Download Classification and Periodic Properties of Elements

I
:+-74 pm ----.:
I
Unit
COVALENT
RADIUS = 37 pm
Classification of Elements and Periodicity
in Properties
••
THE NEED FOR CLASSIFICA·
TION OF ELEMENTS
Before the beginning of eighteenth century, only
a very few elements were known and it was quite
easy to study and remember their individual
properties. In 1800, only 31 elements were known.
This number of elements grew to 63 by 1865. At
present about 114 elements are known of which 94
are naturally occurring. With the discovery of large
number of elements it became difficult to study
individually the properties of these elements and their
compounds. At this stage the scientists felt the need
of some simple method to facilitate the study of the
properties of various elements and their compounds.
After numerous
attempts
the scientists
were
ultimately successful in arranging the elements in
such a way so that similar elements were grouped
together and different elements were separated.
The arrangement of elements in such a way that
the similar elements fall within same vertical group
and the dissimilar elements are separated, is known
as classification
of elements.
The classification of elements led to the formation
of periodic table. Thus :
Periodic table may be defined as the table
giving the arrangement
of all the known elements
according
to their properties
so that similar
elements fall within the same vertical column and
dissimilar elements are separated.
11
EARLIER ATTEMPI'S OF CLASSIFICATION OF ELEMENTS
Earlier attempts on classification of elements
were based on atomic weights. The formulation of a
satisfactory periodic law took place only after 1860.
DOBEREINER'S
TRIADS
The German chemist, Johann Dobereiner (1829)
made the first significant attempt to show a relationship
between atomic weights and the chemical properties of
the elements. He observed that certain similar elements
exist in groups of three elements which he named
triads. An interesting feature of these triads is that
the atomic weight of middle member was the arithmetic
mean of the atomic weights of the other two members
of the triad. Moreover, the properties of the middle
element were in between those of the other two elements
of the triad. For example, lithium, sodium and
potassium constituted one triad. Atomic weights of
lithium, sodium and potassium are 7, 23 and 39. We
can observe that atomic weight of sodium is equal to
the average of atomic weights oflithium and potassium.
Atomic weight of lithium
'·
. ht
Atomic weig
'of sodium
+ Atomic weight of potassium
= -----=---=----2
7+39
=-2-
=23
Some more examples of triads
Table 3. I.
are given in
However, this law of triads could be applied only
to a limited number of elements and was dismissed as
more coincidence.
THE TELLURIC HELIX
In 1862, A.E.B. de Cbancourtois,
a French
geologist, indicated a relation between the properties of
the elements and their atomic weights in the form ofhis
telluric helix. He used a vertical cylinder with 16
equidistant lines on its surface, the lines being parallel
to the axis of cylinder. He drew a helix at 45° to the axis
and arranged the elements on the helix in the order of
their increasing atomic weights. In this way the elements
that differed from each other in atomic weight by 16 or
multiples of 16 fell very nearly on the same vertical line.
This concept did not attract much attention.
THE LAW OF OCTAVES
The distinction' of correlating the chemical
properties of the elements with their atomic weights
goes to J.A.R. Newlands. In 1865, he arranged the
elements in the order of increasing atomic weights and
Table 3.1. Dobereiner's
Triads
_'J1~~:~;~:
~::.ii~¥~
'~~~~~\:~
; ~<~·<l~·.·~:~~~
&'
S
Se
Te
32
79
128
Cl
Br
35.5
80
127
I
Ca
Sr
Ba
40
88
137
32+ 128
=80
2
35.5 + 127
= 81.25
2
40+ 137
=88.5
2
Table 3.2. Newlands'
Li
7
Na
23
K
39
Octaves
Be
9
B
11
C
12
Mg
AI
Si
24
27
29
N
0
F
19
Ca
40
noticed that the eighth element, starting from a given
one, is a kind of repetition of the first, like the eighth
note in an octave of music. Newlands named this
generalization, the law of octaves, due to its similarity
to the musical scale. Although Newlands' law of octaves
met with ridicule, nevertheless Newlands was the first
to publish a list of the elements in the order of increasing
atomic weights. The arrangement of elements as shown
in Table 3.2 illustrates the law of octaves.
However, the law of octaves also could not be
applied beyond calcium. Moreover, with the discovery
of noble gases, the eighth element no longer remains a
similar element.
The credit for the development of periodic table
goes to the Russian scientist Dmitri I. Mendeleev and
to the German scientist Lothar Meyer. In 1869, they
independently proposed that the properties
of the
elements are periodic function of their atomic
weights.
Attempts to find regularities among the elements
led the Russian scientist, Dmitri I. Mendeleev to put
forward a scheme of classification of elements in 1869.
He gave a periodic law known after his name as
Mendeleev's periodic law. This law states that:
The properties of elements are a periodic
function of their atomic weights.
It means that when the elements are arranged
in order of increasing atomic weights, the elements with
similar properties recur after regular intervals. On the
basis of this periodic law, Mendeleev constructed a
periodic table in such a way that the elements were
arranged horizontally in the order of their increasing
atomic weights. However, he also kept in mind the
similarities in the chemical properties of the elements.
The main criterion of thejudgement ofsimilarities
in the properties was valency of the elements.
Mendeleev observed that some of the elements did not
fit in with his scheme of classification if the order of
atomic weights was strictly followed. He ignored the
order of atomic weights and placed the elements with
similar chemical properties together. For example,
iodine having atomic weight 127 was placed after
tellurium (atomic weight 128), together with fluorine,
chlorine and bromine due to similarities in properties.
The general plan of the modified Mendeleev's periodic
table is shown in Fig. 3.1. The Mendeleev's modified
periodic table consists of :
(i) Nine vertical columns, called groups. These
are numbered from I to VIII and zero. (The members of
zero group were. not discovered at the time of
Mendeleev). Each group, from I to VII, is further subdivided into two sub-groups designated as A and B.
Group VIII consists of three sets each one containing
three elements. Group zero consists of inert gases.
(ii) Seven horizontal rows, called series or
periods. These are numbered from 1 to 7. First period
contains two elements. Second and third periods
contain eight elements each. These periods are called
short periods. Fourth and fifth periods contain
eighteen elements each. These periods are called long
periods. Sixth period contains 32 elements and is
called longest period. Seventh period is incomplete.
USES OF MENDELEEV'S
PERIODIC
TABLE
Some important uses of Mendeleev's periodic
table are given below :
1. Classification
of elements. In this periodic
table elements are classified into groups with similar
properties, thus facilitating the study of properties of
elements.
2. Prediction of new elements. Mendeleev left
certain vacant places in his table which provided a clue
for the discovery of new elements. Some of the properties
Modified
Form
of Mendeleev's
Periodic
Table
20.183
. Ar18
39.948
6.939
9.012
10.81
12.001
14.007
15.999
18.998
3
INaIl
22.99
Mg12
24.31
Al13
26.98
Si 14
28.09
p15
30.974
S16
32.06
Cl17
35.453
4
Se21
Mn25
Ca20
IK19
Ti22
v=
Cr24
39.102
40.08
44.96
47.90
50.94
52.00
54.94
Br35
Ge32
As33
• Se34
Cu29
Zn30 Ga31
72.59
74.92
78.96
79.909
63.54
65.37 69.72
5
IRb37
85.47
Ag47
107.87
6
7
va9
88.91
·Sr38
87.62
Cd46 In49
112.40 114.82
zr40
91.22
Sb51
121.75
Sn50
118.69
Hf12
*La57
I CS55
Ba56
178.49
132.90
137.34
138.9
Hg80 Tl81
Pb82
Au79
196.97
200.59 204.37
207.19
IFr87
(223)
Ra88
(226)
*LANTHANIDES
Ce56
140.12
Pr59
*ACTINIDES
Th90
232.04
Pa91
(231)
140.91
Nb41
92.91
Ta73
180.95
Bi83
208.98
Ku104
*Ac89
(227)
M042
95.94
Te52
127.60
Tc43
(99)
153
126.90
WT4
Re75
183.85
186.2
Po84
At85
(210)
(210)
Ru«
101.07
Os76
190.2
C027
58.93
Rh45
102.91
Ir77
192.2
Ni28
58.71
I
1{r36
83.80
I
131.30
I
Rn86
(222)
Pd46
106.4
XeM
Pt78
195
Ha106
(260)
Nd60
144.24
Pm61 Sm62
(145) 150.35
Eu63
151.96
Gd64
157.25
Tb65
158.92
Dy66
lJ92
Np93 Pu94
238.03
(237)
Am95
(243)
Cm96
(247)
Bk97
(247)
(244)
Fe26
55.85
Er88
Tm69
162.50
Ho67
164.93
167.26
Cf98
(251)
ES89
(254)
Fm100
(254)
168.93
Yb70
173.03
Lu71
174.97
Md101
(256)
NOI02
(254)
Lrl03
-
A and B are the sub-groups.
The figures at the top right corner and below the symbols represent
.
atomic number and atomic weights respectively.
- -
-_.-
(257)
-
I
I
~
Table 3.3. Comparison of the properties of Eka-Aluminium and Eka-SiHcon
as Predicted by Mendeleev with those Observed Experimentally
for Gallium and Germanium
Atomic weight
68
70
72
72.6
Density / (glcm3)
5.9
5.94
5.5
5.36
Melting point / K
Low
29.78
High
1231
Formula of oxide
E20S
E02
Ge°2
Formula of chloride
ECl3
ECl4
GeCl4
GaCIs
could be predicted with a fair accuracy. For example,
both gallium and germanium were not discovered at
the time when Mendeleev proposed his periodic table.
Mendeleevnamed these elements as Eka-Aluminium
and Eka·Silicon respectively. Later on, when these
elements were discovered, Mendeleev's prediction
proved remarkably correct. Some of the properties
.predicted by Mendeleev for these elements and those
found experimentally are given in Table 3.3.
3. Determination of correct atomic weights.
With the help of this table, doubtful atomic weights of
certain elements were corrected.For example,beryllium
was assigned an atomic weight 13.5 on the basis of its
equivalent weight (4.5) and valency (calculated as 3).
Beryllium, with this atomic weight (13.5) should have
been placed between carbon (atomic weights 12) and
nitrogen (atomic weight 14). But there was no place
vacant between C and N and moreover the properties
of beryllium did not justify such a position. Hence,
valency 2 was assigned to beryllium which gave it the
atomic weight equal to 9 and it was placed at its proper
place between 7Liand llB.
DEFECTS IN THE MENDELEEV'S PERIODIC
TABLE
Some of the main defects in the Mendeleev's
periodic table are given below:
1. Position of hydrogen. The position of
hydrogen in the table is not certaiB because it can be
placed in 1st as well as in group VII as it resembles
both with alkali metals of I A group and halogens of
VII A group.
2. Anomalous pairs of elements. Certain
elements of higher atomic weight preceed those with
loweratomicweight. For example,iodine(atomicweight
127) was placed after tellurium (atomic weight 128)
although its atomic weight was lower than that of
tellurium.
3. Lanthanides and Actinides. A group of 14
elements (at. No. 58 to 71) called rare' earths or
lanthanides are placed together in one position i.e., in
group III B of 6th period. Similarly, another group of
elements called actinides do not find their proper places
in this periodic table.
4. Position of isotopes. Isotopes of elements
are placed in the same position in the table though
according to their atomic weights, they should have
been placed in different positions.
5. Separation of similar but grouping of
certain dissimilar elements. Certain chemically
similar elements like copper and mercury are placed in
different groups while some other dissimilar elements
like copper,silver and goldhave been placedin the same
group.
6. No place was assigned for the noble gases in
the Mendeleev's periodic table.
'"
MODERN PERIODIC LAW
. A large number of scientists made attempts to
remove drawbacks of Mendeleev's periodic table. In
1913,the English Physicist Henry Moseleystudied the
X-ray spectra of many elements. He observed that a
plot JV (where v is the frequency ofthe X-rays emitted)
against atomic number (Z) gave a straight line and not
the plot of JV against atomic weight. He proposed that
atomic number is a more fundamental property of an
element than its atomic weight. Therefore, the physical
and chemicalproperties ofthe elements are determined
by their atomic numbers instead of their atomic
weights. This observation led to the development of
modern periodiclaw. The modern periodic law states
that:
The physical and chemical properties of the
elements are the periodic function of their
atomic numbers.
It means that if the elements are arranged in
order of increasing atomic numbers, the elements with
similar properties recur after regular intervals. Many
new forms of periodic table have been proposed in recent
times with modem periodic law as guiding principle,
but the general plan of the table remained the same as
proposed by Mendeleev. The most commonly known
periodic table is the long form of the periodic table.
Before discussing the general plan of the long
form periodic table, let us look into the basic cause of
periodic repetition of properties.
control the properties of the atoms. Thus, if the
arrangement of electrons in the outer most shell
(valence shell) of the atoms is same, their properties
will also be similar. For example, the electronic
configurations of alkali metal as given in Table 3.4,
show the presence of one electron in the s-orbital of
their valence shells.
Similar behaviour of alkali metals is attributed
to the similar valence shell configuration of their atoms.
Similarly, if we examine the electronic configurations
of other elements, we shall find that there is a repetition
of the similar valence shell configuration after certain
regular intervals with the regular increase of atomic
number. Thus, it can be concluded that the periodic
repetition of properties is due to the recurrence of
similar valence shell configurations after certain
regular intervals.
CAUSE OF PERIODICITY
It has been pointed out that if the elements are
arranged in order of increasing atomic numbers, the
elements with similar properties are repeated after
regular intervals.
The periodical repetition of elements with
similar properties after certain regular
intervals when the elements are arranged
in order of increasing atomic number is
called periodicity.
In order to understand the cause of periodicity,
let us first answer the question; why do certain elements
exhibit similar properties?
From the knowledge of atomic structure it may
be recalled that atom has a small positively charged
nucleus with electrons distributed around it. The atomic
nucleus does not undergo any change during the
ordinary chemical reactions. Thus, it may be assumed
that the physical and chemical properties of the
elements must be related to the arrangement
of
electrons in their atoms. Since electrons present in the
inner shells do not take part in chemical combination,
it must be the electrons in the outer most shell which
It may further be noted that this type of
similarity in properties is repeated after the intervals
of2, 8, 18 or 32 in their atomic numbers. These numbers
of 2, 8, 18 and 32 after either of which elements with
similar properties start recurring are frequently called
magic numbers.
_
LONG FORM OF PERIODIC
TABLE
The long form of the periodic table is an improved
form of the periodic table which is based upon modem
periodic law. The long form of periodic table is given on
next page. Let us now, study the structural features of
the periodic table.
STRUCTURAL FEATURES
PERIODIC TABLE
OF LONG FORM OF
Description of Periods
A horizontal row of a periodic table is called a
period. A period consists of a series of elements having
Table 3.4. Electronic Configurations of Alkali Metals
182,281
Li
3
Na
11
1s2, 282 2p6, 381
K
19
182, 282 2p6, 382 3p6, 481
Rb
37
182, 282 2p6, 382 3p6 3d10, 482 4p6, 581
Cs
55
182,282 2p6, 382 3p6 3d10, 482 4p6 4d1O, 582 5p6, 681
Fr
87
182,282 2p6, 382 3p6 3d10, 482 4p6 4d10, 5s2 5p6 5d10, 682 6p6, 781
Long Form of Periodic Table
Representative
I..
