This document provides information on the classification of elements and various periodic properties like atomic size, ionization energy, electron affinity and electronegativity. It discusses the trends in these properties across periods and groups and exceptions to trends. It also explains concepts like ionic size, isoelectronic species, ionization energies, electron affinities, Pauling and Mulliken scales of electronegativity and valency. Sample problems are provided at the end to test the understanding of these concepts.

1. Chemistry

2. Classification of elements-II

3. Session Objectives

4. Session Opener

5. Perspective
Understanding of basic properties like atomic size ionisation energy,
electron affinity and electronegativity will help in understanding
general trends in s and p block elements.

6. Session Objectives
Causes of periodicity
Atomic size,ionic radii,trend in groups and periods
Ionisation energy.
Electron affinity
Electronegativity
Valency and its trend
Anomalous behaviour of first element of group
Diagonal relationship

7. Causes of periodicity
Repetition of similar valence shell
configuration after regular interval.
1s2
,2s2
,2p6
,3s2
,3p6
,4s2
,
3d10
,4p6
,5s1
37Rb
1s2
,2s2
,2p6
,3s2
,3p6
,4s1
19K
1s2
,2s2
,2p6
,3s1
11Na
1s2
,2s1
3Li
Electronic
configuration
Atomic no.Element

8. Atomic size
Covalent and van der waal’s radius:
a b
c
Distance between a and b
Covalent radius
2
=
Distance between b and c
van der Waal's radius=
2

9. Trends of atomic size

10. Ask your self
Which element has highest covalent radius?
Cs
Solution:

11. Size of cation

12. Size of cation
Electrons
Protons
Fe
26
26
Fe2+
26
24
Fe3+
23
26
Cl–
17
18
Cl
17
17Protons
Electrons
Size of anion

13. Isoelectronic ions
4 3 2
C N O rF
r r r
no.of electrons 10 10 10 10
nuclear charge 6 7 8 9
− − − −> > >
+ + + +
2 3
Na Mg Al
r r r
no.of electrons 10 10 10
nuclear charge 11 12 13
+ + +> >
+ + +
• Note for isoelectronic series Na+
, Mg2+
, Al3+
, N3-
, O2-
, F-
,
• N3-
> O2-
> F-
> Na+
> Mg2+
> Al3+
• Most positive ion the smallest, most negative the largest

14. Ionisation energy
+
hν
Isolated gaseous atom
IE
-e-
•Minimum energy required to remove an electron from a ground-
state, gaseous atom
•Energy always positive (requires energy)
•Measures how tightly the e-
is held in atom (think size also)
•Energy associated with this reaction

15. Successive ionisation energies
IE3 > IE2 > IE1
IE1
M
– e
M+
– e
M2+ – e M3+
IE2
IE3

16. Factors affecting values of
ionisation energy
1. Size
Ionisation energy α 1
Atomic size
Atomic size
Ionisation energy
KJ/mole
Li
1.23
520
Be
0.89
899

17. Factors affecting values of
ionisation energy
Ionization energy α Effective nuclear charge
2. Effective nuclear charge
Is net nuclear attraction
towards the valence shell
electrons .
Effective nuclear
charge
Ionisation energy
KJ/mole
Li
+3
520
Be
+4
899

18. Factors affecting values
of ionisation energy
3. Screening effect or
shielding effect
Combined effect of attractive
and repulsive forces between
electron and proton.
αIonisation energy
1
Number of inner shells
No. of inner shells
Ionisation energy
KJ/mole
Li
1
520
Na
2
496

19. Factors affecting values
of ionisation energy
4. Penetrating power of orbitals
s>p>d>f
5. Complete octet
Elements having ns2
,np6
configuration have extremely
high ionisation energy.

20. 6. Stable Configuration
αIonisation energy
1
Stability of configuration
Factors affecting values of
ionisation energy
Configuration
Ionisation energy
KJ/mole
Be
2s2
899
B
2s2
2p1
801

21. Trend of ionisation energy
in period and groups
Exceptions
(i) IE > IE
II A III A
ns2
ns2
,np1
(ii) IE > IE
V A VI A
ns2
,np3
ns2
,np4
(iii) Ionisation energy of Al > Ga
Absence of d electrons in Al

22. In a group (column), I1 decreases with increasing Z.
valence e’s
with larger n are further from the nucleus, less tightly held
Variation of I1 with Z

23. Across a period (row), I1 mainly increases with increasing Z.
Because of increasing nuclear charge (Z).
Variation of I1 with Z

24. Illustrative example
First ionisation energy of Be is more than Li but the second
ionisation energy of Be is less than Li. Why?
Solution:
Li Li Be Be
e –
–
IE1
2s1 2s0
+ e –
–
IE1
2s2 2s1
+
Li Li Be Bee –
–
IE2
1s2 1s1
2++ e –
–
IE2
2s1 2s0
2++
IE1Be > IE1 Li ∴ Be has stable (2s2
) configuration.
IE2 Li > IE2 Be ∴ Li acquires stable configuration when
it loses one electron.

