Download CLASSIFICATION OF ELEMENTS AND PERIODICITY IN

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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES
C1(a) Dabereiner : grouped elements in Traids. He pointed out that there were sets of three elements (Triads)
which showed similar chemical properties. He also noted that the atomic weight of the central element of
the Triad was approximately the mean of the atomic weights of the other two members. The properties of
the middle element were in between of those end members. e.g. Li, Na, K; Ca, Sr, Ba or Cl, Br, I etc.
John Newland : He developed a law of octaves.
(b)
He observed that similar elements are repeated at 8th place like the 8th note of music. The elements are
arranged in the order of increase of atomic weights. Similar element means that physical & chemical
properties of element will be same. e.g. Li has the same property as of Na.
Lother Meyer arrangement : He studied the physical properties such as atomic volume, melting point
and boiling points of various elements. On this basis he plotted a graph between Atomic volume (cm3) Vs.
Atomic weights.
(c)
Observation :
(i)
The most electropositive alkali metals (Li, Na, K, Rb and Cs) occupy the peaks on the curve.
(ii)
The less strongly electropositive elements i.e. alkaline earth metals (Be, Mg, Ca, Sr and Ba) occupy the
descending positions on the curve.
(iii)
The most electronegative elements i.e., halogens (F, Cl, Br and I) occupy the ascending positions on the
curve.
On the basis of these observations, Lother Meyer proposed that the physical properties of the elements are
a periodic functions of their atomic weights.
(d)
Mendeleev’s Periodic law and table : Mendeleev arranged all the elements in the order of increase of
atomic weight.
(i)
A table formed with the help of classification of elements is called periodic table.
(ii)
The method of arranging similar elements in one group and seprating them from dissimilar elements is
called classification of elements.
(iii)
He prepared the periodic table on the basis of periodic law i.e.,
“The properties of elements are periodic function of their atomic weight”.
(iv)
Mendeleev’s periodic table conists of seven horizontal rows known as periods and nine vertical column
known as groups.
(v)
Periods : Out of seven periods, first three periods are short periods while the fourth, fifth and sixth periods
are called long periods.
(vi)
There are nine groups in all including 8th group of transition elements and zero group of inert gases.
(vii)
All the group from I to VII (except zero and VIII) were divided into sub-groups.
(viii) The group number represents the valency.
(ix)
The elements of same sub group resemble to each other more closely and differ from other sub groups.
(x)
He predicted the properties of the missing elements from the known properties of the other elements in the
same group. e.g. gallium and germanium were not discovered at that time. He named these elements as
Eka-Aluminium and Eka-silicon.
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Defects in Mendeleev’s Periodic Table :
1.
Position of Hydrogen : Hydrogen resembles both the alkali metal and halogen. Hence its position in
periodic table is undecided.
2.
Position of Isotopes : According to Mendeleev’s periodic table, isotopes should occupy different
positions in the periodic table, but this is not so.
3.
Position of VIII group elements : Nine elements in the VIII group do not fit into the system.
4.
Positions of Lanthanides and Actinides : Their position was not justified according to the periodic law
and cannot be arranged in the order of their increasing atomic weight.
5.
Dissimilar elements placed in the same group : Alkali metals (Li, Na, K etc.) are placed with coinage
metals (Cu, Ag, Au).
6.
Similar elements placed Apart : Chemically similar elements like Cu and Hg, Ag and Ti, Au and Pt have
been placed in different groups.
7.
Anomalous pair of elements : Some elements of higher atomic weight have been placed before the
elements of lower atomic weight. e.g. Argon (At. wt. = 39.9) has been placed before potassium
(At. wt. = 39.1); cobalt (At. wt. = 58.94) is placed before nickel (At. wt. = 58.69); Tellurium (127.5) has
been placed before iodine (126.9).
(e)
Modern Periodic law and the present form of periodic table : As a result of modern researches it is
estabished that atomic number is a fundamental property not the atomic weight.
Thus this led Moseley to change the basis of calssification of elements from atomic weight to atomic
number. The modern periodic law given by Moseley is :
“The properties of elements are periodic functions of their atomic numbers, i.e., if elements are arranged in
the order of their atomic numbers. Similar elements are repeated after regular intervals”. He also gave the
following formulae
i.e.,
 = a(z – b)
;

=
frequency of X-rays
a, b
=
constant for all lines in a given series of X-rays
z
=
atomic number
Now we will discuss why elements with similar properties reoccurs after certain regular intervals.
Cause of Periodicity : The cause of periodicity in properties is the repeatition of similar outer electronic
configuration at certain regular intervals which indeed determines the physical and chemical properties of
the elements and their compounds.
(i)
A modern version of table contains horizontal rows known as periods (which Mendeleev called series).
Elements having similar outer electronic configuration in their atoms are grouped in vertical columns; these
are referred as Groups or Families.
(ii)
According to IUPAC (International Union of pure and Applied Chemistry), the groups are numbered from
1 to 18 replacing the older notation of groups O, IA, II A .......... VIII B.
(iii)
There are seven periods (three short periods and four long ones).
(iv)
The first period contains 2 elements. The subsequent periods consists of 8, 8, 18, 18 and 32 elements
respectively.
(v)
The 7th period is incomplete.
(vi)
Till now elements upto 112 and 114 have been discovered.
(vii)
Elements with z = 113, 115 and beyond are not known.
Practice Problems :
1.
What is the basic theme of organisation in the periodic table ?
2.
Which important property did Mendeleev use to classify the elements in his periodic table and did
he stick to that ?
3.
What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern
Periodic Law ?
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4.
Why do element in the same group have similar physical and chemical properties ?
5.
In the modern periodic table, the period indicates the value of
(a)
atomic number
(b)
atomic mass
(c)
principal quantum number
(d)
azimuthal quantum number.
[Answers : (1) to simplify and systematize the study of elements and their compounds (2) Mendeleev
used atomic weight as the basis for the classification of elements in the periodic table and he did stick
to that. For example, he put gallium after aluminium and germanium as eka-aluminium and
eka-silicon (3) Mendeleev Periodic law states that the physical and chemical properties of elements
are in periodic functions of their atomic weights, while Modern Periodic Law states that the physical
and chemical properties of elements are in periodic functions of their atomic numbers (4) Elements
of the same group have similar electronic configuration (5) The option (c) is correct, other’s are
wrong]
C2
Arrangement of Elements in Periodic Table is according to Electronic Configuration of the
Elements :
Electronic configuration in Periods :
(i)
Each successive period in the periodic table is associated with the filling of next higher principal energy
shell i.e. n = 1, n = 2 etc.