Elements
s-Block
~
3.
1
2
3
Li
1
28
4
m
12
Mg
2
K
4s
1
5.
6.
7
Noble
18
~
13
Transition
3
20
Ca
482
21
Se
37
Rb
1
58
38
Sr
2
58
55
Cs
Ba
1
2
56
Elements
2
3ci48
4
22
Ti
3;482
39
40
Zr
Y
4ci5/
4i582
57
La*
6s
6s
5ci6s 2
87
Fr
88
Ra
89
Ac-
'1/
7s
2
6ci7s
5
23
V
2
3i4s
41
Nb
4d45s1
42
Mo
43
Tc
4d 5s
73
Ta
74
W
2
5168
4
2
5/ at78
5
4
4
2
5/ at78
1
2
5i6s
106
Sg
2
3d 4s
27
Co
28
Ni
29
Cu
2
2
45
Rh
8
1
Ag
4d 5/
4cl°
76
Os
77
Ir
78
Pt
2
2
5/6s
9
108
Hs
10
79
Au
5d106/
1
2
2s 2/
4 7
4
2
5/ 6d 7s2 5/ 6178
81
82
TI
Pb
2
5dlO6s2 68 6/ 6/6/
114
112
Uuq
Uub
-
51
Sb
5/5/
52
Te
2
58 5/
53
I
2
58 5/
64
Xe
2
58 5/
83
Bi
84
Po
85
At
86
Rn
80
Hg
111
Rg
110
Ds
109
Mt
49
48
50
In
Sn
Cd
2
4d105/ 58 5pl 5/5/
47
5d 6s
5i6s
30
Zn
acl°4s2
3cl°4s
46
Pd
4d 5s
5/46d57s2 5t6;7s2
5
15
14
16
17
18
P
Si
S
Cl
Ar
2
2
2
3823pl 38 3/ 3/3p3 38 3/ 3823/ 38 3/
36
31
32
33
34
35
Ge
As
Kr
Se
Ga
Br
2
1
2
6
2
2
4s 4p 4824p2 48 4p3 4s 4p4 48 411 4824/
12
1
3/4s
4d 5/
107
Bh
5/46i~~
11
3i4s
44
Ru
75
Re
2
10
9
7
4i5/
5d 6s
105
Db
104
Rr
2
2
3i48
6
10
Ne
13
Al
8
26
Fe
25
Mn
1
aJl482
72
Hr
4tW6s
7
6
24
Cr
He
182
16
17
8
9
0
F
2
2
28 2/ 282p
7
N
B
2
1 2S22p2
282p
2/2/
d-Block
as
15
14
6
C
5
2S2
l
19
4.
Be
11
Na
as
.
I
Elements
p-Block
gases
1.
2.
Representative
~
6s2
6p3
6/
2
68
6/
2
68
116
Uuh
-
2
6s 6[/
-
-
5/46d97s2 5/46dI07s2
Inner Transition Elements
f-Block
"Lanthanides
4f5dO-I682
""Actinides
5f6dO-I7s2
58
Ce
59
Pr
60
Nd
2
4/5cisl
4(5i6s
4(5i6/
90
Th
91
Pa
92
U
5/6;71
1
5(6d 71
1
5(6d 7i
61
Pm
62
Sm
2
4t5d°6s
93
Np
1
5(6d 71
--
63
Eu
2
4t5d°6s
94
Pu
5t6i7s
2
2
4/5i6s
95
Am
~(;rl7s2
64
Cd
1
4/5d 6s
96
Cm
5/sci71
66
Dy
65
Tb
2
2
4(5doSs
4t5i6s
98
cr
97
5(si7/
2
4/°5l6s
Bk
2
5/°6do7s
1
-
2
4/35i6s
5/ 6do7i
101
Md
5/
a
71
Lu
70
Yb
2
100
Fm
2
5/ 6i78
-
69
Tm
2
4/ 5i6s
99
Es
2
-----
S8
Er
67
Ho
2
s!'!{
4
0
4/ 5d 6s
2
4/45ci6/
103
Lr
102
No
y~S/?S2
l
5/ 6d 71
4
----
~
same valence shell. There are seven periods in all
which are numbered as 1,2,3,4,5,6
and 7.
'
There is a close connection between the electronic
configurations of the elements and the long form of the
periodic table. As pointed out earlier in Unit 2 that the
principal quantum number n defines the main energy
level of the electron also called main energy shell. Each
period of the periodic table begins with the filling of
new energy shell. In fact, the number of the period
also represents
the highest principal quantum
number of the elements present in it. The number
of elements in each period is equal to the number of
electrons which can be accommodated in the orbitals
belonging to that electron shell.
The first period corresponds to the filling of
electrons in first energy shell (i.e., n
1). Now this
energy level has only one orbital ti.e., Is) and, therefore,
it can accommodate two electrons. This means that
there can be only two elements in the first period.
The second period starts with the electrons
beginning to enter the second energy shell (n = 2). There
are only four orbitals (one 2s and three 2p orbitals) to
be filled which can accommodate eight electrons. Thus,
second period has eight elements in it.
The third period begins with the electrons
entering the third energy shell (n = 3). It may be recalled
that out of nine orbitals of this energy level (one s, three
p and five d), the five 3d orbitals have higher energy
than 4s orbitals. As such only four orbitals (one 3s and
three 3p) corresponding to n = 3 are filled before the
fourth energy level begins to be formed. Hence, there
are only eight elements in the third period.
The fourth period corresponds to n 4. It starts
with the filling of 4s-orbitals. However, after the 4s but
=
=
Table 3.5. Number
First
Second
Third
Fourth
Fifth
Sixth
Seventh
(incomplete)
*At
n=l
n=2
n=3
n=4
n=5
n=6
n=7
before the 4p orbitals, there are five 3d orbitals also to
be filled. Thus, in all, nine orbitals (one 4s, five 3d and
three 4p) have to be filled and as such there are
eighteen elements in fourth period. It may be noted
that the filling of 3d-orbitals starts from Sc (Z = 21).
The elements from Sc (Z = 21) to Zn (Z = 30) are called
3d-transition series.
The fifth period beginning with 5s orbital
(n = 5) is similar to fourth period. There are nine
orbitals (one 5s, five 4d and three 5p) to be filled and,
therefore, there are eighteen elements in fifth period
as well.
The sixth period starts with the filling of 6s
orbital in. 6). There are sixteen orbitals (one 6s, seven
4f, five 5d and three 6p) in which filling of electrons
takes place before the next energy level starts. As such
there are thirty two elements in sixth period. The
filling up of 4f orbitals begins with cerium (Z = 58)
and ends at lutetium (Z = 71). It constitutes the first
f-transition series which is called Ianthanoid series.
The seventh period begins with 7s-orbital
(n 7). It would also have contained 32 elements
corresponding to the filling of7s, 5f, 6d and 7p orbitals.
But it is still incomplete. The filling up of 5f orbitals
begins with thorium (Z = 90) and ends up at lawrencium
(Z = 103). It constitutes secondf-transition series which
is called actinoid series. It mostly includes man-made
radioactive elements. In order to avoid undue expansion
of the periodic table the 4f and 5f-transition elements
have been placed separately.
The relationship between number of elements in
a period and electron filling of orbitals have been
summed up in Table 3.5.
=
=
of Elements
in Different
Is
2s,2p
38,3p
4s, 3d, 4p
5s, 4d, 5p
Ss, 4{, 5d, 6p
7s, 5(, 6d, 7p
present 7th period is incomplete. It contains about 32 elements.
Periods
2
2+6
2+6
2 + 10 + 6
2+10+6
2 + 14 + 10 + 6
2 + 14 + 10 + 6
2
8
8
18
18
32
32*
It may be noted that periods 2 and 3 contain 8
elements each and are called short periods. There are
18 elements each in 4th and 5th periods and they are
called long periods. Sixth period containing 32
elements is called the longest period.
Description of Groups
I
A vertical column of the periodic table is called a
group. A group consists of a series of elements having
similar configuration of the outer energy shell. For
example, all the group 1 elements have ns) valence shell
electronic configuration. There are eighteen vertical
columns in the long form of the periodic table. According
to the recommendation of the International Union of
Pure and Applied Chemistry (lUPAC), these groups are
numbered from 1 to 18.
It may be noted that the elements belonging to
same group are said to constitute a family. For example,
elements of group 17 constitute halogen family.
Similarly, elements of group 16 constitute chalcogen
family or oxygen family.
Kurchatovium. In order to avoid such controversies, it
was decided by IUPAC that until a new element's
discovery is proved, and its name is officially recognized,
a systematic nomenclature be derived directly from the
atomic number of the element using the Latin words
for their numbers : nil for zero, un for one and so on.
The Latin words for various digits of the atomic
number are written together in the order of which
makes the atomic number and 'ium' is added at the
end. For example, the element with atomic number 104
was named unnilquadium and was assigned symbol
Unq. The latin words for various digits are given in
Table 3.6.
Table 3.6. Latin Words Roots for Various Digits
0
1
2
3
4
5
NOMENCLATURE OF ELEMENTS WITH ATOMIC
NUMBERS> 100
6
7
8
9
The elements beyond uranium (Z = 92) are known
as transuranium
elements. The elements beyond
fermium (Z = 100) are known as transfermium
elements. These elements have atomic numbers 101
onwards.
The elements fermium (Z = 100), mendelevium
(Z 101), nobelium (Z 102) and lawrencium (Z 103)
are named after the names of famous scientists. When
elements with Z > 103 were synthesized, the syntheses
of some of these elements were reported almost
simultaneously (in 1970) by scientists in United States
and in erstwhile Soviet Union and each group proposed
different names. For example, the element with atomic
number 104 was named Rutherfordium by American
scientists
while
Soviet
scientists
named
it
=
=
=
nil
un
hi
tri
quad
pent
hex
sept
oet
enn
n
u
b
t
q
p
h
s
0
e
Similarly, the elements with atomic numbers
105-112 were named as given in Table 3.7.
Thus, when a new element is discovered, it first
gets a temporary name, with symbol consisting of three
letters. Later, a permanent name and symbol are given
by a vote of IUPAC representatives from member
countries. The permanent name might reflect the
country (or state of the country) in which the element
was discovered, or pay tribute to a famous scientist. As
of now, elements with atomic numbers up to 112, 114
and 116 have been discovered. However, their official
IUP AC names are yet to be announced. Elements with
atomic numbers 113, 115, 117 and 118 are not yet
known.
Table 3.7. Names and Symbols in Current Use (or proposed)
for Elements with Atomic Number Above 100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
Unnilunium
Unnilbium
Unniltrium
Unu
Unb
Mendelevium
Md
Nobelium
Unt
Lawrencium
Rutherfordium
No
Lr
Unnilquadium
Unq
Unnilpentium
Unnilhexium
Unp
Unh
Dubnium
Db
Seaborgium
Sg
Unnilseptium
Uns
Bohrium
Unniloctium
Unnilennium
Uno
Hassium
Bh
Hs
Une
Meitnerium
Mt
Ununnilium
Unununium
Uun
Darmstadtium
Roentgenium
Ds
Rg
Ununbium
Uub
Ununtrium
Uut
Uuu
Ununquadium
Uuq
Ununpentium
Ununhexium
Uup
Uuh
Ununseptium
Ununoctium
Uus
119
Ununennium
Uue
120
Unbinilium
Ubn
Uuo
DIVISION OF PERIODIC TABLE
INTO S·, p., d- AND f·BLOCKS
ON THE BASIS OF ELEC·
TRONIC CONFIGURATIONS
The long form of periodic table can be divided
into four main blocks. These are s-, po, d- and f-blocks.
The division of elements into blocks is primarily based
upon their electronic configuration as shown in Fig. 3.3.
1. s-BLOCK ELEMENTS
The elements in which the last electron enters the
s-sub-shell of their outermost energy level are called
s-block elements. This block is situated at extreme
left of the periodic table. It contains elements of groups
1 and 2. Their general configuration is nsl-2, where
n represents the outermost shell. The elements of group
1 are called alkali metals whereas the elements of
group 2 are called aJkaline earth metals.
Rf
1
.----__
---1
I
I
d-Block elements
(transition
elements)
p-Block elements
(representative
elements)
: ~
: gj
: ~
::0
(n_1)d1-10nsO-2
2
ns np
1-6
I 0
:Z
I
I
I
r
f-Block elements
(inner transition elements)
(n_2)f1-14 (n_1)dD-1 ns2
Some of the general characteristics
of s-bloek
elements are:
(i) They are soft metals.
(ii) They have low melting and boiling points.
I
tiii) They have low ionization enthalpies and hence
are highly electropositive.
Civ)They are very reactive and do not occur in
native state in nature.
(u) They show oxidation states of + 1 (in case of
alkali metals) or + 2 (in case of alkaline earth
metals).
(vi) They are good reducing agents.
(vii) The compounds of s-block elements are
predominantly ionic. However, lithium and
beryllium form covalent compounds.
2. p-BLOCK ELEMENTS
The elements in which the last electron enters the
p-sub-shell of their outermost energy level are called
p-bloek elements. The general configuration of their
outermost shell is ns2 npl-8. The only exception is
helium (1s2). Strictly, helium belongs to the s-block but
its positioning in the p-block along with other group 18
elements is justified because it has a completely filled
valence shell (ls2) and as a result, exhibits properties
characteristic of other noble gases. This block is situated
at the extreme right of the periodic table and contains
elements of groups 13, 14, 15, 16, 17 and 18 of the
periodic table. Most of these elements are non-metals,
some are metalloids and a few others are heavy
elements which exhibit metallic character. The nonmetallic character increases as we move from left to
right across a period and metallic character increases
as we go down the group. Some of the general
characteristics
of p-block elements are:
(i) They show variable oxidation states.
tii) They form ionic as well as covalent compounds.
tiii) They have relatively higher values of ionization energy.
(iv) Most of them are non-metals.
(e) Most of them are highly electronegative.
(vi) Most of them form acidic oxides.
3. d·BLOCK ELEMENTS
The elements in which the last electron enters the
d-sub-shell of the penultimate energy level are called
d-block elements.
Their general valence shell
configuration is (n _l)d1-10, ns0-2,where n represents
the outer most energy level. d-Block contains three
complete rows of ten elements in each. The fourth row
is incomplete. The three rows are called first, second
and third transition series. They involve the filling of
3d, 4d and 5d orbitals respectively. The d-block contains
elements of groups 3 to 12 of the periodic table. The
general characteristics
of d-bloek elements are:
(i)
(ii)
They are hard, high melting metals.
They show variable oxidation states.
.(iii) They form coloured complexes.
(iv)
They form ionic as well as covalent compounds.
Cv)Most of them exhibit paramagnetism.
(vi) Most of them possess catalytic properties.
(vii) They form alloys. For example, brass is an
alloy of copper and zinc.
tuiii) They are good conductors of heat and electricity.
On one side of the d-block metals there are
reactive elements of groups 1 and 2, and on the right
hand side there are less reactive elements of groups 13
and 14. Thus, transition elements form a bridge
between reactive metals of groups 1 and, 2, and less
reactive elements of groups 13 and 14. That is why
d-block elements are called transition elements.
