, they have stronger
interatomic interactions and hence, stronger
bonding between atoms. Thus, they have
high enthalpies of atomization.
• 10) Why do Zr and Hf exhibit similar
properties?
• Ans: Due to lanthanoid contraction, Hf and Zr
have almost similar size and therefore, their
properties are similar.