(ii)
Number of atomic orbitals in each period is twice the number of atomic orbitals available in the energy
level is being filled.
(iii)
The first period starts with the filling of the lowest energy level (1s) and has the two elements – hydrogen
(1s1) and helium (1s2).
(iv)
Second period starts with lithium.
(v)
The third period begins with sodium and added electrons enters a 3s orbital. This shell has nine orbitals
(one 3s, three 3p and five 3d) but 3d orbital are of higher energy than 4s according to (n + l) rule. Therefore
3d orbitals are only filled after filling the 4s-orbitals. Thus it contains only eight elements.
(vi)
In the fourth period, the filling of electrons is in the fourth energy level i.e., n = 4. It starts with 4s. But in
this filling of 4d and 4f orbital does not takes place. After filling 4s orbital, filling of 3d orbital and then 4p
orbitals takes place. 4d and 4f are of higher energy than 5s. Thus it contains 18 elements.
(vii)
In the fifth period the filling of electrons starts with 5s orbital and then 4d orbitals are filled and then three
5p orbitals and filled.
Sixth Period : Corresponds to filling of sixth energy level i.e. n = 6. In this period filling takes place in
(one 6s, seven 4f, five 5d and three 6p) orbitals.There are 16 orbitals are available, thus 32 elements can be
there.
(viii) Filling up of the 4f orbital begins with cerium (z = 58) and ends at lutetium (z = 71) to give the 4f-inner
transition series called the Lanthanoide Series. They are placed at the bottom of the periodic table.
(ix)
Similar to the sixth period, the seventh period corresponds to the filling of seventh energy shell i.e. n = 7. It
is also expected to contain thirty two elements corresponding to filling of sixteen orbitals i.e. one 7s,
seven 5f, five 6d and three 7p.
(x)
Filling up of 5f orbitals after actinium (z = 89) gives the 5f inner tansition series known as Actinoid Series.
Thus 4f, 5f transition series of elements are placed seprately in the periodic table to maintain its structure.
Group wise electronic configuration
(xi)
Elements in the same vertical column or group have similar electronic configuration, have same number of
electrons in the outer orbitals and have similar properties.
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Practice Problems :
1.
On the basis of quantum numbers, justify that the sixth period of the periodic table should have
32 elements.
[Answers : (1) in 6th period, electrons can be filled in only 6s, 4f, 5d and 6p-subshells whose energies
increase in the order as 6s < 4f < 5d < 6p]
Types of elements : s-, p-, d-, f- blocks
C3
-s Block elements : These elements contain 1 or 2 electrons in the s-orbital of their respective outer most
shell.
(i)
The elements of group 1 having outer-most electronic configuration ns1 are called as alkali metals.
(ii)
The elements of group 2 having outer-most electronic configuration ns2 are are called alkaline earth metals.
(iii)
Properties of s- block elements : They are reactive metals with low ionization energy.
(iv)
They lose the outermost electron readily to form +1 oxidation state in case of alkali metals.
(v)
Alkaline earth metals can loose two electrons to aquire +2 oxidation state very easily.
(vi)
Metallic character and reactivity increases as we move down the group.
(vii)
The compounds of s-block are predominantly ionic with exception of Be (beryllium)
P-block Elements : These elements contain 1-6 electrons in the P-orbital of their respective outermost
shells.
(i)
General electronic configuration of outermost shell is ns2np1-6, where n  2-7.
(ii)
These include elements belonging to group of 13, 14, 15, 16, 17 and 18 excluding helium.
Properties of P-Block Elements :
(i)
All the orbitals in the valence shell of the noble gases are completely filled by electrons. Thus it is difficult
to alter this stable arrangement by addition or removal of electrons.
(ii)
Thus noble gases exhibit low chemical reactivity.
(iii)
Group - 17 elements are known as halogens and then have high negative electron gain enthalpy.
(iv)
Similarly group - 16 elements i.e., oxygen family are also known as chalcogens, have high tendency to gain
electrons.
(v)
The non-metallic character increases as we move from left to right in a period.
(vi)
Metallic character increases as we move down the group.
(vii)
Their ionization energies are higher than s-block elements.
(viii) They mostly form covalent compounds.
(ix)
Some of them show more than one oxidation states in their compounds.
d-Block Elements : These elements characterise by filling of inner d-orbital by electrons and therefore
referred to d-Block elements.
(i)
General electronic configuration of d-Block elements is (n – 1) d1-10 ns0-2.
(ii)
They are the elements belonging to 3 to 12 groups.
Properties of d-Block Elements :
(i)
There ionization energies are between s and p-block elements.
(ii)
They show variable oxidation states.
(iii)
They form both ionic and covalent compounds
(iv)
Their compounds are generally coloured and paramagnetic.
(v)
Most of transition metals form alloys.
(vi)
Most of transition elements are used as catalyst i.e., V, Cr, Mn, Fe, Co, Ni, Cu etc.
Exception : Zn, Hg, Cd have (n – 1)d10ns2 electronic configuration, do not show most of properties of
transition elements as they have complete d subshell and in them last electron enters the s-subshell not the
d-subshell.
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Similarly of Zn, Cd, Hg with transition elements are :
(i)
they form complexes like d-block elements.
(ii)
they form covalent compounds.
(iii)
first ionization energies are much higher
f-Block Elements : The two rows at the bottom of periodic table called Lanthanoids and Antinoids have
outer most electronic configuration as : (n – 2)f1-14(n – 1)d0-1ns2
(i)
Last electron enters the f-subshell.
(ii)
Thus two series elements are called f-Block elements or inner-transition elements.
Properties of f-block elements :
(i)
They are all metals.
(ii)
They show variable oxidation states.
(iii)
There compounds are generally coloured.
(iv)
Most of the elements of actinide series are radioactive. In this series after uranium elements are called as
transuranium elements.
(v)
Classification of periodic table can be the basis of metals and non-metals.
Practice Problems :
1.
Write the general outer electronic configuration of s-, p-, d-, and f-block elements.
2.
Assign the position of the element having outer electronic configuration.