It may be mentioned here that Zn, Cd and Hg do
not exhibit most of the characteristic properties of
transition elements because they have fully filled
d-subshell.
4. f·BLOCK ELEMENTS
The elements in which the last electron enters the
f-sub-leuel of the anti-penultimate
(third to the outer
most shell) shell are called f-bloek elements.
Their general configuration is (n - 2)f1-14 (n - 1)
dO-I, ns2, where n represents the outer most shell. They
consist of two series of elements placed at the bottom of
the periodic table. The elements of first series follow
lanthanum (57La) and are called lanthanoids. The
elements of second series follow actinium (80c) and are
called actinoids. Actinoid elements are radioactive.
Many of them have been made only in nanogram
quantities or even less by nuclear reactions. Chemistry
of the actinoids is complicated and is not fully studied.
The general characteristics off·block elements are:
(i) They show variable oxidation states.
(ii) They are high melting metals.
(iii) They have high densities.
(iv) They form complexes, most of which are
coloured.
(u) Most of the elements of actinoid series are
radioactive.
It may be noted that:
1. The element hydrogen, having electronic
configuration IS1, can be placed in s-block along with
alkali metals but is placed separately at the top centre
of the periodic table as shown in Fig. 3.1. This is due to
the reason that hydrogen, like halogens, can gain an
electron to achieve a noble gas configuration and hence
it can behave like halogens also and can be grouped
with them. Due to its resemblance with alkali metals
as well as halogens, hydrogen is placed separately at
the top of the periodic table.
2. The elements of group 18, the last column of
p-block, are known as inert gases or noble gases or
metals, non-metals and metalloids. More than 78% of
the elements are metals. These elements are present
on the left side and the centre of the periodic table.
Metals are the elements which are malleable and
ductile, possess lustre, are good conductors of heat and
electricity and have high densities. Metals usually have
high melting and boiling points, and are generally solids
at room temperature. Mercury is the only metal which
is liquid at room temperature. Gallium (303 K) and
caesium (302 K) also have very low melting points.
aerogens.
3. The elements
and p-blocks are called
only
outermost shell incomplete. They are also known as
representative
ofs
elements. They have their
main group elements.
Non-metals are much less in number than
metals. There are only about 20 non-metals. Non-metals
are located at the top right hand side of the Periodic
Table. Non-metals have low melting and boiling points.
They usually solids or gases at room temperature. Nonmetals are neither malleable nor ductile. They are poor
conductors of heat and electricity. In a period, the nonmetallic character increases as we move from left to
right. In a group, the non-metallic character decreases
and metallic character increases on going down a group.
There is no sharp line dividing metals from non-metals.
A zig-zag line separates metals from non-metals as
shown in Fig. 3.4. The borderline elements such as
silicon, germanium, arsenic, antimony and tellurium
exhibit characteristic properties of metals as well as
non-metals. These elements are called semi-metals or
4. The elements of d-block are called d-transition elements. The name is derived from the fact that
they represent transition (change) in character from
reactive metals (elements of group 1 and 2) on one side
and less active metals of groups 13 and 14 on the other
side.
5. The elements of{-block are called (-transition
or inner transition elements. In this group of
elements outer three principal shells are incomplete.
6. The {-block elements after uranium are called
transuranium elements.
METALS, NON-METALS AND
METALLOIDS
S-, p-,
metalloids.
In addition to the classification of elements into
d- and {-blocks, it is possible to divide them into
Group
Period
2
2
3
3
4
5
6
7
1
1
1
1
11
1
1
reml~
1
1
1
1
1
As is evident from the electronic configurations:
(i) leA receives last electrons in 3p sub-shell, therefore,
its:
period
block
= 3rd
=p
10 + Number of valence electrons
group.
= 10 + 6 = 16.
(ii)
37E
receives last electrons in 5s-orbitals. Hence its:
period
- 5th
block
=
group
= Number
= 1.
8
of valence electrons
(iU) 30G receives last electrons
in 3d-orbitals.
Hence,
its:
block-
d
. Since its valence shell is 4th. Therefore,
period
group
= 4th
= Number of n s-electrons
+ Number of (n - 1) d-electrons
= 2 + 10 = 12.
Example :3.:3. Elements A, B, C, D and E have the
following electronic configurations:
A : 1s2 2s2 2pI
B : 1s22s22p63s23pI
C : 1s22s22p63s23p3
D : 1s22s22p63s23p5
E : 1s22s22p63s23p64s2
which among these will belong to the same group in the
periodic table?
The element with atomic number 119 has
not been discovered. What would be the IUPAC name and
symbol of this element? Also predict the electronic configuration,
group number and period of this element.
Solution.
The IUPAC name for the element with atomic
number 119 would be ununennium
and its symbol would
be Uue.
The electronic configuration of this element would be
[Uuo] Ssl.
This element would belong to group-1 and period-So
Example
:3.1.
Solution. The elements A and B have similar valence shell
electronic configuration (ns2np1). Therefore, they belong to
same group in the periodic table.
M"i1iM,. Predict the position of the element in the
Periodic Tii5le satisfying the electron configuration (n - l)dI
ns2 forn =4.
Solution. The electronic configuration of the element is 3d1
4s2• Thus, the element belongs to d-block, fourth period and
group 3.
Write the electronic configuration of the
Example ::'2.
elements given below and also predict their period, group and
block.
A (At. No. 16), E (At. No. 37) and G (At. No. 30).
Solution.
The electronic configurations
A •• ls2 , 2S2, 2;p6, 3s2 , 3p4
1{Y'.a.
of the elements are:
: ls2, 2s2, 2p6, 3s2, 3p6, 3dlO, 4s2, 4p6, 5s1
30G : 1S2, 2S2, 2p6, 3s2, 3p6, 3d10, 4s2.
37E
SET - 3.1
1. What is modem periodic law? What is periodicity
and what is its cause?
2. What is the total number of groups in the long form
of periodic table?
3 •. An element has atomic number
period, group and block.
34. Deduce its
4. Write the general outer electronic configuration of
s-, p-, d- and {-block elements.
5. Which groups constitute p-block of the periodic
table?
6. An element
has
valence
shell
electronic
configuration as ns2, np3 . To which group does this
element belong?
7. In what group of the periodic table is each of the
following elements found?
(0 1s2, 2s2, 2p6, 3s2, &f1, 482 (ii) [Xe] 4f4 5d4 6s2
(iiO [Ar] 3d10 4s2
(iv) [Ar] 3d10, ~, 4p4.
8. Assign the position of the element having outer
electronic configuration (i) ns2 np" for n = 3
(ii) (n - 1) d2 ns2 for n = 4 and (iii) (n - 2) [1(n - 1)
d1 ns2 for n = 6, in the periodic table.
9. What would be the IUPAC name and symbol for
the element with atomic number 120?
(NCERT Solved Example)
10. How would you justify the presence of 18 elements
in the 5th period of the periodic table?
(NCERT Solved Example)
2. 18.
Atomic volume is defined as the volume
occ.upied by one mole atoms of the element
at its melting point in solid state.
A.tomic volume can be obtained by dividing the
gram atomic mass of the element by its density.
.
I
Gram atomic mass
Atomic vo ume
D'
ensity
=
Variation of Atomic Volume in a Period
Atomic volume decreases along the
reaches a minimum in the middle and then
increasing. Alkali metals have the maximum
volume in a period. A plot of atomic volume
atomic number is shown in Fig. 3.5.
period,
starts
atomic
versus
3. Period-4, group-16, block-t».
5. Groups-13, 14, 15, 16, 17 and 18.
6. Group. 15.
7. (i) 3
(ii) 6
8.
(iii) 12
(i) Period-S, Group-16
(iv) 16
(ii) Period-s, Group-4
(iii) Period-B. Group-S, Lanthanoid.
9. Unbinillium, Ubn.
Variation in a Group
In moving down the group, the atomic volume
goes on increasing gradually.
Most of the properties of the elements such as
atomic volume, atomic size, ionization enthalpy, electron
gain enthalpy and electronegativity are directly related
to the electronic configuration of the atoms. These
properties undergo periodic variation with the change in
the atomic number within a period or a group. These
properties indirectly control the physical properties
such as melting point, boiling point, density, etc. Let
us now proceed to study the variation of some of the
atomic properties in the periodic table.
Example :l.5.
The atomic mass of germanium is 72.6
and its density is 5.47 g cm:". What is the atomic volume of
germanium?
Solution.
Atomic volume
=
Gram atomic maSs
Density
72.6g
= 5.47gcm- 3 13.27 cm3•
=
The atomic size is very important property ofthe
atoms because it is related to many other chemical and
physical properties. In dealing with atomic size, the
atom is assumed to be a sphere and its radius
determines the size. In general, atomic radius
is
defined as the distance from the centre of nucleus
of the atom to the outermost shell of electrons.
However, it is not possible to find precisely the radius
of the atoms because of the following reasons:
1. Atom is too small to be isolated. .
2. Wave mechanical model of the atom does not
allow us to have its well-defined boundary because probability of finding the electrons is
never zero even at a very large distance from
the nucleus.
3. The probability distribution of an atom is affected by the other atoms present in its neighbourhood.
4. Size of an atom also changes from one bonding state to another.
Due to the above mentioned factors, size of atom
varies from one kind of environment to another. It is
for this reason that atomic size is expressed in terms of
different types of radii. Some of these are being
discussed below:
1. COVALENT
Similarly internuclear
distance
between
two
chlorine atoms in chlorine
molecule
is
198
pm.
Therefore, covalent radius of
chlorine is 19812 = 99 pm. In
case the covalent bond is
between two unlike atoms
such as in Hel the covalent
radius
is taken
as the
distance between the nucleus
of the atom and the mean
position of shared pair of electrons.
+74 pm"":
,,
,,
''
,
,i+--+' '
: Covalent
,
radius = 37 pm
2. VAN DER WAAL'S RADIUS
It may be defined as half of the internuclear
distance between two adjacent atoms of the same
element belonging to two nearest neighbouring
molecules of the same substance in solid state.
Covalent radius of the elements is shorter
than its van der Waal radius. The formation of
covalent bond involves overlapping of atomic orbitals.
As a result of this, the internuclear distance between
the covalently
bonded atoms is less than the
internuclear distance between the non-bonded atoms.
This has been shown in Fig. 3.7. Thus,
r{cov.)< r(van der Waal)
Adjacent Molecules in Solid State
, r(COv),
:+--+:
, ,
,,
RADIUS
'Phe approximate
radii of atoms can be
determined by measuring the distance between the
atoms in a covalent molecule by X-ray diffraction and
other spectroscopic techniques. This radius of an atom
is referred to as covalent radius. It may be defined as
,,
,
,,
,,
~
r(~ander W~)
one-half of the distance between the centres of the
nuclei of two similar atoms bonded by a single
covalent bond.
In case of homonuclear
rcovalent=
1
2"
3. METALLIC
bonds,
[Internuclear
two bonded
distance
between
atoms]
For example, the internuclear distance between
two hydrogen atoms in H2 molecule is 74 pm (Fig. 3.6).
Therefore, the covalent radius of hydrogen atom is
37 pm.
OR CRYSTAL RADIUS
The term is applied for metals only. A metallic
lattice is considered to be consisting of closely packed
atoms which are spherical in shape. Metallic radius
may be defined as halfofthe internuclear distance
between two adjacent atoms in the metallic lattice.
It is measured
in angstrom
units. Internuclear
distances are measured by X-ray studies. The metallic
radius of sodium is found to be 186 pm while that of
copper is 128 pm.
The metallic radius of an atom is always
larger than its covalent radius because metallic
bond is weaker than covalent bond. Therefore, two
atoms held by covalent bond are closer to each other.
For example, the metallic radius of sodium is 186 pm,
whereas its covalent radius as determined from its
vapours which exist as Na2 molecules is 154 pm.
Similarly, metallic radius of potassium is 231 pm
whereas covalent radius is 203 pm.
VARIATION OF ATOMIC RADD IN TIlE
PERIODIC TABLE
Atomic radii usually depend upon nuclear charge
and number of main energy levels of an atom. The
periodic trends in atomic radii have been described as
follows:
(a) Variation in a Period
In general, the atomic radii decrease with the
increase in the atomic number in a period. For example,
atomic radii decrease from lithium to fluorine in second
period. The variation of atomic radii from lithium to
fluorine has been shown graphically in Fig. 3.8.
The decrease of atomic radii along a period can
be explained on the basis of nuclear charge. In moving
from left to right across the period, the nuclear charge
increases progressively by one unit but the additional
electron goes to the same principal shell. As a result,
the electron cloud is pulled closer to the nucleus by the
increased effective nuclear charge. This causes the
decrease of atomic size. The atomic radii of elements of
second period are given in Table 3.8.
The values given in Table 3.8, show abrupt
increase in the atomic size of Ne. This is due to the
reason that the values for other elements are covalent
radii whereas that for Ne it is van der Waal's radius
because it does not form covalent bond. Therefore inert
gas radii should be compared with the van der Waal's
radii of other elements and not with the covalent radii.
(b) Variation in a Group
In general, the atomic radii increase from top to
bottom within a group of the periodic table. For example,
atomic radii increase from lithium to cesium among
alkali metals and the similar trend is followed by
halogens from fluorine to iodine as shown graphically
in Fig. 3.9 and in tabular form in Table 3.9.
300
160
E
250
140
~200
120
~ 150
100
j
Q.
en::J
'5
1\1
:>
o
Cl:
0
'E
0
<:
100
50
80
Atomic number (Zl
Table 3.8. Atomic Radii of Elements
of Second Period
+3
+4
+5
+6
+7
+8
+9
+ 10
152
111
88
77
75
74
72
160
Table 3.9. Variation of Atomic Radii of Group-I
and Group-17 elements
':
t1
,..:l~.I4:"~~
,
Radius Of
cation'
,
,,
,,
Li
Na
K
Rb
Cs
152
186
231
244
262
F
Cl
Br
I
At
72
99
114
133
140
In moving down a group, from top to bottom, the
nuclear charge increases with increase in atomic
number but at the same time, there is a progressive
increase in the principal energy levels. The number of
electrons in the outermost shell, however, remains the
same. Since, the effect of additional energy level is more
pronounced than the effect of increased nuclear charge,
therefore, the effective nuclear charge decreases.
Consequently, the distance of the outermost electron
from the nucleus increases on going down the group.
In other words, the atomic size goes on increasing as we
move down a group.
,"
.:,
,,
,
Internuclear
distance
The study of ionic radii leads to two very
important generalisations:
(a) The size of the cation is smaller as compared
with that of the parent atom.
(b) The size of the anion is larger as compared
with that of the parent atom.
(a) The radius of cation is smaller than that
of the parent atom.
In short,
Atomic radii increase down the group.
Atomic radii decrease across the period.
flltJIONIC
11 Electrons 10
11 N. charge +11
186 pm
size
98 pm
IfllJJ_~WJ!IIj1lfllljfllll.I.~JJ~fl~
RADIUS
Ions are formed when the neutral atoms lose or
gain electrons. Apositive ion or cation is formed by the
loss of one or more electrons by the neutral atom
whereas a negative ion or anion is formed by the gain
of one or more electrons by the atom. The term ionic
radii refers to the size of the ions in the ionic crystals.