(i) ns2np4 for n = 3, (ii) (n – 1) d2ns2 for n = 4, and (iii) (n – 2) f7 (n – 1) d1ns2 for n = 6, in the periodic
table.
[Answers : (1) s-block elements : ns1 – 2, n = 2 to 7, where n represents period in all the cases, p-block
elements : ns2np1 – 6, n = 2 to 6, d-block elements : (n – 1)d1 – 10 ns0 – 2, n = 4 to 7, f-block elements :
(n – 2)f1 – 14(n – 1) d0 – 1 ns2; n = 6 to 7 (2) (i) The group number of the element = 10 + number of
electrons in the valence shell = 10 + 6. The elements with atomic number 16 is sulphur and the
electronic configuration is 1s2, 2s22p53s23p4 (ii) the group number of the element = Number of
d-electrons + number of s-electrons = 2 + 2 = 4. Thus, the element belongs to group 4 and 4th period
i.e, Titanium with atomic number 22. The electronic configuration is 1s2, 2s2, 2p6, 3s23p63d2, 4s2
(iii) The complete electronic configuration of the element is [Xe]4f7, 5d1, 6s2. Thus, the atomic
number of the element = 54 + 7 + 1 + 2 = 64. The element is gadolinium with atomic number 64]
C4
Properties of Metals
(i)
They are good conductors of heat and electricity
(ii)
They are usually solids at room temperature
(iii)
They are malleable (can be flattened into thin sheets by hammering) and ductile (can be drawn into wires)
Properties of Non-Metals
(i)
They are bad conductor of heat and electricity
(ii)
They are usually solids or gases at room temperature
(iii)
They cannot be drawn into thin sheets or wires.
*
Metallic character increases down the group.
*
Metallic character decreases along the period.
*
The change from metals to non-metals in periodic table is not abrupt. The elements on the border line are
semi-metals or metalloids e.g. (Si, As, Sb, Te).
Practice Problems :
1.
What are the major differences between metals and non-metals ?
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C5
Prediction of period, group and block of given elements :
(a)
Period of an element corresponds to principal quantum number of the valence shell.
(b)
The block of an element is type of orbital which receives the last electron.
(c)
Group is predicted as follows :
(i)
for s-block elements : group number is equal to the number of valence electrons.
(ii)
for p-block elements : group number is equal to 10 + number of valence electrons.
(iii)
for d-block elements : group number is equal to number of electrons in (n – 1) d subshell +
number of electrons in valence shell (nth shell)
Practice Problems :
1.
In terms of period and group, where would you locate the element with Z = 114.
2.
Write the atomic number of the element present in the third period and seventeenth group of the
periodic table.
[Answers : (1) The filling of the 6th period ends at 86Rn (1s2, 2s2 2p6, 3s2 3p6 3d10, 4s2 4p6 4d10 4f14, 5s2
5p6 5d10, 6s2 6p6). Thereafter the filling of 7th period starts and is filling according to the Aufbau
principle in increasing order of energies as : 7s < 5f < 6d < 7p. Therefore, after 86Rn, the next two
elements with Z = 87 and Z = 88 are s-block elements, the next fourteen i.e., z = 90 – 103 are f-block
elements, the next ten i.e., Z = 104 – 112 are d-block elements and last six i.e. Z = 113 – 118 are p-block
elements. From this, it is concluded that Z = 114 is the second p-block element i.e., group 14 of the 7th
period (2) In the third period, the filling up of only 3s- and 3p-orbital occurs, the element which will
lie in seventeenth group will have Z = 12 + 5 = 17]
C6
Periodic trends in properties of elements :
1.
Atomic radii : is the distance from the centre of nuclei to the point upto which the density of electron cloud
is maximum.
It is of four types :
(i)
covalent radii
(ii)
vander wall radii
(iii)
metallic radii
(iv)
Ionic radii
Covalent radii : rcovalent
=
½ [Internuclear distance between two covalently bonded atoms of
same molecule]
=
½ [bond length]
Vander wall radii : rvander wall =
Metallic radii :
=
½ [Internuclear distance between two non-bonded atoms] of
different or neighbouring molecules
½ [internuclear distance between two adjacent atoms in the metallic
lattice]
Variation of Atomic radii in periodic table :
1.
Along the period it decreases as nuclear charge increases.
Exception :
The size of atoms of inert gases are however larger than the halogen elements.
Reason :
As we move along the period charge increases while the number of shells remain the same thus along the
period size decreases.
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2.
Along the group the atomic radii of elements increases with increase in atomic number as we move from
top to bottom of group.
Reason : As we move down the number of shell increases thus distance between the outer electrons and
nucleus increases. Also with increase of atomic number the nuclear charge down the group increases. Thus
atomic radii should decreases, but effect of increased nuclear charge is reduced due to screening or sheilding
effect on the valence electrons by the electrons present in the inner shells.
–
Comparison of the ionic radii with atomic radii of same atom – Mg > Mg2+
Reason : In
Mg
np = 12, n e   12
Mg2+
np = 12, n e   10
Thus in Mg 12 protons are attracting by the 12 electrons outer nuclear part. Whereas in Mg2+ the 12 protons
are attracting 10 e– outside the nucleus thus 12 protons will attract 10 e– electrons more strongly thus size
of Mg2+ decreases as compared to Mg.
*
Atoms or ions with same electronic configuration are called isoelectronic. If we consider a series of
isoelectronic species (atoms or ions) then size decreases with increase of atomic number.
e.g.
Atomic number
*
O–2,
I– ,
Ne,
Na+,
Mg2+
(1.40)
(1.36)
(1.31)
(0.95)
(0.65)
8
9
10
11
12
The size of same cation decreases with increase of magnitude of positive charge i.e. Fe3+ is smaller than
Fe2+, Cu2+ is smaller than Cu+ so on.....
Ionisation Energy
First I.E. : Amount of energy required to remove a single electron from the outer shell of a neutral gaseous
atom.
M(g)  M+(g) + e– .... IE1
M+  M2+ + e–.....IE2
(IE2) Second Ionisation Energy : Amount of energy required to remove the second electron from an atom
who has already lost one electron.
IE3 > IE2 > IE1
As the number of electrons in outer shell decreases so attraction of nucleus for remaining electrons
increases thus I.E. increases.
Trends of first I.E. along the period and down the group is as follows :
(a)
First I.E. decreases as we move down the group. Reason is that as we move down the group no. of shell
increases thus nuclear attraction decreases. Although nuclear charge also increases but its effect is weakend
by sheilding supplied by the inner shells to the outer most shell.
(b)
As we move along the period ionisation energy increases. The reason for this is because of decrease of size
of atom along the period. As the number of electrons are added to same shell along the period thus nuclear
attraction for the outer electrons increases with increase of nuclear charge.
Exception :
(1)
I.E. (B) < I.E. (Be)
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Electronic Configuration : B = 1s22s22p1,
Electronic configuration Be = 1s22s2
As in Be e– is to be removed from completely filled shell thus its I.E. is more than B. Here is extra stability
of subshell is the cause of this irregularity
(2)
I.E. of O < I.E. N
Electronic configuration
N = 1s2 2s2 2p3
O = 1s2 2s2 2p4
The extra stability of half-shell is cause of irregularity in IE pattern.
(3)
I.E. of Al < I.E. of Mg
Electronic configuration
Al = 1s2 2s2 2p6 3s2 3p1
Mg = 1s2 2s2 2p6 3s2
*
The electrons in different orbitals (s, p, d, f) beloging to the same energy level experience different pull of
the nucleus.
The I.E. for pulling out an s-electron is maximum and it decreases in pulling out p-electron.
Hence we can say that I.E. for pulling out an electron from a given energy level decreases in the order s >
p > d > f orbitals.
Electron Affinity : It is defined as energy given out when an extra electron is taken up by a neutral gaseous
atom.
X(g) + e–  X–(g)
E. A measures the tightness with which an atom binds an extra electron to it.
E.A. decreases down the group, because an atom gets larger, the attractions of positive nucleus for an
outside electron decreases.
Exception : E.A. of F < E.A. of Cl
Reason : It is due to extremely small size of F atom as compare to Cl atom.
Extra electron create strong electron-electron repulsion among all the electrons.
Along the period : Along the period electron affinity increases as we move from left to right.
(i)
E.A. of noble gases are zero.
(ii)
E.A of N and Be atoms are quite low due to extra stability of half filled orbitals (p3 in N & s2 in Be)
(iii)
After taking up an extra electron an atom becomes negatively charged (anion) and now second electron is
to be added to it. The anion will repel the incoming of an electron and an additional energy will be required
to add it to the anion.
First E.A.

negative (energy released)
Second E.A.

positive (energy released)
Electronegativity : It is the measure of the ability of an atom in a combined state (i.e. in a molecule) to
attract itself to the electrons within a chemical bond or tendency to attract bond pair towards itself.
*
Non-metals have high value of electronegativity than metals.
F, O, N & Cl are highly electronegative than [K, Rb, Cs] (metals) which are electropositive in nature.
*
Electronegativity along the period increases & down the group decreases.
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*
It is measured with the help of pauling scale & mulliken scale.
Screening (sheilding) effect :
In d-block elements (transition element) while writing electronic configuration of elements, it is seen that
new electrons are added to inner shells i.e. penultimate shells. Thus nuclear attraction for out electons gets
affected. As the new electrons enter the inner shells they tend to sheild or screen the outer shell electrons
from nucleus and thus decreases the nuclear attractive force. This is called as screening effect. Due to this
effect, the atomic size of transition elements remain same as we move along the period which should
decrease as we move from left to right. Thus ionisation energy, electron affinity and other properties
remains nearly same as we move along the period.
Practice Problems :
1.
What does atomic radius and ionic radius really mean to you ?
2.
How do atomic radius vary in a period and in a group ? How do you explain the variation ?
3.
What do you understand by isoelectronic species ? Name a species that will be isoelectronic with
each of the following atoms or ions.
(i)
4.
F—
(ii)
Ar
(iii)
Mg2+
(iv)
Rb+
Consider the following species : N3–, O2–, F–, Na+, Mg2+ and Al3+.
(i)
What is common in them ?
(ii)
Arrange them in the order of increasing ionic radii.
5.
Explain, why cations are smaller and anions are larger in radii than their parent atoms ?
6.
What is the significance of the terms – ‘isolated gaseous atom’ and ‘ground state’ while defining the
ionization enthalpy and electron gain enthalpy ?
7.
Among the second period elements the actual ionization enthalpies are in the order :
Li < B < Be < C < O < N < F < Ne
Explain why ?
(i)
Be has higher iH than B
(ii)
O has lower iH and N and F ?
8.
How would you explain the fact that the first ionization enthalpy of sodium is lower than that of
magnesium, but its second ionization enthalpy is higher than that of magnesium ?
9.
What are the various factors due to which the ionization enthalpy of the main group elements tends
to decrease down a group ?
10.
The first ionization enthalpy values (in kJ mol–1) of group 13 elements are :
B
Al
Ga
In
Tl
801
577
579
558
589
How would you explain this deviation from the general trend ?
11.
Which of the following pairs of elements have a more negative electron gain enthalpy ?
(i)
O or F
(ii)
F or Cl
12.
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative
than the first ? Justify your answer.
13.
What is the basic difference between the terms electron gain enthalpy and electronegativity ?
14.
How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the
nitrogen compounds ?
15.
Describe the theory associated with the radius of an atom as it (a) gains an electron (b) loses an
electron.
16.
Would you expect the first ionization enthalpies of two isotopes of the same element to be the same or
different ? Justify your answer.
17.
Use the periodic table to answer the following questions.
(a)
Identify an element with five electrons in the outermost subshell.
(b)
Identify an element that would tend to lose two electrons.
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CCP – 10
(c)
Identify an element that would tend to gain two electrons.
(d)
Identify the group having metal, non-metal, liquid as well as gas at the room temperature.
18.
The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs; whereas that
among group 17 elements is F > Cl > Br > I. Explain.
19.
The first (iH1) and the second (iH2) ionisation enthalpies (in kJ mol–1) and the (egH) electron gain
enthalpy in (kJ mol–1) of a few elements are given below :
Elements
 iH 1
 iH 2
rgH
I
520
7300
–60
II
419
3051
–48
III
1681
3374
–328
IV
1008
1846
–295
V
2372
5251
+48
VI
738
1451
–40
Which of the above element is likely to be :
20.
(a)
the least reactive element
(b)
the most reactive metal
(c)
the most reactive non-metal
(d)
the least reactive non-metal
(e)
the metal which can form a stable binary halide of the formula MX2 (X = halogen).