Ionic radius may be defined as the effective
distance from the nucleus of the ion to the
point up to which it has an influence in
the ionic bond.
The equilibrium distance between the nuclei of
the two adjacent ions can be determined by X-ray
analysis of ionic crystals. Assuming ions to be spheres,
the internuclear distance can be taken as the sum of
the ionic radii of the adjacent ions (Fig. 3.10). Knowing
the ionic radius of one of the ions, the ionic radius of
other can be calculated.
Cation is formed by the loss of one or more
electron from the gaseous atom. Now, in the cation the
nuclear charge remains the same as that in the parent
atom but the number of electrons becomes less. As a
result of this, the nuclear hold on the remaining
electrons increases because of the increase in the
effective nuclear charge per electron. This causes a
decrease in the size.
In many cases the formation of cation also
involves the removal of the valence shell completely.
For example, formation of Na" ion from Na atom
involves the removal ofthird shell completely. This also
results in the decrease in the size of the ion.
Na
1s2, 2s2, 2p6, 3s1
The comparative sizes of certain atoms and their
corresponding cations are given in Table 3.10.
Table 3.10. Atomic and Ionic Radii
of Some Elements
3Li
uNa
19K 12Mg 20Ca
IsM
152
186
231
143
Li+
Na+
K+
78
98
133
160
197
Mg2+ Ca2+ Al3+
78
106
57
(b) The radius of anion is larger than that
of parent atom. Anion is formed by the gain of one or
more electrons by the gaseous atom. In the anion, the
nuclear charge is the same as that in the parent atom
but the number of electrons has increased. Since same
nuclear charge now acts on increased number of
electrons, the effective nuclear charge per electron
decreases in the anion. The electron cloud is held less
tightly by the nucleus. This causes increase in the size.
The relative sizes of chlorine atom and chloride ion have
been shown in Fig. 3.12.
arrangement. For example, each one. of sulphide (S2-),
chloride (CI-) and potassium (K+) ion has eighteen
electrons but they have different nuclear charge, + 16,
+ 17 and + 19 respectively.
Variation of size among iso-electronic ions.
Within the series of iso-electronic ions, as the nuclear
charge increases, the attractive
force between the
electrons and nucleus also increases. This results in
the decrease of ionic radius. In other words, size of the
iso-electronic ions decreases with the increase in the
magnitude of nuclear charge. For example, N3-, 02-,
F-, Na+, Mg2+, Al3+ are iso-electronic and have 10
electrons each. The sizes of these ions are in the order:
Mg 2+< Na" < F - < 02- < N3Mg2+having the highest nuclear charge (12 units)
has the smallest size whereas N3 - ion having the
smallest nuclear charge (7 units) has the largest size.
Variation of size among these ions has been shown in
Table 3.12.
Table 3.12. Variation of Size Among
lso-electronic Ions
17 Electrons 18
+ 17 N. charge + 17
99 pm
size 181 pm
N3-
02-
F-
Na+
Mg2+ Al3+
+7
+8
+9
+11
+ 12
+ 13
10
10
10
10
10
10
171
140
136
98
78
57
The comparative sizes of some atoms and their
corresponding anions are given in Table 3.11.
Table 3.11. Atomic and Ionic Radii
of Some Elements
17CI
35Br
531
sO
7N
99
114
133
74
75
Cl-
Br-
1-
02-
N3-
181
195
220
140
171
ISO-ELECTRONIC
different
o,er,
IONS
elements
s«,
Solution. In Na", F-, 02-, Mg2+,Al3+,the nuclear charges
are 11, 9, 8, 12 and 13 respectively. Among isoelectronic
species, greater the nuclear charge smaller is the size.
Therefore, the sizes of the above ionic species are in the order:
Al3+< Mg2+< Na+ < F- < ()2-.
Select from each group the species which
Example :1.7.
has the smallest radius stating appropriate reason.
(i)
()2(ii) K+, Sr2+ ,Ar
The ions having Bame number of electrons
but different magnitude of nuclear charge are
called iso-eleetronie ions. In fact, these are the ions
of
The following species are isoelectronic
Example a.f).
with the noble gas neon. Arrange them in order of increasing
size:
F-, ()2-, Mg2+, AJ3+.
having
same
electronic
(iii) Si, P, Cl
Solution. (i) The species 0 has the smallest radius because
the radius of anion is always larger than the radius of the
atom from which it is formed. 0- and 02- are anions of oxygen.
(ii) K+has the smallest radius. In K+and Ar the outermost shell is third whereas in Sr2+it is fourth. Out of K+and
Ar, K+has smaller size because it has greater nuclear charge.
(iii) Cl has the smallest radius. Si, P and Cl belong to
same period. In a period atomic radius decreases with increase
in atomic number due to increase in effective nuclear charge.
Example 3.8.
Name a species that will be isoelectronic
with each of the following atoms or ions:
(i) Ne
Solution.
(ii) Cl-
(iii) Ca2+
(iv) Rb
Sodium ion, Na+
(ii) Potassium ion, K+
(UO Sulphide ion, S2(iv) Sr+
(i)
Example 3.9.
and why?
Out of Na: and Na which has smaller size
Solution. Na" has smaller size than Na. Na" is formed
by removal of one electron from Na. Therefore, Na" has one
electron less than Na. However, Na and Nat have same
nuclear charge. Therefore, electrons in Na" are more tightly
held than in Na. Moreover, removal of one electron from Na
leads to complete removal of the third shell so that in Na+,
the outermost shell is second. Hence, Nat has smaller size
than Na.
Example 3.10. Give examples of three cations and three
anions which are isoelectronic with argon.
Solution.
Cations
Anions
K+,
: CI-,
Ca2+,
S2-,
Sc3+
p3-.
SET- 3.2
1. Out of metallic radius and covalent radius of an
2.
3.
4.
5.
6.
7.
element, which is larger and why?
Why van der Waal's radius of an element is always
larger than the covalent radius?
Out of the following ions which has smallest ionic
size?
u-, Na+, K+
Arrange the following ions in the increasing order
of their sizes.
CI-, p3--,S2-, F Arrange the following sets of ions in the decreasing
order of their sizes
(i) A}3-, Mg2+,Na+, 02-, F (ii) Na+, Mg2+,K+.
Out of I and 1+which has larger size and why?
Which ofthe following species will have the largest
and the smallest size?
Mg, Mg2+,Al, Al3+. (NCERT Solved Example)
It is a well known fact that the electrons in an
atom are attracted by the positively charged nucleus.
In order to remove electron from an atom, energy has
to be' supplied to it to overcome the attractive force.
This energy is referred to as ionization enthalpy, ~H.
Thus,
Ionization enthalpy of an element may be
defined as the amount of energy required to
remove the most loosely bound electron from
its isolated gaseous atom in the ground
state.
A (g) ----+ A+ (g) + eAB = a.H
Ionization enthalpy is a very important property
which gives an idea about the tendency of an atom to
form a gaseous positive ion.
Units. Ionization enthalpy is expressed either
in terms of electron volts per atom (eV/atom) or kilo
Joules per mole of atoms (kJ mol=)
1 e V per atom = 96.49 kJ mot".
SUCCESSIVE
IONIZATION ENTHALPIES
Once the first electron has been removed from
the gaseous atom, it is possible to remove second and
successive electrons from positive ions one after the
other. For example,
A (g)
----+ A+(g) + eA+(g)
----+ A2+(g) + eUnipositive ion
A2+ (g)
----+
A3+(g)
+ e-
Dipositive ion
The amounts of energies required to remove most
loosely bound electron from unipositiue, dipositive,
tripositive ..... ions of the element in gaseous state are
called second, third, fourth ...... ionization enthalpies
respectively.
The second (L\i~)'
ionization
enthalpies
third (L\iHa), fourth (L\iH4)' etc.
are collectively
known as
successive ionization enthaIpies. It may be noted
that:
~Ha>~~>~Hl
The variation in the values of successive
ionization enthaIpies can be explained, in general,
as follows:
After the removal of first electron, the atom
changes into monopositive ion. In the ion, the number
of electrons decreases but the nuclear charge remains
the same as in the parent atom. As a result of this,
effective nuclear charge per electron increases. The
remaining electrons are, therefore, held more tightly
by the nucleus. Thus, more energy is required to remove
the second electron. Hence, the value of second
ionization enthalpy (L1iH2) is greater than the
first (~Hl).
Similarly, the removal of second electron results
in the formation of divalent positive ion and the
attraction between the nucleus and the remaining
electrons increases further. This accounts for the
progressive increase in the values of successive
ionization enthalpies.
If the removal of first, second or third electron
also results in the removal ofthe valence shell, the next
electron (which belongs to lower energy shell) will
require very large amount of energy for its removal.
For example, in sodium, the removal of first electron
leads to the removal ofthird shell completely.Therefore,
for sodium, L1iH2» L1iHr
Na(g) - e- -----+ Na+(g); L1iHl= 495 kJ mol=!
(2, 8, 1)
(2, 8)
Na+(g)- e-
----7
(2,8)
Na2+(g); L1iH2
= 4581
kJ mol=!
(2,7)
The first, second and third ionization enthalpies
of some elements are given in Table 3.13.
Table 3.13. ~Hl' ~~ and ~1Is Values
of Some Elements
Li
520
7297
11810
Be
899.5
1757
14850
B
800.6
2427
3650
C
1086.4
2352
4619
N
1402
2860
4577
0
1314
3389
5301
FACTORS ON WIllCD IONIZATION ENTHALPY
DEPENDS
The ionization enthalpy depends upon the
following factors:
1. Size of the atom;
2. Magnitude of nuclear charge;
3. Screening effect of the inner electron;
4. Penetration effect of the electrons; and
5. Electronic configuration.
1. Size of the Atom. The ionization enthalpy
depends upon the distance between the electron and
the nucleus, i.e., size of the atom. The attractive force
between the electron and the nucleus is inversely
proportional to the distance between them. Therefore,
as the (3izeof the atom increases, the outermost electrons
are less tightly held by the nucleus. As a result, it
becomes easier to remove the electron. Therefore,
ionization e,!-thalpy decreases with increase in
atomic size.
2. Magnitude of Nuclear Charge. The
attractive force between the nucleus and the electrons
increases with the increase in nuclear charge provided
their main energy shell remains the same. This is
because, the force of attraction is directly proportional
to the product of charges on the nucleus and that on
the electron. Therefore, with the increase in nuclear
charge, it becomes more difficult to remove an electron
and, therefore, ionization enthalpy increases.
3. Screening Effect of the Inner Electrons.
In multi-electron atoms, the electrons present in the
outermost shell do not experience the completenuclear
charge because of repulsive interaction of the
intervening electrons. Thus, the outermost electrons are
shielded or screened from the nucleus by the inner
electrons. This is known as screening effect.
If the number of electrons in the inner shells is
large, the screening effectwill be large. As a result, the
attractive interactions between the nucleus and
outermost electrons will be less. Consequently,
ionization enthalpy will decrease. Thus, if other factors
do not change, an increase in the number of inner
electrons tends to decrease the ionization enthalpy.
4. Penetration Effect of the Electrons. It is a
well known fact that in case of multi-electron atoms,
the electrons in the s-orbital have the maximum
probability of being found near the nucleus and this
probability goes on decreasing in case of p, d and f
orbitals. In other words, s-electrons are more
penetrating towards the nucleus than p-electrons. The
penetration power decreases in a given shell (same
value ofn) in the order: s > p > d > f.
Now,ifthe penetration ofthe electron is more, it
experiences less shielding effect by the inner electrons
and will be held firmly. Consequently, ionization
enthalpy will be high. This means that ionization
enthalpy increases with increase in penetration power
of the electrons. Thus, for the same shell, it is easier to
remove the p-electrons in comparison to the s-electrons.
5. Electronic Configuration. It has been
noticed that certain electronic configurations are more
stable than the other. The atom having a more stable
configuration has less tendency to lose the electron and
consequently, has high value of ionization enthalpy. For
example:
(i) The noble gases have stable configuration
(ns2np6). They have highest
enthalpies within their respective
(ii) The elements like N (ls2, 2s2, 2p,/,
and P (ls2, 2s2, 2p6, ar, 3p,/, 3p/,
ionization
periods.
2p/, 2p/)
3p/) have
configurations in which orbitals belonging to
same sub-shell are exactly half-filled. Such
configurations
are
quite
stable
and
consequently, require more energy for the
removal of electron. Hence, their ionization
enthalpies are relatively high.
(iii) The elements like Be (ls2, 2S2) and Mg (1s2,
2s2, 2p6, 3s2) have all the electrons paired.
Such configurations being stable also result
in the higher values of ionization enthalpies.
VARIATION OF IONIZATION
PERIODIC TABLE
ENTHALPY
IN THE
Let us now, study the variation of ionization
enthalpies across the period and along the group of
representative elements on the basis of above factors.
Variation Across the Period
In general, the value of ionization enthalpy
increases with the increase in atomic number across the
period. This can be attributed to the fact that moving
across the period from left to right,
(i)
nuclear charge increases regularly;
(ii) addition of electrons occurs in the same energy
level;
Ciii) atomic size decreases.
Thus, due to the gradual increase in nuclear
charge and simultaneous decrease in atomic size, the
attractive force between the nucleus and the electron
cloud increases. Consequently, the electrons are more
and more tightly bound to the nucleus. This results in
the gradual increase in ionization enthalpy across the
period. The ionization enthalpies of the elements of 2nd
periods are given in Fig. 3.13.
2500 ....------------,
2000
(5
E
~
W
1500
Atomic number, Z
On carefully exammmg the values given in
Fig. 3.13, we find some exceptions within the period.
These can be explained on the basis of other factors
governing ionization enthalpy. For example, let us
consider the ionization enthalpies for the elements of
2nd period.
(i) There is an increase in ionization enthalpy
from Li to Be. This is due to increased nuclear charge
and smaller atomic size.
(ii) There is a decrease in the value of ionization
enthalpy from Be to B inspite of increased nuclear
charge. The effect of increased nuclear charge is
cancelled by (a) greater penetration of 2s electron as
compared to 2p electron; (b) better shielding of 2p
electrons by the inner electrons; and (c) relatively stable
configuration of Be due to completely filled orbitals.
Thus, 2p electron of boron is relatively less tightly held
by its nucleus in comparison to 2s electrons of Be.
tiii) There is a regular increase in ionization
enthalpy from B to C to N. It is again due to gradual
increase of nuclear charge and decrease of atomic size.
(iv) There is slight decrease in ionization enthalpy
from N to O. It is attributed to the relatively stable configuration of the nitrogen due to a half filled 2p-orbital.
In the nitrogen atom, three 2p-electrons are present in
different atomic orbitals (Hund's rule) whereas in the
Table 3.14.Ionization Enthalpies of the Elements of 3rd Period
1
2
13
14
15
16
17
18
495
737
577
785
1062
999
1254
1579
oxygen atom, two of the four 2p-electrons are present in
the same 2p-orbital resulting in an increased electronelectron repulsion. Therefore, it is easier to remove one
of the 2p-electron from oxygen than it is, to remove one
of the 2p-electrons from nitrogen.