(f)
the metal which can form a predominantly stable covalent halide of the formula
MX (X = halogen) ?
Predict the formulas of the stable binary compounds that would be formed by the combination of
the following pairs of elements.
(a)
Lithium and oxygen
(b)
Magnesium and nitrogen
(c)
Aluminium and iodine
(d)
Silicon and oxygen
(e)
Phosphorus and fluorine
(f)
Element 71 and fluorine
[Answers : (1) Atomic radius is one half of the distance between the nuclei of two identical atoms in
a molecule bonded by a single bond. Ionic radius is the distance between the cations and anions in
the ionic crystals (2) The atomic radii of elements decreases in a period from left to right with an
increase in atomic number. In a period electrons, enter one by one in the same energy level. On the
addition of each electron, the nuclear charge increases by one unit. The result is that the electrons
are attracted more and more strongly towards the nucleus. This result in decrease of atomic radii.
The atomic radii increases in a group while moving from top to bottom. In a group from top to
bottom, new shells are added with increasing atomic number. Nuclear size increases and hence atomic
size should decrease with increase in nuclear charge, but effect is dominated by increase in number
of shells and hence over all size increases in moving from top to bottom (3) Ions of different elements
having the same number of electrons but different magnitude of the nuclear charge are termed as
isoelectronic ions. (i) F– has (9 + 1) = 10 electrons. The other isoelectronic species with 10 electrons
are : N3– (7 + 3); O2– (8 + 2); Na+(11 – 1); Mg2+(12 – 2) and Al3+(13 – 3). (ii) Ar has 18 electrons. The
other isoelectronic species with 18 electrons are : S2–(16 + 2), Cl–(17 + 1), K+(19 – 1), Ca2+(20 – 2). (iii)
Mg2+ has (12 – 2) = 10 electrons. The other isoelectronic species with 10 electrons are : N3–, O2–, Na+,
F–, Mg2+, Al3+. (iv) Rb+ has (37 – 1) = 36 electrons. The other isoelectronic species with 36 electrons are
: Br(35 + 1), Kr(36), Sr2+ (38 – 2) (4) (i) Each one of these have 10 electrons and hence all are
isoelectronic (ii) N3– > O2– > F– > Na+ > Mg2+ > Al3+ (5) in the formation of cations their is a loss of one
or more electrons which increases the effective nuclear charge. As a result, the force of attraction of
the nucleus for the electrons increases and hence the ionic radii decreases. The ionic radii of an anion
is always larger than its parent atom because the addition of one or more electrons decreases the
effective nuclear charge. As a result, the force of attraction of the nucleus for the electrons decreases
and hence the ionic radii increases (7) (i) The ionization enthalpy, beside other things, depends upon
the type of electron to be removed from the same principal shell. In Be (electronic configuration; 1s2,
2s2), the outermost electron is present in 2s-orbital while in B (electronic configuration : 1s2, 2s22p1)
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it is present in 2p-orbital. Since 2s-electrons are more strongly attracted to the nucleus as compared
to 2p-electrons, therefore, lesser energy is required to remove out a 2p-electron, as compared to
2s-electron. Because of this, H of Be is higher than that of iH of B. (ii) Nitrogen has the electronic
configuration : 1s2, 2s2 2px1 2py1 2pz1 in which 2p-orbital are half filled is more stable than the oxygen
which has the electronic configuration 1s2, 2s2 2px2 2py1 2pz1 in which the 2p-orbitals are neither half
filled nor completely filled. Therefore, it is difficult to remove electron from nitrogen than oxygen.
Because of this iH of nitrogen is higher than that of oxygen. Further the electronic configuration of
F is 1s2, 2s2 2px2 2py2 2pz1. Because of high nuclear charge, that the first ionization enthalpy of F is
higher than of O. Also, the effects of increased nuclear charge outweights the effect of stability due to
exactly half-filled orbitals. Because of this iH of fluorine is higher than that of oxygen. (As a general
rule if the electron is to be removed from same orbital then the one having half-filled or completely
filled have higher ionization enthalpy as compared to one having neither half-filled nor completely
filled orbital. Also, ionization enthalpy of half-filled orbital is < that of ionisation enthalpy of
completely filled orbital) (8) The sodium has electronic configuration 1s2, 2s22p6,3s1 and magnesium
has electronic configuration 1s2, 2s22p6, 3s2. Here, in both the cases first electron is to be removed
from 3s-orbital, but the nuclear charge of Na (+11) is lower than that of Mg (+12). Therefore, the
first ionization enthalpy of sodium is lower than magnesium. After the loss of one electron from
sodium, the electronic configuration of sodium becomes completely filled i.e., 1s2, 2s22p6 and after
the loss of one electron from magnesium, the electronic configuration of magnesium becomes 1s2,
2s22p6, 3s1 i.e., half-filled. Hence iH of Na > iH of Mg (9) (i) Atomic size : On moving down the
group, the atomic size gradually increases with the addition of one energy shell to each succeeding
element. As a result, the distance of valence electrons from the nucleus increases. Consequently, the
force of attraction of the nucleus from the valence electrons decreases and hence the ionization
enthalpy decreases on moving down the group. (ii) Screening effect : On moving down the group,
with the addition of new shells, the number of inner electron shells which shield the valence electrons
increases regularly, thereby increasing the shielding effect or screening effect. This means, that the
force of attraction of the nucleus for the valence electrons further decreases and hence the ionization
enthalpy further decreases (10) On moving down the group from Boron to Thalium via aluminium,
indium and thallium the ionization enthalpy decreases (although not regular). This can be explained
on the basis of increasing atomic size and screening effect. The abnormal behaviour of Al, Ga, In and
Tl can be explained as follows : Al follows immediately after s-block elements while Ga and In follow
after d-block elements, whereas Tl after d- and f-block elements. These d- and f- electrons do not
shield the outer-shell electrons from the nucleus effectively. The result is that, the valence electron
remain more tightly held by nucleus and thus larger amount of energy is needed for their removal.