(u) There is an expected increase in ionization
enthalpy from 0 to F to Ne.
It may be noted that among the representative
elements (elements of s and p-block) similar periodic
trends are observed as among the elements of 2nd
period. For reference the ionization enthalpies of the
elements of 3rd period are given in Table 3.14.
Variation in a Group
The values of ionization
enthalpies
of elements
decrease regularly with the increase in atomic number
within a group. The values of ionization enthalpies of
the elements of group 1 have been represented
graphically in Fig. 3.14.
550 ....---------------,
The decrease in the value of ionization enthalpy
within the group can be explained on the basis of net
effect of the following factors:
As we move down the group there is:
(i) A gradual increase in the atomic size due to
progressive addition of new energy shells;
(ii) Increase in the shielding effect on the outermost electron due to increase in the number of inner
electrons.
The nuclear charge also increases but the effect
of increased nuclear charge is cancelled by the increase
in atomic size and the shielding effect. Consequently,
the nuclear hold on the valence electron decreases
gradually and ionization enthalpy also decreases. The
variation of first ionization enthalpy as a function of
atomic number for elements with atomic number upto
60 have been shown in Fig. 3.15. The maxima in the
curve represents noble gases which means that the
ionization enthalpies of noble gases are highest within
their periods. The minima in the curve represent alkali
metals which implies that ionization enthalpies of alkali
metals are the lowest within their respective periods.
2500
He
Ne
500
--
2000
'i
"0
E
~ 450
@'
__ 1500
'i
(5
E
Xe
~ 1000
400
500
350
0
30
40
Atomic number, Z
50
Na
K
10
20
Rb
60
0
30
40
Atomic number, Z
50
60
Example ;~.l:t
Out of Nar and Ne which has higher
ionization enthalpy ? Explain why.
Example :U 1.
Consider
configurations given below:
the ground
state electronic
X: (ls2, 2s2, 2p5); Y: (ls2, 2s2, 2p4); Z: (1s2, 2S1); Q: (ls2, 2s2,
2p6, 3sI); R: (ls2, 2s2, 2p6). Pick up the correct answers.
(i) Which of the above configuration is associated with
highest and which is associated with lowest ionization
enthalpy?
(ii) Arrange these configurations in order of increasing
ionization enthalpies.
Solution.
Elements X, Y, Z and R have 2nd shell as the
valence shell. Thus, they belong to same period. Arranging
their configurations in order of increasing atomic number,
we have
Z: 1s2, 2s1; Y: 1s2, 2s2, 2p4; X: 1s2, 2s2, 2p5;
R: 1s2, 2s2, 2p6.
Among these elements, the nuclear charge increases
from (Z ~ Y ~ X ~ R). The ionization enthalpies would also
increase in the same order. Thus, among these R should have
highest value of ionization enthalpy and Z should have the
least.
Now, Z (1s2, 2s1) and Q (182, 2s2, 2p6, 3s1) belong to
same group and Q lies below Z.
Therefore, ionization enthalpy of Q is less than the
ionization enthalpy of Z.
The above arguments prove that:
(i) Element with highest ionization enthalpy is
R (182, 282, 2p8)
Element with lowest ionization enthalpy is
Q (lsi, 2s2, 2p8, 381)
(ii) Arrangement
enthalpy is:
in order of increasing
ionization
Q<Z<Y<X<R.
Examp\p :t l~.
From each set, choose the atom which has
the largest ionization enthalpy and explain your answer
(i) F, 0, N.
(ii) Mg, P, Ar.
(iii) B, Al, Ga.
Solution. (i) F has the highest ionization enthalpy among F,
o and N because it has smallest size and highest nuclear
charge. In general, ionization enthalpy increases as we go
from left to right in a period.
(ii) Ar (a noble gas) has the highest ionization enthalpy
among the elements Mg, P and Ar because it has stable
electronic configuration and maximum nuclear charge.
(iii) B, Al and Ga belong to group-13. Among these
elements B has the largest ionization enthalpy because on
moving down a group, from top to bottom, ionization enthalpy
decreases. B is the first element of group 13.
Solution. Na" has higher ionization enthalpy than Ne. Na+
and Ne are isoelectronic species. However, the nuclear charge
in Na" is more than in Ne. Hence, the electrons are more
tightly held in Nat and it has higher ionization enthalpy.
",iiiiliW'.
The
520 kJ moi+. Calculate
ionization enthalpy of lithium is
the amount of energy required to
convert 70 mg of lithium atoms in gaseous state into Li: ions.
Solution.
Mass of lithium
= 70 mg = 70 x 10-3 g
=7xlO-2g
Moles of lithium = 7 x ~0-2 = 1 x 10-2 mol.
Energy required to convert 1 x 10-2 mol atoms of
lithium in gaseous state into Li" ions
= 520 x 1 x 10-2 = 5.2 IW.
Exnmp!o :1.I:>. The ionization potential of hydrogen is
13.6 ev' Calculate the energy required to produce H+ ions from
0.5 g of hydrogen atoms.
Solution.
H(g) + 13.6 eV ~
H+(g)
1 eV = 96.49 kJ/mole
13.6 eV = 13.6 x 96.49 kJ/mole
= 1312.3 kJ/mole
. . Energy required to convert 1 g hydrogen atoms
into ions = 1312.3 kJ
Energy required to convert 0.5 g hydrogen atoms into
ions = 1312.3 x 0.5 = 856.2 lW.
Exam pip :t 1H.
How do you explain that 31Ga has slightly
higher ionization enthalpy than laM, although it occupies lower
position in the group ?
1aAI: 1s2, 2s2, 2p6, 3s2, 3p1
1aGa: 1s2, 2s2, 2p6, 3s2, 3p6, adlO, 4s2, 4p1
In Ga, the 10 electrons present in ad-sub-shell do not
shield the outer electrons from the nucleus effectively. As a
result, effective nuclear charge in Ga increases. This explains
why ionization enthalpy of Ga is slightly more than that of
1aAl.
Solution.
SI:'], - 3.3
1. Which element has the highest ionization enthalpy?
2. Among the alkali metals which element has the
highest ionization enthalpy?
3. Arrange each of the following sets of elements
in the increasing
order of their ionization
enthalpies:
(i) 0, N, S
(ii) C, N,
(iii) Li, Be, Na
(iv) Ne, He, Ar
(v) Li, K, Ca, S, Kr.
°
4. Out ofNa (Z = 11) and Mg (Z = 12), which has higher
second ionization enthalpy and why?
5. Explain why
(i)
per mole of atoms. For example, electron affinity of
chlorine is - 348 kJ mol:", i.e.,
Cl(g) + e" ------t CI-
(g)
Ae~
=-
348 kJ.
B has lower ionization enthalpy than Be
(ii) 0 has lower ionization enthalpy than N and F?
6. The first ionization enthalpy (~iH) values of the third
period elements, Na, Mg and Si are respectively 496,
737 and 786 kJ mol='. Predict whether the first ~iH
value for AI will be more close to 575 or 760 kJ
mor? ? Justify your answer.
(NCERT Solved Example)
",,
ELECTRON GAIN ENTHALPY
We have already learnt that ionization enthalpy
is a measure of the tendency of the atom to form cation.
In the similar way, the tendency of a gaseous atom to
form anion is expressed in terms of electron gain
enthalpy.
Electron gain enthalpy of an element may
be defined as the enthalpy change taking
place when an isolated gaseous atom of the
element accepts an electron to form a
monovalent gaseous anion.
The process can be expressed as :
X(g) + e- ------t X-(g) MI = AegH
Depending on the element, the process of adding
an electron can be either exothermic or endothermic.
The magnitude of electron gain enthalpy
measures the tightness with which the atom can hold
the additional electron. The large negative value of
electron gain enthalpy reflects the greater
tendency of an atom to accept the electron. For
example, the elements of group-17 (the halogens) have
very high negative electron gain enthalpies because
they can attain stable noble gas electronic configuration
by gaining an electron. Thus, halogens have great
tendency to accept an electron. On the other hand, the
elements of group-18 (the noble gases) have large
positive electron gain enthalpies because the additional
electron goes to the next principal shell resulting in a
very unstable electronic configuration.
Units. The values of electron gain enthalpy are
expressed either in electron volt per atom or kilo joules
FACTORS AFFECTING ELECTRON GAIN
ENTHALPY
Some of the important factors which affect
electron gain enthalpy are discussed below:
1. Nuclear Charge. Greater the magnitude of
nuclear charge greater will be the attraction for the
incoming electron and as a result, larger will be the
negative value of electron gain enthalpy.
2. Atomic Size. Larger the size of an atom is,
more will be the distance between the nucleus and the
additional electron and smaller will be the negative
value of electron gain enthalpy.
3. Electronic
Configuration.
Stable the
electronic configuration of an atom is, lesser will be its
tendency to accept the electron and larger will be the
positive value ofits electron gain enthalpy. For example,
the elements having completely filled sub-levels of the
valence shell have relatively stable configurations and
consequently, possess large positive values of electron
gain enthalpy.
Table 3.1S. Electron
1
2
3
Gain Enthalpies
H
He
+ 48
Li
-60
Be
+ 240
0
-141
Na
Mg
+ 230
F
Ne
- 328
+ 116
Cl
-349
Ar
-200
S
+ 96
.K
-48
Se
Br
Kr
-195
-325
+ 96
5
Rb
-47
Te
-190
I
-295
Xe
+77
6
Cs
-46
-174
-270
4
VARIATION OF ELECTRON GAIN ENTHALPY IN
THE PERIODIC TABLE
Since experimental determination of electron
gain enthalpy is not as easy as that of ionization
enthalpy, the sufficient data regarding electron gain
enthalpies is not available. Consequently, the varying
trends of electron gain enthalpies are not well defined.
The electron gain enthalpies of some elements are given
in Table 3.15.
in a Period
On moving across the period, the atomic size
decreases and nuclear charge increases. Both these
factors result into greater attraction for the incoming
electron, therefore, electron gain enthalpies tend to
become more negative as we go from left to right across
a period. However, some irregularities are observed
among the elements of group 2, group 15 and group 18.
These can be explained on the basis of electronic
configuration. Group 2 elements have filled ns subshell,
group 15 elements have a half filled np subshell and
group 18 elements have all sub shells filled. These
electronic configurations are relatively stable and hence
these elements have positive or very low negative
electron gain enthalpies.
Variation
(kJ mol-I)
-73
- 53
Variation
of Some Elements
Down a Group
On moving down a group, the atomic size as well
as nuclear charge increases. But the effect of increase
Rn
.
+ 68
in atomic size is much more pronounced than that of
nuclear charge and thus, the additional electron feels
less attraction. Consequently, electron gain enthalpy
becomes less negative on going down the group. Let us
now examine the values of electron gain enthalpies of
halogens as shown in Table 3.15.
It may be noted that the electron gain enthalpy
becomes less negative as we go from chlorine to bromine
to iodine. However, as we move from fluorine to chlorine
the electron gain affinity becomes more negative
whereas reverse was expected. This is because when
an electron is added to fluorine atom, it goes to the
relatively compact second energy level. As a result, it
experiences significant repulsion from the other
electrons present in this shell. On the other hand, in
•
chlorine atom, the added electron goes to the third
energy shell which is relatively larger. Hence, it
experiences less electron-electron repulsion. Therefore,
electron gain enthalpy of fluorine is less negative as
compared with chlorine. The unexpected trend is
observed in case of many other elements of third period
as their electron gain enthalpies are more negative than
those of the elements of second period.
SUCCESSIVE ELECTRON GAIN ENTHALPIES
When the first electron is added to the gaseous
atom, it forms a uninegative ion and the enthalpy
change during the process is called first electron gain
enthalpy. Now, if an electron is added to the uninegative
ion, it experiences a repulsive force from the anion. A1?,
a result, the energy has to be supplied to overcome the
repulsive force. Thus, in order to add the second
electron, the energy is required rather than released.
Therefore, the value of second electron
gain
enthalpy is positive. Similarly, addition of third,
fourth electrons, etc., also requires energy. Hence, the
values ofsuccessive electron gain enthalpies are positive.
For example, let us study the addition of electrons to
oxygen atom
O(g) + e" -----7
O-(g)
8H
= (AegH)l = -
141 kJ
(i) Which element has highest negative .1egH?
(ii) Which element has lowest negative .1eJI?
6. In each of the following sets, arrange the elements
in the increasing order of their negative electron
gain enthalpies:
°
. (i) C, N,
(ii) 0, N, S
(iii) S, Cl, Ar
(iv) F, CI,_Br.
7. Which of the following will have the most negative
electron gain enthalpy
and which the least
negative?
P, S, Cl, F.
Explain your answer. (NCERT Solved Example)
02-(g)
8H
= (A~)2 = + 780 kJ
altj ELECTRONEGATIVITY
How much energy, in kJ, is produced when
7.1 g of chlorine, in the form of Cl atoms, is converted to Clions in gaseous state ? Electron gain enthalpy of chlorine is
- 3.7 eV.
Example
Solution.
:U7.
71 mol = 0.2 mol
35.5
1 eV = 96.49 kJ mol!
7.1 g of chlorine
=
Energy produced during conversion of 1mole of CI(g)
to CI- = 3.7 x 96.49 kJ = 357.01 kJ
Energy produced during conversion of 0.2 mole of Cl(g)
to Cl- = 357.01 x 0.2 = 71.4 kJ.
S/~'T - :1.1
1. Why halogens have highest negative electron gain
enthalpies in their respective periods ?
a
Why noble gases have largest positive electron gain
enthalpies in their respective periods?
3. Comment on the statement
that "all elements
having high ionization enthalpies also have high
negative electron gain enthalpies."
4. Out of oxygen and sulphur, which has greater
negative electron gain enthalpy and why?
5. Consider the electronic configurations
elements X, Y and Z and answer.
X: 1s2, 2s2, 2p/, 2p/
Y : 182, 282 ''Yx'
2n
2
2n
1 2n 1
'Yy'
'Yz
Z : 182,282, 2p/, 2p/, 2p/
of the
Electronegativity may be defined as the
tendency of an atom in a molecule to attract
towards itself the shared pair of electrons.
It may be mentioned here that unlike ionization
enthalpy and electron gain enthalpy, electronegativity
is not a measurable quantity. However, in order to
compare the electronegativity values of different
elements
a number
of numerical
scales of
electronegativity of elements have been proposed. The
most common and widely used scale of electronegativity
is the Pauling scale.
Pauling's scale of electronegativity is based on
excess bond energies. It is observed that the energy of
a heteropolar bond A - B is generally greater than the
average arithmetic mean value of the homopolar bond
energies for A - A and B - B bonds.
DA_B
1
= "2 IDA-A
+ DB_B] +
AAB
The excess bond energy AAB is related to the
electronegativities of elements A and B (XA and XB) as
XA - XB = 2.208 ~AAB
By assigning arbitrary value of electronegativity
to one of the elements, the electronegativity ofthe other
can be calculated. Electronegativity of fluorine, the most
electronegative element, is arbitrarily taken as 4.0 in
this scale.
The main factors on which the electronegativity depends are effective nuclear charge
and atomic radius.
...,
Table 3.16. Electronegativity
Values of Elements
Li
Be
B
C
N
o
F
1.0
1.5
2.0
2.5
3.0
3.5
4.0
Na
Mg
AI
Si
p
S
Cl
0.9
1.2
1.5
1.8
2.1
2.5
3.0
• Greater the effective nuclear charge greater is
the electronegativity.
• Smaller the atomic radius greater is the
electronegativity.
The electronegativity of any given element is not
constant but varies depending on the element to which
it is bound.
In a period electronegativity increases in
moving from left to right. This is due to the reason that
nuclear charge increases whereas atomic radius
decreases as we move from left to right in a period.
Halogens have the highest value of electronegativity
in their respective periods. The eleetronegativity values
(on Pauling scale) of elements of second and third
periods are listed in Table 3.16.
In a group electronegativity decreases on
moving down the group. This is due to the effect of
increased atomic radius. For example, among halogens
fluorine has the highest electronegativity. In fact,
fluorine has the highest value of electronegativity among
all the elements.
The electronegativity values (on Pauling scale) of
group 1 and group 17 elements are given in Table 3.17.
Table 3.17. Electronegativity
Values of Group-1
and Group-17 Elements
Li
Na
K
Rb
Cs
1.0
0.9
0.8
0.8
0.7
F
Cl
Br
I
At
of Second and Third Periods
4.0
3.5
2.8
2.5
2.2
Relationship between Electronegativity
and Nonmetallic (or Metallic) Character of an Element
Non-metallic elements have strong tendency to
gain electrons. Therefore, electronegativity is directly
related to the non-metallic character of elements. We
can also say that the electronegativity is inversely
related to the metallic character of elements. Thus, the
increase in electronegativities
across a period is
accompanied by an increase in non-metallic character
(or decrease in metallic character) of elements.
Similarly, the decrease in electronegativity down a
group is accompanied by a decrease in non-metallic
character (or increase in metallic character) of elements.
_
ELECTROPOSITIVITY OR
METALLIC CHARACTER
Tendency of atoms of an element to lose
electrons and form positive ion is known as
electropositivity.
A more electropositive element has more metallic
character.
In a period, from left to right electropositivity
decreases. This is due to increase in ionization enthalpy
along a period which makes loss of electrons difficult.
For example, in the second period, lithium and
beryllium are metals. Boron is a semimetal whereas
carbon, nitrogen, oxygen and fluorine are non-metals.
In a group, from top to bottom electropositivity
increases. This is due to decrease in ionization enthalpy
on going down a group.
If we consider the elements of group 14, we find
clear increase in metallic character (electropositive
character) on moving down the group. In this group
the first two elements (Carbon and Silicon) are nonmetals, the third element (Germanium) is a metalloid
whereas the next two elements (Tin and Lead) are
metals.
The periodic trends of various physical properties
are given in Fig. 3.16.
---~
'".•._/-';:..,
~
....•.~ _=... ~~
-".
+
~
r
e--
r-
I- I-I- I--
SET - 3..)
Fxample s.rs. Arrange the following
irtcreasing order of metallic character:
.
B,At, Mg, K
elements
in the
Solution. Metallic character increases on moving down the
group and decreases on going across a period from left to
~ght. Hence, the order of increasing metallic character is
';,
B < AI < Mg< K.
Example 3.19. Arrange the following
i1l{;reasing order of non metallic character:
i" ,
B, C, Si, N, F
elements
in the
~lution.
Non-metallic character increases across a period
frOlll left to right and decreases on moving down the group
frt)m top to bottom. Hence, the order of increasing non~tallic character is
Si < B < C < N < F.
1. What is the trend of metallic character on going
down from top to bottom in a group?
2. (i) The element with highest electronegativity
belongs to group
and period
.
(ii) In the third period, the element with highest
electronegativity is
.
3. Define electronegativity. How does it vary along a
period and along a group?
4. What is the basic difference between the terms
electron gain enthalpy and electronegativity?
5. How would you react to the statement that the
electronegativity of N on Pauling scale is 3.0 in all
the nitrogen compounds?
Example 3.20.
(i)
(ii)
(iii)
(iv)
Solution.
enthalpy.
(ii)
enthalpy.
(iii)
Among the elements B, Al, C and Si
Which has the highest first ionization enthalpy?
Which has the most negative electron gain enthalpy?
Which has the largest atomic radius?
Which has the most metallic character?
(i) Carbon (C) has the highest first ionization
Carbon (C) has the most negative electron gain
Aluminium (Al) has the largest atomic radius.
(iv) Aluminium (Al) has the most metallic character.
The chemical properties of elements depend on
the electronic configuration and the various atomic
properties such as atomic size, ionization enthalpy,
electron gain enthalpy. In the following sections we will
study the variation of some of the chemical properties
in the periodic table.
The valence of an element may be defined
aB the combining capacity of element.
Valence is generally expressed in terms of the
number of hydrogen atoms' or the number of chlorine
atoms or double the number of oxygen atoms that
combine with an atom of the element. The chemical
properties of the elements usually depend upon the
number of electrons present in the outermost shell of
their atoms. The electrons present in the outermost shell
are called valence electrons and these electrons
determine the valence of the atom.
In case of representative elements, the valence is
generally equal to either the number of valence
electrons or eight minus the number of valence
electrons. However, the transition elements, exhibit
variable valence.
Table 3.18 shows the valencies of the elements of 2nd
and 3rd periods. The valencies of the elements have
been shown in brackets.
Variation in a Group
On moving down a group, the number of valence
electrons remains same and, therefore, all the elements
in a group exhibit same valence. For example, all the
elements of group 1 have valence equal to 1 and those
of group 2 have valence equal to 2.
Example :t21. Predict the formulas of the stable binary
compounds that would be formed by the following pairs of
elements:
(i) silicon and oxygen
(ii) aluminium
and bromine
(iii) calcium and iodine
(iv) element 114 and fluorine
VARIATION OF VALENCE IN THE PERIODIC
TABLE
(v) element 120-and oxygen.
Variation in a Period
The number of valence electrons increases from
1 to 8 on moving across a period, the valence to the
elements with respect to hydrogen and chlorine
increases from 1 to 4 and then decreases from 4 to zero.
Solution. (i) Silicon belongs to group 14. Its valence is 4.
The valence of oxygen is 2. Thus, the binary compound
between silicon and oxygen would have formula Si02•
(ii) Aluminium belongs to group 13. Its valence is 3.
The valence of bromine is 1. Thus, the binary compound
between aluminium and bromine would have formula AlBrs.
Table 3.1S. Variation of Valence of Elements of Second and Third Periods
Element of second period
B
Li
Be
C
N
°
F
Valence with respect to H
LiH
BeH2
BHs
CH4
NHs
H2O
HF
(1)
(2)
(3)
(4)
(3)
(2)
(1)
Valence with respect to Cl
LiCI
BeCl2
BCIs
CCl4
NCIs
Cl20
CIF
(1)
(2)
(3)
(4)
(3)
(2)
(1)
Si
P
8
Cl
Elements of third period
Mg
Al
Na
Valence with respect to H
NaH
MgH2
AlHs
8iH4
PHs
~8
HCI
(1)
(2)
(3)
(4)
(3)
(2)
(1)
8i02
P205
8°3
Cl207
(6)
(7)
Valence with respect to 0
MgO
Al20S
Na20
(1)
(2)
(3)
(4)
(5)
tiii) Calcium belongs to group 2. Its valence is 2. The
valence of iodine is 1. Thus, the compound between calcium
and iodine would have formula C~.
The element (M) with atomic number 114 belongs
to group 14. It will exhibit valence of2 (due to inert pair effect)
and 4. The stable binary compound between this element and
fluorine would have formula MF 2'
(iv)
(u) The element (X) with atomic number 120 belongs
to group 2. Its valence would be 2. The valence of oxygen is 2.
Thus, the formula of the compound between this element and
oxygen would have formula XO.
The normal oxides of the elements at the extreme
left of the periodic table are most basic in nature. On
the other hand, the elements at the extreme right react
with oxygen to form most acidic oxides. Oxides of the
elements in the centre are amphoteric or neutral. For
example, Na20 is strongly basic whereas Cl207 is
strongly acidic. Al20S is amphoteric while CO, NO and
N20 are neutral.
A basic oxide when dissolved in water gives basic
solution whereas an acidic oxide gives an acidic solution.
For example,
PERIODIC TRENDS AND
CHEMICAL REACTIVITY
N~O(s)
We have already studied the periodic trends in
the various fundamental properties such as atomic and
ionic radii, ionization enthalpy, electron gain enthalpy,
electronegativity and valence. The periodic trends in
these properties can be explained on the basis of
electronic configuration of the elements. The chemical
reactivity of elements can be related to the fundamental
properties of elements.
As already discussed the ionization enthalpy is
least for the element at the extreme left of the period
and the electron gain enthalpy is most negative for the
element at the extreme right of the period (For group17 elements), the elements of group-18 have positive
electron gain enthalpies due to their stable electronic
configurations. As a result the chemical reactivity is
maximum at the two extremes and lowest in the centre.
The extreme reactivity of group-1 elements is due to
the ease with which these elements can lose an electron
leading to the formation of corresponding cation. On
the other hand the reactivity of halogens is due to the
ease with which these elements can gain an electron to
form the corresponding anion. Thus, elements at the
extreme left exhibit strong reducing behaviour whereas
the elements at the extreme right exhibit strong
oxidizing behaviour.
+
H20(l) ~
Basic oxide
2NaOH(aq)
A strong base
C1207(l) + ~O(l)
2HCIOiaq)
~
Acidic oxide
A strong acid
An amphoteric oxide exhibits acidic behaviour
in the presence of bases and basic behaviour in the
presence of acids. A neutral oxide exhibits neither acidic
nor basic properties.
The amphoteric nature of aluminium
evident from the following reactions :
2AlCls(aq) + 3~O(l)
Al20is) + 2NaOH(aq) + 3H20(l)
~
2Na[Al(OH)41 (aq)
Al20S(s) + 6HC1(aq) ~
The acid-base nature of oxides of third period
elements is listed in Table 3.19.
From Table 3.19 it is clear that as the metallic
character of the elements decreases from left to right
across the period, their oxides change from basic to
amphoteric to acidic. Oxides of metals are usually basic
while that of non-metals are acidic. It may be noted
that the metallic character of elements increases from
top to bottom in a group of representative elements,
therefore, the basic character of oxides increases on
descending a group.
Table 3.19. Acid-base Nature of Oxides of the Third Period Elements
MgO
Basic
Basic
oxide is
Amphoteric
Acidic
Acidic
Acidic
Acidic
Example
:"t22.
basic oxide while
Show by a chemical reaction that MgO is
is an acidic oxide.
P4010
Solution. MgO reacts with aqueous HCI to form salt and
water thus exhibiting it basic character
MgO(s) + 2HCl(aq) ~
MgCliaq) + H20(l)
P40lO reacts with water to form phosphoric acid
PPlO(s) + 6H20(1) ~
4H3POiaq)
The solution turns blue litmus red.
ANO~OUSPROPERTmSOF
SECOND PERIOD ELEMENTS
AND DIAGONAL RELATION·
SIDP
It is observed that the first member of each group
(the element in the second period from lithium to
fluorine) differs in many respects from the rest of the
members of the same group. For example, lithium
unlike other alkali metals forms predominantly
covalent compounds. The other members of group-l
form predominantly ionic compounds. Similarly,
beryllium forms compounds which have predominant
covalent character while the other member of group-2
form compounds having predominant ionic character.
The difference in the behaviour of the first
member of a group in the s- and p-blocks compared to
the other members in the same group can be attributed
to the following factors:
(i) Small atomic size of the first element "
tii) Large charge/radius ratio
(iii) High electronegativity
(iv) Absence of d-orbitals in the valence shell of the
first element.
The first element in each group of s- and p-blocks
has second energy level as the valence shell and hence
has only four valence orbitals available for bonding. As
a result, the maximum covalency of the first member
of each group is 4. On the other hand the other members
of each group also have vacant d-orbitals in the valence
shell. Hence these members can expand their covalency
beyond 4, through participation of d-orbitals in bond
formation.
(v) Ability to form pn-ptt multiple bonds.
The first member of each group of p-block
elements has great tendency to form ptt-pt: multiple
bonds to itself and to the other second period elements.
For example, carbon forms bonds of the type, C = C,
C == C, C = 0, C == N, etc. This property ofthese elements
is due to their small size. The higher members of the
group have little tendency to form ptt-pt: bonds.
DIAGONAL RELATIONSIDP
Another trend in the chemical behaviour of the
representative elements is the diagonal relationship.
It is observed that an element of the second period
exhibits certain similarities with the" second element
of the following group.
For example, lithium resembles magnesium and
beryllium resembles aluminium in some respects. This
type of similarity is referred to as diagonal
relationship.
Li
Be
B
C
N~M~Al~Si
Thus, diagonal relationship is the similarity
between a pair of elements in different groups and
different periods and located diagonally in the
periodic table.
In moving along the period from left to right, the
electronegativity increases while in descending a group
electropositivity of the elements increases. These two
effects tend to cancel each other in moving diagonally
from top left to bottom right. Therefore, the elements
diagonally related in this way tend to have similar
properties. They form similar compounds though, the
valency is different.
Sometimes the diagonal relationship is explained
Ionic charge ]
in terms of polarizing power [ (I·
di )2.
OIDC ra lUS
On
moving along a period, the charge on the ions increases
while ionic size decreases hence polarizing power
increases. On moving down the group the ionic size
increases and hence polarizing power decreases. On
moving diagonally these two effects cancel each other
to some extent and hence properties remain similar.
SHT - :1.(;
1. How does valence vary in a period for representative
elements?
2. Predict the formulae of the stable binary compounds
that would be formed by the combination of the
following pairs of elements:
(i)
Lithium and oxygen
(ii) Magnesium and nitrogen
(iii) Aluminium and iodine
Silicon and oxygen
(v) Phosphorus and fluorine
(vi) Element 71 and fluorine.
3. Explain why the reactivity of elements of group-1
increases on descending the group while that of
group-17 elements increases.
4. Predict the formulas of compounds which might be
formed by the following pairs of elements;
(i) silicon and bromine (ii) aluminium and sulphur.
(NCERT Solved Example)
5. Show by a chemical reaction with water that N~O
is basic oxide and C~07 is an acidic oxide.
(NCERT Solved Example)
(iv)
6. Give examples of three neutral oxides.
7. Give examples of three amphoteric oxides.
8. How does an amphoteric oxide differ from a neutral
oxide?
9. Explain why the first member of each of the groups
of representative
elements exhibits anomalous
properties.
10. Explain why properties of lithium are somewhat
different from the properties of the other members
of the group-L
11. Name the element
relationship with Be.
which
shows
diagonal
3.1. What is the basic
periodic table?
theme
of organisation
in the
Solution. The basic theme of organisation of elements in the
periodic table is to facilitate the study of the properties of all
the elements and their compounds. On the basis of similarities
in chemical properties, the various elements have been divided
into different groups. This has made the study of elements
simple because their properties are now studied in the form of
groups rather than individually.