This explains why Ga has higher ionization enthalpy than Al. Further moving down the group from
Ga to In, the shielding effect increases due to the presence of additional 4d-electrons, outweighs the
effect of increased nuclear charge, and hence iH of In is lower than that of Ga. Thereafter, the effect
of increased nuclear charge outweight the shielding effect due to the additional 4f and 5d-electrons
and thus the iH of Tl is higher than that of In (11) (i) Both oxygen and fluorine belongs to 2nd
period. As we move from left to right i.e., from O to F, the atomic size decreases with the increase of
nuclear charge. Both these factors are responsible to increase the attraction of the nucleus for the
incoming electron and thus electron gain enthalpy becomes more negative. Further, gain of one
electron by O gives O– ion does not have stable inert gas configuration, whereas, gain of one electron
by F gives F– ion which has stable inert gas configuration. Because of this, the energy released is
much higher in going from F to F– than in going from O to O–, i.e., the electron gain enthalpy of
F(–328 kJ mol–1) is much more negative than that of oxygen (–141 kJ mol–1). (ii) In general, the
electron gain enthalpy becomes less negative on moving down the group, but the electron gain
enthalpy of chlorine (–349 kJ mol–1) is little more negative than that of fluorine (–328 kJ mol–1). This
is because fluorine has a small atomic size (only two shells) and hence the incoming electron is not
accepted with the same ease as in the case of large chlorine atom. Because of this electron gain
enthalpy of chlorine is more negative than that of fluorine (12) The second electron gain enthalpy of
oxygen is positive which can be explained as given : When an electron is added to O atom to form O–
ion, energy is released. O(g) + e–(g)  O–(g), iH = –141 kJ mol–1. Because of this, first electron gain
enthalpy of oxygen atom is positive. But when another electron is added to O– to form O2– ion, energy
is released to overcome the strong electrostatic repulsion between the negatively charged O– ion and
the second electron which is being added. O–(g) + e–(g)  O2– (g); iH = +780 kJ mol–1. Because of
this, the second electron gain enthalpy is positive (13) The electron gain enthalpy refers to the
tendency of an isolated gaseous atom to accept an additional electron to form a negative ion, whereas
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electronegativity refers to the tendency of the atom of an element to attract the shared pair of
electrons towards it in the formation of a covalent bond (14) The statement seems to be absurd. In
fact, the electronegativity of any given atom is not constant. It increases as the percentage of
s-character of a hybrid orbital increase or the oxidation state of the element increase. For example,
the electronegativity of N increases as we move from sp3-sp2-sp hybrid orbitals (15) (a) When a
neutral atom gains an electron to form an anion, the number of electrons in the anion increases
whereas its nuclear charge remains the same as that of the parent atom. Since the same nuclear
charge now attracts larger numbers of electrons, therefore, the force of attraction of the nucleus on
the electrons of all shells decreases. This means that the effective nuclear charge decreases and the
expansion of the electron cloud occurs. Because of this, the distance between the centre of the nucleus
and the last shell increases thereby increasing the ionic radius. (b) When a neutral atom loses an
electron to form a cation, the number of electrons in the cation decreases whereas its nuclear charge
remains the same as that of the parent atom. Since the same nuclear charge now attracts lesser
numbers of electrons, therefore, the force of attraction of the nucleus on the electrons of all shells
increases. This means that the effective nuclear charge increases and the contraction of the electron
cloud occurs. Because of this distance between the centre of the nucleus and the last shell decreases
thereby decreasing the ionic radius (16) Two isotopes of the same element have the same atomic
number; which means that they have the same number of electrons and hence they have the same
ionisation enthalpy (17) (a) The element with five electrons in the outermost sub-shell belongs to
seventeenth group i.e. fluorine, chlorine, bromine, etc. (b) The element that would tend to loose two
electrons belongs to alkaline earth metals i.e., magnesium, calcium, etc. (c) The element that would
tend to gain two electrons belongs to sixteenth group i.e., oxygen and sulphur. (d) A metal which is
liquid at room temperature is bromine belongs to fifteenth group (18) Reactivity of an element
depends upon the ease with which it can lose the outermost electron. The tendency to lose electron, in
turn, depends upon the ionization enthalpy. The ionization enthlapy decreases down the group,
therefore, the reactivities of group 1 element increase in the same order i.e., Li < Na < K < Rb < Cs.
In contrast group 17 elements, have seven electrons in their respective valence shells and thus have
a strong tendency to accept one electron to get the nearest inert-gas configuration. The tendency to
accept electrons, in turn, depends upon their electrode potentials. The electrode potential of
seventeenth group decreases in order i.e., F > Cl > Br > I (19) (a) The element V has highest first
ionization enthalpy (iH1) and positive electron gain enthalpy (egH) and hence it is the least reactive
element. Element V must be inert gas because inert gases have positive egH. The values of iH1, iH2
and egH match that of Helium. (b) The element II which has the least first ionization enthalpy (iH1)
and a low negative electron gain enthalpy (egH) is the most reactive metal. The values of iH1, iH2
and egH match that of potassium. (c) The element III which has high first ionization enthalpy (iH1)
and very high negative electron gain enthalpy (egH) is most reactive non-metal. The values of iH1,
iH2 and egH match that of fluorine. (d) The element IV has high negative electron gain enthalpy
(egH) but not so high first ionization enthalpy (iH1) is the least reactive non-metal. The values of
iH1, iH2 and egH match that of Iodine. (e) The element VI has low first ionization enthalpy (iH1)
but higher than that of alkali metals. Therefore, it appears that the element is an alkaline earth
metal and hence will form binary halide of the formula MX2 (where X = halogen). The values of iH1,
iH2 and egH match that of Magnesium. (f) The element 1 has low first ionization (iH1) but a very
high second ionization enthalpy (iH2), therefore, it must be an alkali metal. Since the metal form a
predominantly stable covalent halide of the formula MX (X = halogen), therefore, the alkali metal
must be least reactive. The values of iH1, iH2 and egH match that of Lithium (20) (a) Li2O (Lithium
oxide) (b) Mg3N2 (Magnesium nitride) (c) AlI3 (Aluminium iodide) (d) SiO2 (Silicon dioxide) (e) PF3
(Phosporus tri-fluoride) or PF5 (Phosphorus penta-fluoride) (f) LuF3 (Lutetium fluoride)]
C7
Hydration and Hydration Energy :
Hydration energy is the enthalpy change that accompanies the dissolving of one mole of gaseous ions in
water.
Li+(g) + H2O  [Li(H2O)]+, H = –806 kJ mol–1
(i)
Size of the ion and its charge determines extent of hydration.