3.2. Which important
classify the elements
stick to that?
property did Mendeleev use to
in his periodic table and did he
Solution. Mendeleev used atomic weight as the basis of
classification of elements in the periodic table. He arranged
the then known elements in order of increasing atomic weights
grouping together elements with similar properties. Mendeleev
relied on the similarities in the properties of the compounds
formed by the elements. He realized that some of the elements
did not fit in with his scheme of classification if the order of
atomic weight was strictly followed. He ignored the order of
atomic weights and placed the elements with similar properties
together for example, iodine with lower atomic weight than
tellurium (Group VI) was placed in group VII along with
fluorine, chlorine, etc., because of similarities in properties.
Thus, Mendeleev did not stick strictly to arrangement of
elements in the increasing order of atomic weight.
3.3. What is the basic difference in approach
Mendeleev's
Periodic Law and the Modern
Law?
between
Periodic
Solution.
Mendeleev
Periodic
Law states that the
properties of the elements are a periodic function of their atomic
weights whereas Modern Periodic
Law states that the
Properties of elements are a periodic function of their atomic
numbers. Thus, the basic difference in approach between
Mendeleev's Periodic Law and Modem Periodic Law is the
change in basis of arrangement of elements from atomic weight
to atomic number.
3.4. On the basis of quantum numbers, justify that the
sixth period
of the periodic
table should
have
32 elements.
Solution. In the modem periodic table, each period starts
with the filling of a new principal energy level. Thus, the sixth
period begins with the filling of 6s orbital and continues till
the filling of seventh energy level starts. According to Aufbau
rule the subshells which follow 6s are 4f, 5d, 6p, 7s. Therefore,
in the 6th period,electrons can be filled in only 6s, 4{, 5d, and
6p-subshells. Now s-subshell has two, p-subshell has three,
d-subshell has five and f-subshell has seven orbitals. Hence,
in all there are 16 orbitals that can be filled in this period
which, at the maximum, can accommodate, 32 electrons and
therefore, sixth period has 32 elements.
3.5. In terms of period and group,
locate the element with Z = 114?
where
would
you
Solution. The element with atomic number 114 would have
the electronic configuration 861Rn]5f46d1O, 7s2 6p2. Thus, the
element belongs to : Period 7, group-14 and block p.
3.6. Write the atomic number of the element in the third
period and seventeenth
group of the periodic table.
Solution. The elements of group-17 (halogens) have ns2 np5
valence shell electronic configuration. The element of group17 belonging to third period would have valence shell electronic
configuration 3s23p5. The complete electronic configuration of
this element would be lr, 2s2, 2p6, 382, 3p5.
There are 17 electrons in it. Hence, the atomic number
of the element in the third period and group-17 is 17.
3.7. Which element do you think would have been named
by
(i) Lawrence Berkeley Laboratory
(ii) Seaborg's group?
Solution.
(i) Berkelium (Bk)
(ii) Seaborgium (Sg)
3.S. Why do elements in the same group have similar
physical and chemical properties?
Solution. Elem~nts in the same group have similar valence
shell electronic configuration and hence have similar physical
and chemical properties.
3.9. What does atomic
you?
radius
or ionic radius
mean to
Solution. Atomic radius gives us idea about size of the atom.
Atomic radius may be taken as the distance between the centre
of the nucleus and the outermost shell of the atom. It can be
measured either by X-ray or by spectroscopic methods.
The ionic radius tells us about size of the ion. It is defined as
the distance between centre of the nucleus and the point upto
which the ion has influence in the ionic bond.
3.10. How do atomic radii vary in a period
group? How do you explain the variation?
and in a
Solution. Within a group, the atomic radius increases down
the group. This is because a new energy shell is added at each
succeeding element while the number of electrons in the
valence shell remains the same. In other words, the electrons
in the valence shell of each succeeding element lie farther and
farther away from the nucleus. As a result, the force of
attraction of the nucleus for the valence electrons decreases
and hence the atomic size increases.
On the other hand, the atomic size decreases as we move
from left to right in a period. This is due to the reason that
within a period the valence shell remains the same shell but
the nuclear charge increases by one unit at each succeeding
element. Due to this increased nuclear charge, the attraction
of the nucleus for the outer electrons increases and hence the
atomic size decreases.
3.11. What do you understand by isoelectronic species?
Name the species that will be isoelectronic with each
of the following atoms or ions.
(i) F(ii) Ar
(iii) Mg2+
(iv) Rb+
Solution. The species which have the same number of electrons
but different magnitude of the nuclear charge are called
isoelectronic
species.
(i) F- has 10 (9 + 1) electrons. The species nitride ion,
N3-(7 + 3); oxide ion, 02- (8 + 2); neon, Ne (10 + 0);
sodium ion, Na" (11 - 1) ; magnesium ion, Mg2+
(12 - 2); aluminium ion, Al3+(13 - 3), are isoelectronic with it.
(ii) Ar has 18 electrons. The species phosphide ion, p3(15 + 3), sulphide ion; S2- (16 + 2); chloride ion, CI(17 + 1), potassium ion, K+ (19 - 1), calcium ion,
Ca2+(20 - 2) are isoelectronic with it.
(iii) Mg2+has 10 (12 - 2) electrons. The species nitride
ion, N3- (7 + 3); oxide ion, 02- (8 + 2) and sodium
ion, Na" (11 - 1) are isoelectronic with it.
(iv) Rh" has 36 (37 - 1) electrons. The species bromide
ion, Br" (35 + 1), krypton, Kr (36 + 0) and strontium
Sr2+ (38 - 2) each one of which has 36 electrons,
are isoelectronic with it.
3.12. Consider the following
Mgz+ and AI3+-.
(a)
species. :N3-, ()2-, F-, Na+,
What is common in them?
Arrange them in order of increasing
ionic
radii?
Solution. (a) Each one ofthese ions contains 10 electrons and
hence these are isoelectronic ions.
(b) The ionic radii of isoelectronic ions decrease with
the increase in the magnitude of the nuclear charge. Among
(b)
the isoelectronic ions: N3-, 02-, F-, Na", Mg2+and Al'", nuclear
charges increase in the order:
N3- < 02- < }r < Na" < Mg2+< Al3+.Therefore, their
ionic radii decrease in the order:
N3- > Q2- >}r > Na" > Mg2+> Al3+.
3.13. Explain why cations are smaller and anions
larger in radii than their parent atoms.
are
Solution. The ionic radius of a cation is always smaller than
the parent atom because a cation is formed by loss of one or
more electrons by the neutral atom. The loss of one or more
electrons increases the effective nuclear charge. As a result,
the force of attraction of nucleus for the electrons increases
and hence the ionic radii decrease. In contrast, the ionic radius
of an anion is always larger than its parent atom because an
anion is formed by addition of one or more electrons to the
neutral atom. 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 increase.
3.14. What is the significance
of the terms 'isolated
gaseous atom' and 'ground state' while defining the
ionization enthalpy and electron gain enthalpy?
Solution. Ionization enthalpy is the minimum amount of
energy required to remove the most loosely bound electron from
an isolated gaseous atom in ground state, so as to convert it
into a gaseous cation. Electron gain enthalpy is the energy
released when an isolated gaseous atom in the ground state
accepts an extra electron to form the gaseous negative ion. The
force with which an electron is attracted by the nucleus of an
atom is appreciably affected by presence of other atom within
its molecule or in the neighbourhood. Therefore, for the purpose of determination of ionization enthalpy and electron gain
enthalpy it is essential that these interatomic forces of attraction should be minimum. Since in the gaseous state, the atoms are widely separated, therefore, these interatomic forces
are minimum. It is because of these reasons, that the term
isolated gaseous atom has been included in the definition of
ionization enthalpy and electron gain enthalpy.
The term ground state here means that the atom must
be present in the most stable state, i.e., the ground state. The
reason being that when the isolated gaseous atom is in
the excited state, less amount of energy is required to
remove the outer most electron. Similarly, less amount of
energy is released when an electron is added to the atom in
excited state. Therefore, for comparison purposes, the
ionization enthalpies and electron gain enthalpies of gaseous
atoms must be determined in their respective ground states.
3.15. Energy of an electron in the ground state of the
hydrogen
atom is - 2.18 x 10-18 J. Calculate
the
ionization
enthalpy of atomic hydrogen in terms of
kJmol-1•
Solution. The energy required to remove an electron in the
ground state of hydrogen atom
= E~- El = 0 - El
= - (- 2.18 X 10-18 J) = 2.18 x 1018 J.
..
Ionization enthalpy per mole of hydrogen atoms
18
= 2.18 x 10
=
x 6.02 x 1023 kJ mol-l
1000
1312.36 kJ mol';
3.16. Among the second period elements, the actual
ionization energies are in the order: Li < B < Be < C < 0
<N <F<Ne.
Explain why
(i)
Be has higher ~H than B
(ii) 0 has lower ~H than N and F ?
Solution. (i) In case of Be (ls2 2s2) the outermost electron is
present in 2s-orbital while in B(1s22s2 2pl) it is present in
2p-orbital. Since 2s-electrons, due to greater penetration, are
more strongly attracted by the nucleus than 2p-electrons,
therefore, more amount of energy is required to knock out a
2s-electron than a 2p-electron. Consequently, t1;H of Be is
higher than that t1!l of B.
. (ii) The electronic configuration
2p; 2p!)
in which
is more stable
than
2p-orbitals
the
are
electronic
of N(1s2 282 2p!
exactly
half-filled
configuration
of
0(1s2 2s2 2p! 2p; 2p;) in which the 2p-orbitals are neither
half-filled nor completely filled. Therefore, it is difficult to
remove an electron from N than from O. As a result, i1;H of 0
is less than that ofN. Because of higher nuclear charge (+ 9)
and smaller atomic size, the first ionization enthalpy of F is
higher than that of O. Therefore, ionization enthalpy of 0 is
less than that of N as well as that of F.
3.17. 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?
Solution.
Na
Na"
1s2, 282, 2p6, 3s1
1s2, 2s2, 2p6
+ e-
,\H1
Mg
Mg+
+e
2
2
2
2
1
18 ,28 , 2p6, 3s
1s , 2s2, 2p6, as
The first ionization enthalpy ofMg is higher than that
of Na because it has greater nuclear charge, smaller atomic
size and stable electronic configuration. However, the second
ionization enthalpy of sodium is much higher than that of
magnesium because Nar has stable noble gas configuration.
Moreover in Na" the outermost shell is second whereas in Mg+
it is third.
3.18. What are the various factors due to which the ionization enthalpy of the main group elements tends to
decrease down the group?
Solution. Within the main group elements, the ionization
enthalpy decreases regularly as we move down the group due
to the following two factors.
. (0 Atomic size. On moving down the group, the
atomic size increases progressively due to the addition of one
new energy shell at each succeeding element. As a result, the
distance of the valence electrons from the nucleus increases.
Consequently, the force of attraction of the nucleus for the
valence electrons decreases.
(ii) Screening effect. With the addition of new shells,
the shielding effect or the screening effect increases. As a result,
the force of attraction of the nucleus for the valence electrons
further decreases and hence the ionization enthalpy decreases.
3.19. The first ionization enthalpy
of group 13 elements are:
values (in kJ mof+)
B
AI
Ga
In
TI
801
577
579
558
589
How will you explain this deviation from the
general trend?
Solution. On moving down the group 13 from B to Al, the
ionization enthalpy decreases as expected due to an increase
in atomic size and screening effect which outweigh the effect
of increased nuclear charge. However, i1;Hl of Ga is slightly
higher than that of Al while that of Tl is higher than those of
Al, Ga and In. These deviations can be explained as follows:
Al follows immediately after s-block elements while Ga
and In follow after d-block elements and Tl after d- and f-block
elements. These extra d- and f-electrons do not shield the
valence shell-electrons from the nucleus effectively. As a result,
the valence electrons are more tightly held by the nucleus and
hence larger amount of energy is needed for their removal.
This explains why Ga has higher ionization enthalpy than Al.
Further on moving down the group from Ga to In, the increased
shielding effect due to the presence of additional 4d-electrons
outweighs the effect of increased nuclear charge and hence
the t1iHl ofIn is less than, that of Ga. Thereafter, the effect of
increased nuclear charge predominates and hence the t1iHl of
Tl is higher than that of In.
3.20. Which of the following pairs of elements would
have a more negative electron gain enthalpy? Explain.
(i) 0 or F,
(ii) F or Cl.
Solution. (i) F has more negative electron gain enthalpy due
to its smaller size and greater effective nuclear charge.
(ii) Chlorine (Cl) has more negative electron gain
enthalpy than fluorine (F). F has less negative electron gain
enthalpy because in it the added electron goes to the smaller
energy level (n = 2) and hence suffers significant repulsion
from the electrons already present in this shell.
3.21. Would you expect the second electron gain enthalpy of 0 as positive, more negative or less negative
than the first. Justify your answer.
Solution. The second electron gain enthalpy of 0 is positive
as explained below:
When an electron is added to 0- to form 02- ion, energy
is required to overcome the strong electrostatic repulsion
between the negatively charged 0- ion and the electron being
added.
0- (g) + e- (g) ~
02- (g) ;
.:ill;:;.:leg~;:;+ ve
Therefore, the second electron gain enthalpy of oxygen
is positive.
3.22. What is the basic difference between the terms
electron gain enthalpy and electronegativity?
Solution. Both electron gain enthalpy and electronegativity
are measure ofthe tendency ofthe atom ofan element to attract
electrons. Whereas electron gain enthalpy refers to the
tendency of an isolated gaseous atom to accept an electron to
form a negative ion, electronegativity refers to the tendency
of the atom to attract the shared pair of electrons towards
itself in a molecule.
3.23. How would you react to the statement that the
electronegativity
of N on Pauling scale is 3.0 in all the
nitrogen compounds.
Solution. The electronegativity of any given atom is not
constant; it varies depending on the element to which it is
bound and also on the state of hybridization of the atom of the
element. Therefore, the statement, that the electronegativity
ofN on Pauling scale is 3.0 in all nitrogen compounds, is wrong.
3.24. Describe the theory associated with the radius of
an atom as it (a) gains electron (b) loses electron.
Solution. (a) Gain of electron. When a neutral atom gains
one electron to form an anion, its radius increases. The reason
being that the number of electrons in the anion increases while
its nuclear charge remains the same as the parent atom. Since
the same nuclear charge now attracts greater number of
electrons, therefore, effective nuclear charge decreases and
hence the electron cloud expands and the atomic radius
increases.
(b) Loss of electrons. When a neutral atom loses one
electron to form a cation, its atomic radius decreases. The
reason being that the number of electrons in the cation
decreases while its nuclear charge remains the same as the
parent atom. Since the same nuclear charge now attracts
smaller number ofelectrons, therefore, effective nuclear charges
increases and hence the atomic radius decreases.
3.25. Would you expect the first ionization enthalpies
of two isotopes of the same element to be same or
different? Justify your answer.
Solution. Two isotopes of the same element have same
number of electrons, same nuclear charge, same atomic radii.
Therefore, they are expected to have same ionization
enthalpies.
3.26. What are major differences between metals and
non-metals?
Solution. Elements which have a strong tendency to lose
electrons to form cations are called metals while those which
have a strong tendency to accept electrons to form anions are
called non-metals.