(ii)
Greater the charge, smaller the size of the ion, greater the attraction for the lone pair of O of H2O, hence
greater the extent of hydration and hence greater the hydration energy.
–
size of the hydrated ions increases,
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–
ionic mobility decreases [heavier (hydrated) ions moves slower]
Practice Problems :
Arrange the following ions in increasing Na+, Mg2+, Al3+
1.
(i)
extent of hydration
(iii)
size of the hydrated ions
+
(ii)
3+
(iv)
+
ionic mobility
[Answers : (1) (i) Na < Mg < Al (ii) Na < Mg < Al3+ (iii) Na+ < Mg2+ < Al3+ (iv) Al3+ < Mg2+ < Na+]
C8
2+
hydration energy
2+
Acid-Base character of oxides :
On moving across a period, the basic character of the oxides gradually changes first into amphoteric and
finally into acidic character.
(i)
On moving down the group, reverse behaviour is observed, i.e., from more acidic to more basic.
Oxides of the element M in H2O produce MOH
–
If electronegativities difference of M and O is greater than that of H and O in H2O then MOH is acidic due
to formation of H3O+
M — O — H + H2O  H3O+ + MO–
–
If electronegativities difference of M and O is less than that of H and O in H2O then MOH is basic due to
formation of OH–
M — O — H + H — O — H  [MOH2]+ + OH–
(ii)
Stability of oxides decreases across a period.
(Min.  Max)
(iii)
Oxides of the following elements are amphoteric
H, Be, Al, Ga, In, Tl, Sn, Pb, Sb, Bi, Po
–
H2O is amphoteric (also called amphiprotic)
H2O + H2O  H3O+ + OH–
Since it is H+ acceptor (base) as well as H+ donor (acid).
–
BeO, Al2O3, SnO2, PbO2,.... are amphoteric since they form salts with acid as well as with base
Al 2 O 3  6HCl 
 2AlCl 3  3H 2 O
base
Al 2 O 3  2NaOH  3H 2 O 
 2Na[ Al(OH ) 4 ]
acid
base
–
oxide is acidic if it reacts with a base.
–
oxide is basic if it reacts with an acid.
Practice Problems :
1.
Arrange the following oxides in order of increasing molecular (acidic) character :
SO3, Cl2O7, CaO and PbO2
2.
Arrange following oxides in increasing acidic nature Li2O, BeO, B2O3.
3.
Which oxide is more basic, MgO or BaO ? Why ?
[Answers : (1) CaO < PbO2 < SO3 < Cl2O7 (2) Li 2O 
basic
BeO
amphoteric
 B 2 O 3 (3) BaO]
acidic
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SINGLE CORRECT CHOICE TYPE
1.
The following acids have been arranged in the
order of decreasing acid strength. Identify the
correct order
ClOH(I)
2.
4.
5.
6.
IOH(III)
(a)
I > II > III
(b)
II > I > III
(c)
III > II > I
(d)
I > III > II
The statement is not true for the long form of the
periodic table
(a)
it reflects the sequence of filling the
electrons in the order of the sub-energy
shells s, p, d and f
(b)
it helps to predict the stable valency states
of the elements
(c)
it reflects trends in physical and
chemical properties of the elements.
(d)
3.
BrOH(II)
it helps to predict the relative inicity of
the bond between any two elements.
7.
8.
9.
10.
Ionic radii of
(a)
Ti4+ > Mn7+
(b)
35
Cl— < 37Cl—
(c)
K+ > Cl—
(d)
P3+ < P5+
Which of the following ions has the smallest
radius ?
(a)
Ti2+
(b)
Pt2+
(c)
Ni2+
(d)
Zr2+
When the following five anions are arranged in
order of decreasing ionic radius, the correct
sequence is
(a)
Se2—, I—, Br—, O2—, F—
(b)
I—, Se2—, O2—, Br—, F—
(c)
Se2—, I—, Br—, F—, O2—
(d)
I—, Se2—, Br—, O2—, F—
Which of the following ions has the largest heat of
hydration ?
The electronic configuration
(a)
Na+
(b)
Al3+
1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2, 4p6, 4d10, 5s2 is for
(c)
F—
(d)
Sr2+
(a)
f-block element
11.
(b)
d-block element
Which of the following anions is most easily
polarized ?
(c)
p-block element
(a)
Cl—
(b)
Se2—
(d)
s-block element
(c)
Br—
(d)
Te2—
Which of the following statements is false :
(a)
elements of I-B and II-B groups are
transition elements
(b)
elements of V-B group do not contain
metalloids
(c)
elements of I-A and II-A groups are
normal elements
(d)
elements of IV-B group are neither
strongly electronegative nor strongly
electropositive
12.
Of the four H values needed to calculate a lattice
energy using the Born-Haber cycle, the one that is
most difficult to measure is
(a)
the heat of sublimation of the metal
(b)
the heat of formation of gaseous atoms
of the non-metal
(c)
the ionization energy of the metal
(d)
the electron affinity of the non-metal
Transition metals are characterised by the
properties :
The melting point of RbBr is 6820C while that of
NaF is 988 0C. The principal reason that the
melting point of NaF is much higher than that of
RbBr is that
(a)
variable valency
(a)
(b)
coloured compounds
the molar mass of NaF is smaller than
that of RbBr.
(c)
high melting and boiling points
(b)
the bond in RbBr has more covalent
character than the bond in NaF
(d)
all of the above
(c)
the difference in electronegativity
between Rb and Br is smaller than the
difference between Na and F.
(d)
the internuclear distance, rc + ra is greater
for RbBr than for NaF.
13.
The atomic number of an element is 16. This
element in Periodic Table belongs to
(a)
group VI and period II
(b)
period III and group VI
(c)
group IV and period II
(d)
period III and IV
14.
Going down in a group from F to I, which of the
following properties does not decreases
(a)
ionic radius
(b)
ionisation energy
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15.
16.
17.
18.
19.
20.
21.
22.
23.
(c)
oxidising power
24.
(d)
electronegativity
(a)
Ge2+ < Sn2+ < Pb2+
(b)
Ge4+ > Sn4+ > Pb4+
1s22s22p6
(c)
Sn4+ > Sn2+
1s22s22p63s23p6
(d)
Pb2+ < Pb4+
Which of the following will have maximum
electron affinity ?