Metals are highly electropositive whereas non-metals
are highly electronegative. Metals are good reducing agents
whereas non-metals are good acidizing agents. Metals are
generally solids whereas non-metals are generally gases.
Oxides of metals are basic or amphoteric whereas
oxides of non-metals are acidic or neutral.
Metals are generally good conductors of heat and
electricity whereas non-matals are generally bad conductors.
3.27. Use the periodic table to answer the following
questions.
(a) Identify an element with five electrons in the
outer sub-shell.
(b) Identify the element that would tend to lose
two electrons.
(c) Identify the element that would tend to gain
two electrons.
Solution. (a) Fluorine (F). Its electronic configuration is
1S2, 2s2 2p5.
1S2,
(b) Magnesium (Mg). Its electronic configuration is
2s2 2p6, 3s2 by losing 2 electrons it attains noble gas
configuration.
(c) Oxygen (0). Its electronic configuration is 1S2, 2s2
2p4 by gaining two electrons it attains stable noble gas configuration.
.
3.28. The increasing order of reactivity among group 1
elements is Li < Na < K < Rb < Cs whereas that of group
17 is F > Cl > Br > I. Explain.
Solution. The elements of group-1 have a strong tendency to
lose electron. The tendency to lose electrons, in turn, depends
upon the ionization enthalpy. Since the ionization enthalpy
decreases down the group, therefore, the reactivity of group-1
elements increases in the order: Li < Na < K < Rb < Cs. On
the other hand, the elements of group-17, have a strong
tendency to accept electron. The tendency to accept electrons,
in turn, depends upon their electrode potential. Since the
electrode potentials of group-17 elements decrease on
descending the group, therefore, their re activities decrease in
the order:
F > Cl> Br> I.
3.29. Write the general electronic configuration
PO, do, and f-block elements:
Solution.
s-Block elements: nsl-2
p-Block elements: ns2npl-6
of
d-Block elements: (n - 1) d1-10ns0-2•
f-Block elements: (n - 2) [1-14 (n -1)
~l
ns2•
SO,
3.30. Assign the position
electronic
configuration,
(i) ns2 np4 for n
of the element
having
(c) The element
III which has high first ionizatlo,n
enthalpy and a very high negative electrongain
enthalpy is likely to be the most reactive non-metal.
(d) The element IV has a high negative electron ~
enthalpy but not so high first ionization enthalpy.
Therefore, it is the least reactive non-metal.
outer
=3
(ii) (n - 1) d2 ns2 for n = 4 and
(iii) (n - 2)(7 (n -1) di ns2 for n = 6 in the periodic
table.
Solution. (i) n-= 3 indicates that the element belongs to third
period. Since the last electron enters the p-orbital, therefore,
the given element is a p-block element. For p-block elements,
group number = 10 + number of electrons in the valence shell
:.
Group number of the element = 10 + 6 = 16.
(ii)
n
= 4 indicates
(e) The element VI has low values for first and second
ionization enthalpies. Therefore, it appears that the
element is an alkaline earth metal and hence will
form binary halide of the formula ~.
(f) The element I has low first ionization enthalpy a
very high second ionization enthalpy therefore, it
must be an alkali metal with smaller atomic number and is likely to form a predominantly stable
covalent halide of the formula MX.
that the element lies in the 4th
period.
Since the d-orbital is incomplete, therefore, it is d-block
element.
The group number of the element = number of (n - 1)
d-electrons + number of (n) s-electrons
=2+2=4(iii) n = 6 means that the element lies in the sixth
period. Since the last electron goes to the {-orbital, therefore,
the element is a {-block element. All (-block elements lie in
group 3.
3.31. The first (~ Hi) and the second (~ ~) ionization
enthalpies
(in kJ molr") and the (Aeg H) electron gain
enthalpy
(in kJ mol-i) of a few elements
are given
below:
I
11
III
IV
V
VI
520
419
1681
1008
2372
738
7300
3051
3374
1846
5251
1451
-60
-48
-328
-295
+48
-40
Which of the above element is likely to be:
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 ~
(X = halogen).
the metal which can form predominantly
stable covalent halide of the formula MX (X = halogen)?
Solution.
(a) The element V has highest first ionization enthalpy
and positive electron gain enthalpy and hence it is
likely to be the least reactive element.
(b) The element 11which has the least first ionization
enthalpy and a low negative electron gain enthalpy
is the most reactive metal.
(a)
(n
3.32. Predict
the formulas
of the stable binary compounds that would be formed by the combination
ot
the following pairs of elements:
;
(a) Lithium
and oxygen; (b) Magnesium
and
nitrogen;
(c) Aluminium
and iodine; (d) Silicon. aiIld
oxygen; (e) Phosphorus
and fluorine; if) Element 71 apd
fluorine.
Solution.
Lithium is an element of group-1. It has only o~e
electron in the valence shell, therefore, its valence is
1. Oxygen is a group-16 element with a valence.of
2. Therefore, formula of the compound fomied
would be Lip.
(b) Magnesium is an alkaline earth metal (Group-2)
and hence has a valence of2. Nitrogen is a group-l5
element with a valence of 8 - 5 = 3. Thus, the
formula of the compound formed would be Mg~.
(c) Aluminium is group-13 element with a valence of
3 while iodine is a halogen (group-17) with a valence
of 1. Therefore, the formula of the compound formed
would be AlI3.
(d) Silicon is a group-14 element with a valence of 4
while oxygen is a group-16 element with a valence
of2. Hence, the formula of the compound formed is
(a)
Si02•
Phosphorus is a group-15 element with a valence
of 3 or 5 while fluorine is a group-17 element with
a valence of 1. Hence the formula of the compound
formed would be PF 3 or PF 5'
(f) Element with atomic number 71 is a lanthanoid
called lutetium (Lu), Its common valence is 3.
Fluorine is a group-17 element with a valence of 1.
Therefore, the formula of the compound formed
would be LuF 3'
(e)
3.33. In the modern
the value of:
(a)
quantum
periodic
table, the period
atomic number (b) mass number
number (d) azimuthal
quantum
indicates
(c) princilPal
number.
Solution. In the modem periodic table, the period number is
same as the value of'prineipal quantum number. Thus, option
(c) is correct.
(b)
The greatest increase in ionization enthalpy
is experienced on removal of electrons from
core noble gas configuration.
3.M. Which of the following statements
modern periodic table is incorrect?
(c)
End of valence electrons is marked
jump in ionization enthalpy.
related
to the
by a big
{d) Removal of electron from orbitals bearing
lower n value is easier than from orbital
having higher n value.
(a)
The p-block has six columns, because a
maximum of 6 electrons can occupy all the
orbitals in p-subshell,
(b)
The d-block has 8 columns,
because
a
maximum of 8 electrons can occupy all the
orbitals in a d-subshell.
Solution. Statement (d) is incorrect because an electron with
lower value of n is more tightly held as it is closer to the
nucleus.
(c)
Each block contains
equal to the number
occupy that subshell.
3.38. Considering
the elements B, AI, Mg and K, the
correct order of their metallic character is:
(d)
The block indicates
value of azimuthal
quantum number (I) for the last subshell that
received
electrons
in building
up the
electronic configuration.
a number of columns
of electrons that can
Solution.
Statement (b) is incorrect. The d-block has
10 columns, because a maximum of 10 electrons can occupy
all the orbitals is a d-subshell.
3.35. Anything that influences the valence electrons will
affect the chemistry of the element. Which one of the
following factors does not affect the valence shell?
(a)
Valence principal
(b) Nuclear
charge
quantum
number
(n)
(Z)
(c)
Nuclear
mass
(d)
Number of core electrons.
Solution. Nuclear mass does not affect the valence shell. Thus,
option (c) is the correct answer.
3.36. The size of isoelectronic
is affected by
charge
species - F -, Ne and Na+
(a)
nuclear
(b)
valence principal
(c)
electron-electron
orbital
(d)
none of factors because their size is the same.
(Z)
quantum
number
interaction
(n)
in the
outer
Solution. The size of the isoelectronic ions depends upon the
nuclear charge (Z). AB the nuclear charge increases the ionic
size decreases. Therefore, statement (a) is correct.
3.37. Which of the following statements
relation to ionization enthalpy?
(a)
Ionization
successive
enthalpy increases
electron.
is incorrect
for each
in
(a)
B > AI > Mg > K
(b) AI > Mg > B > K
(c)
Mg> AI > K > B
(d)
K> Mg > AI > B
Solution. Across a period, metallic character decreases as we
move from left to right. However, within a group, the metallic
character, increases from top to bottom. Therefore, K is
expected to be most metallic and B the least. Hence, the choice
(d) is correct.
3.39. Considering the elements B, C, N, F and Si, the
correct order of their non-metallic character is:
(a)
B> C > Si > N > F
(b)
(c)
F > N > C > B > Si
(d) F > N > C > Si > B
Si > C > B > N > F
Solution.
Across a period, the non-metallic character
increases from left to right. Therefore, among B, C, N and F,
non-metallic character decreases in the order: F > N > C > B.
However, within a group, non-metallic character decreases
from top to bottom. Thus, C is more non-metallic than Si.
Therefore, the correct sequence of decreasing non-metallic
character is: F > N > C > B > Si, and the choice (c) is correct.
3.40. Considering
the elements F, Cl, 0 and N, the
correct order of their chemical reactivity in terms of
oxidising property is:
(a)
F > Cl> 0 > N
(b)
F > 0 > Cl > N
(c)
Cl> F > 0 > N
(d)
0 > F > N > Cl
Solution. Across a period, the oxidising character increases
from left to right. Therefore, among F, 0 and N, oxidising power
decreases in the order: F> 0> N. However, within a group,
oxidising power decreases from top to bottom. Thus, F is a
stronger oxidising agent than Cl. Thus, overall decreasing
order of oxidising power is: F > Cl > 0 > N and the choice (a) is
correct.
••
What is the group number, period and block of the element with atomic number 40?
Solution. The electronic configuration of the element with
atomic number 40 is:
~ : 182, 2s2, 2p6, &2, 3p6, 482, 3dlO, 4p6, 582, 4cJ2
or ~:
lKr] 582, 4cJ2
Since the last electron enters d-subshell, the element
belongs to d-block.
Since the outermost shell of the element is fifth, the
element belongs to period 5
Group number of the elements of d-block is equal to
sum of the electrons in outermost s-subshell and d-subshell
of penultimate shell.
Hence, group number of the element
=2+2=4.
••
Explain why the first ionization enthalpy of carbon is
more than that of boron but the reverse is true for the second
ionization enthalpy.
Solution. The first ionization enthalpy of carbon is more than
that of boron because effective nuclear charge of carbon is
more and atomic size is smaller than that of boron.
After losing one electron the boron atom changes into
B+ which has stable electronic configuration.
B+ : Is2, 2s2
C+ : Is2, 2s2, 2px 1
As a result more energy is required to remove an
electron from B+ than from C+. Hence, second ionization
enthalpy of boron is higher than that of carbon.
••
Arrange the following ions in order of their increasing
radii:
Lir, Mg2+ , K+, AI3+
Solution.
Li" < Al3+ < Mg2+ < K+
Li" is isoelectronic with helium.
Mg2+ and Al3+ are isoelectronic with neon and have
nuclear charge 12 and 13.
K+ is isoelectronic with argon.
Li" is smallest because in it the outermost shell is first.
K+ is largest because in it the outermost shell is third.
Mg2+ and Al3+ have second shell as the outermost.
Al3+ is smaller than Mg2+ because it has greater
nuclear charge.
••
Arrange the following as stated:
(i) ~ , Na', F: , ()2-, Mg2+
..... Increasing ionic size
(ii) H, F, Cl
..... Increasing electronegativity
tiii) H, He, Li
..... Increasing ionization energy.
Solution.
(i) Mg2+ < Na+ < F- < 02- < N3-(ii) H < Cl < F
(iii) Li < H < He.
••
The ionic size of Cl- is greater than that of K+ though
the two ions are isoelectronic. Give reason.
Solution. The nuclear charge in CI- is 17 whereas that in K +
is 19. Therefore, electrons in K+ are more tightly held than in
CI-. Hence, ionic size of K+ is smaller. In other words, ionic
size of CI- is greater than that of K".
••
size.
Arrange the following in the increasing order of their
F -, u-, Na+ , Cl
Solution. Li+ < Na+ < F- < CILi" and Na" belong to same group. Ionic size increases
on descending the group. Therefore, Li" < Na". Na+ and Fare isoelectronic ions with nuclear charge 11 and 9. Na'
having greater nuclear charge is smaller than F-. Therefore
Na +< F-. F- and CI- belong to same group. Ionic size increases
on descending the group. Therefore, F- < CI-. Therefore, the
overall increasing order of sizes is Li" < Na" <F- < CI-.
••
The first ionization enthalpy of N is more as compared
to those of C and O.
Solution.
N has smaller atomic size and greater nuclear
charge than C and hence has higher first ionization enthalpy
than that of C.
N has higher ionization enthalpy than that of 0 due
to its stable electronic configuration which has exactly half
filled p-subshell
N: 1s2, 2s2, 2p/, 2p/, 2p/.
••
Out of Al+ and Mg+ which
enthalpy?
has higher
ionization
Or
Out of AI and Mg which has higher second ionization
enthalpy?
Solution.
In both Al+ and Mg" the outermost
removed from 3s-orbital.
Al"
: 1s2, 2S2, 2p6, 3s2
Mg"
: ls2, 2S2, 2p6, 3s1
electron
is
Al" has higher ionization enthalpy than Mg+ because
nuclear charge in Al+ (13 units) is higher than in Mg+ (12
units). Moreover, Al" has stable configuration (fully filled 3s
sub-shell).
••
Arrange C, N, 0, and F in the decreasing order of their
second ionization enthalpies and explain briefly.
Solution.
Second ionization enthalpies of C, N, 0 and F
correspond to the ionization enthalpies of C+, N+, 0+ and F+
respectively.
sC+ : ls2, 2s2, 2p/
_N+
. ls2 , 2s2 '''''x'
2n 1 2n
1
I
•
'l""y
0+ .. ls2 , 2S2''Yx'
2n
8
1
2n
1 2p 1
ry'
z
~+
: ls2, 2s2, 2Px2, 2p/, 2p/.
All these ions have second shell as the outermost shell.
The nuclear charge decreases in the order F+ > 0+ > N+ > C+.
Therefore, the expected order of ionization enthalpies is:
F+ > 0+ > N+ > C+
But ifwe observe the electronic configurations, we find
that 0+ has stable configuration (exactly half-filled p-subshell) and hence would have higher ionization enthalpy than
that of'F", Therefore, the correct order of ionization enthalpies
is:
0+ > F+ > N+ > C+.
Therefore, decreasing
order
enthalpies of C, N, 0 and F is:
of second
ionization
O>F>N>C.
IlDI
Explain, why electron gain enthalpies of noble gasses
are positive.
Solution.
All noble gases have ns2 np6 valence shell
configuration (except helium where it is 182). In the valence
shell of noble gases s and p sub-shells are fully filled, therefore,
these configurations are highly stable. Hence, noble gases do
not have any tendency to accept more electrons because this
would disturb their stable configuration. Hence, energy has
to be supplied to add an electron to the noble gases and thus
their electron gain enthalpies are positive.