(a)
1s22s22p5
(c)
1s22s22p63s23p5 (d)
(b)
Among the following stability of ions of Ge, Sn and
Pb will be in order except
Size of cation is smaller than that of the atom
because of
(a)
the whole of the outer shell of electrons
removed
(b)
increase in effective nuclear charge
(c)
due to gain of electrons
(d)
statement, that cation is smaller than
atom, is wrong
25.
26.
The sizes of the second and third row transition
elements are almost the same. This is due to
(a)
d- and f-orbitals do not shield the nuclei
charge very effectively
(b)
lanthanide contraction
(c)
both true
(d)
none is true
27.
Which is true statement
(a)
Tl3+ salts are better oxidising agents
(b)
Ga+ salts are better reducing agents
(c)
Pb4+ salts are better oxidising agents
(d)
all of above
The first element of a group in many ways differs
from the other heavier members of the group. This
is due to
(a)
the small size
(b)
the high electronegativity
(c)
the unavailability of d-orbitals
(d)
all of above
Catenation properties of C, Si, Ge, Sn, Pb in order
(a)
C >> Si > Ge  Sn >> Pb
The factors that influence the ionisation energies
are
(b)
C < Si < Ge < Sn < Pb
(c)
C > Si > Sn > Ge > Pb
(a)
the size of the atom
(d)
none is correct
(b)
the charge on the nucleus
(c)
how effectively the inner electron shell
screen the nuclear charge
(d)
all the above
28.
Following are the values of the electron affinity of
the formation of O— and O2— from O
(a)
–142, –702
(b)
–142, 702
(c)
142, 702
(d)
–142, –142
29.
Covalency is favoured in the following cases
(a)
a smaller cation
(b)
a larger anion
(c)
large charges on catio or anion
(d)
all the above
30.
A molecule H—X will be 50%
electronegativity difference of H and X is
(a)
1.2 eV
(b)
1.4 eV
(c)
1.5 eV
(d)
1.7 eV
s- and p-block
(b)
d-block
(c)
d- and f-block
(d)
f-block
Inert pair effect is shown by
(a)
s-block
(b)
p-block
(c)
d-block
(d)
f-block
(a)
NaI < NaBr < NaCl < NaF
(b)
NaF < NaCl < NaBr < NaI
(c)
NaBr < NaF < NaCl < NaI
(d)
NaCl < NaI < NaF < NaBr
Which is amphoteric oxides
(a)
BeO
(b)
SnO
(c)
ZnO
(d)
All
Atomic number 64 will have electronic
configuration
(a)
(b)
(c)
(d)
ionic
31.
Representative elements belong to
(a)
Melting points of NaCl, NaBr, NaI and NaF will be
in order
32.
[Xe]54 6s24f8
[Xe]54 6s2 4f7 5d1
[Xe]54 4f10
[Xe]54 6s2 4f7 5p1
The correct order of second ionisation potential of
carbon, nitrogen, oxygen and fluorine is
(a)
C>N>O>F
(b)
O>N>F>C
(c)
O>F>N>C
(d)
F>O>N>C
The element with the highest first ionisation
potential is
(a)
boron
(b)
carbon
(c)
nitrogen
(d)
oxygen
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33.
The hydration energy of Mg2+ is larger than that of
(a)
(c)
34.
35.
36.
37.
38.
39.
41.
(b)
Na
2+
(d)
Mg3+
Be
42.
+
The first ionisation potential in electron volts of
nitrogen and oxygen atoms are respectively given
by
(a)
14.6, 13.6
(b)
13.6, 14.6
(c)
13.6, 13.6
(d)
14.6, 14.6
43.
Which of the following has the maximum number
of unpaired electron ?
(a)
Mg2+
(b)
Ti3+
(c)
V3+
(d)
Fe2+
The incorrect statement among the following is
(a)
The first ionisation potential of Al is less
than the first ionisation potential of Mg
(b)
The second ionisation potential of Mg is
greater than the second ionisation
potential of Na
(c)
The first ionisation potential of Na is less
than the first ionisation potential of Mg
(d)
The third ionisation of Mg is greater than
third ionisation potential of Mg
Atomic radii of fluorine and neon in Angstrom units
are respectively given by
(a)
0.72, 1.60
(b)
1.60, 1.60
(c)
0.72, 0.72
(d)
None of these
The electronegativity of the following elements
increases in the order
(a)
C, N, Si, P
(b)
N, Si, C, P
(c)
Si, P, C, N
(d)
P, Si, N, C
The first ionisation potential of Na, Mg, Al and Si
are in the order
(a)
Na < Mg < Al < Si
(b)
Na > Mg > Al > Si
(c)
Na < Mg < Al > Si
(d)
Na > Mg > Al < Si
Which one of the following is the smallest in size ?
(a)
N3–
(b)
O2–
(c)
–
(d)
Na+
F
Amongst the following elements (whose electronic
configurations are given below), the one having the
highest ionisation energy is
(a)
40.
Al
3+
[Ne] 3s2 3p1
2
3
(b)
[Ne] 3s 3p
(c)
[Ne] 3s2 3p2
(d)
[Ar] 3d10 4s2 4p3
The statement that is not correct for the periodic
classification of elements, is
(a)
The properties of elements are the
periodic functions of their atomic
numbers
(b)
Non-metallic elements are lesser in
number than metallic elements.
(c)
The first ionisation energies of elements
along a period do not vary in a regular
manner with increase in atomic number.
(d)
For transition elements the d-subshells
are filled with electrons monotonically
with increase in atomic number.
Which has most stable +2 oxidation state ?
(a)
Sn
(b)
Pb
(c)
Fe
(d)
Ag
44.
45.
Which of the following compounds is expected
coloured ?
(a)
Ag2SO4
(b)
CuF2
(c)
MgF2
(d)
CuCl
The correct order of radii is
(a)
N < Be < B
(b)
F– < O2– < N3–
(c)
Na < Li < K
(d)
Fe3+ < Fe2+ < Fe4+
A N S W E R S (SINGLE CORRECT
CHOICE TYPE)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
a
d
b
d
d
b
a
c
d
b
d
d
d
a
c
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
b
a
d
b
d
d
a
b
d
d
d
a
a
d
b
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
45.
Einstein Classes, Unit No. 102, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road,
New Delhi – 110 018, Ph. : 011-65643806, 9312629035
c
c
b
a
a
c
a
d
b
d
b
d
b
b
b