25. Electron affinity
Successive affinities
e–
Isolated gaseous atom
EA
•Electron affinity is energy change when an e-
adds to a
gas-phase, ground-state atom
•Positive EA means that energy is released, e- addition is
favorable and anion is stable!
•First EA’s mostly positive, a few negative
( )
( )
1
2
EA
EA 2
A(g) e A g energy released
A (g) e energy supplied A g
− −
− − −
+ → +
+ + →

26. α Effective nuclear
charge
Li Be
E.N.C 1.23 0.89
EA kJ/mol -57 66
Factors affecting electron
affinity
Electron affinity
Li Na
Inner shells 1 2
EA kJ/mol -57 -21
α
1
Screening effect
α
1
atomic size
Li Na
At. size 1.23 1.57
EA kJ/mol -57 -21
α Penetrating
power
s>p>d>f
Li Be
Config. 2s1
2s2
EA kJ/mol -57 66
α
1
Stable configuration

27. Trends in electron affinities
•Decrease down a group and increase across a period in general but
there are not clear cut trends as with atomic size and I.E.
•Nonmetals are more likely to accept e-s than metals. VIIA’s like to
accept e-s the most.
Exceptions
1. EA of Cl > EA of F
2. Group II A have almost zero
electron affinities due to
stable ns2
configuration of
valence shell.
3. Group V A have very low
values of electron affinities
due to ns2
,np3
configuration
of valence shell.

28. Do you know?
More the value of electron affinity
greater is the oxidising power.

29. Electronegativity
It is the relative tendency to attract
shared pair of electrons towards itself.
Factors effecting electronegativity
α1. Electronegativity
1
Atomic size
2. Electronegativity is higher for nearly filled
configuration e.g. O(3.5) and F(4.0).
χ
H H
.. H Cl
.
x x x
x x x
x

30. Periodic variation
(i) In period
Li Be B C N O F
Valence shell
configuration
2s1
2s2
2s2,
2p1
2s2
,2p2
2s2
,2P3
2s2
,2P4
2s2
,2P5
Electronegativity 1.0 1.5 2.0 2.5 3.0 3.5 4.0
(ii) In groups-decreases down the group

31. Pauling scale of electronegativity
AB AA BB AB
2
AB A B A B AB
A B
Pauling's Electronegativity
E = 1/2(E + E ) +
= 96.49(X - X ) or |X - X | =0.102
where X and X are constants
characteristic of the atoms A and B.
∆
∆ ∆

32. Mulliken’s scale of electronegativity
Electronegativity represents an average of the binding energy of the
outermost electrons over a range of valence-state ionizations (A+
 A
A-
in A-B)
In other words, the average of the ionization energy and
the electron affinity.
M
P M
IE EA
X
2
Relation with Pauling's electronegativity
X = 0.336(X - 0.615)
+
=

33. Do you know
1. Smaller atoms have more
electronegativities
2. F is most electronegative element.
3. Decreasing order of electro negativity
F > O > Cl ≈ N > Br > C > I > H

34. • The valency of an element is
decided by number of electrons present in
outermost shell.
• All the elements of a group have same
valency.
E.g.- All the group I elements show 1
valency.
Valency
1s2
,2s2
,2p6
,3s2
,3p6
,4s2
,
3d10
,4p6
,5s1
37Rb
1s2
,2s2
,2p6
,3s2
,3p6
,4s1
19K
1s2
,2s2
,2p6
,3s1
11Na
1s2
,2s1
3Li

35. • Valency of s block elements is
same as their group number.
Examples:
Ca is member of group 2
∴ its valency is 2
• K is member of group 1
∴its valency is 1
Valency

36. Valency
•Valency of p block elements is
equal to number of electrons in
valence shell.
e.g.- Al has 3 electrons in valence shell.
Therefore, its valency is 3.
Or
8-number of electrons in valence shell.
e.g- valency of oxygen is 8 – 6 =2

37. Valency
• Valency of d and f block elements
variable.
Iron shows the valence 2 and 3

38. Valency

39. Valency in period
01234321Valency
87654321Number of
electrons in
valence shell
NeFONCBBeLiElement

40. Anomalous behaviour of first element
of group
• Smallest size in group.
• Highest value of ionisation energy in the
group.
• Absence of vacant d orbitals.
Causes:

41. Examples of anomalous behaviour
of first element of group
• Carbon forms multiple bonds but rest of the members form
only single bonds.
• Nitrogen does not form NCl5 but phosphorous forms PCl5.

42. Diagonal relationship
2nd
period
3rd
period
CLi
Na Mg
Be
Al Si
B

43. Causes of diagonal relationship
• Similarity in size.
• Similarity in ionisation energy.
• Similarity in electron affinity.

44. Class Test

45. Class Exercise - 1
Which has the smallest size?
(a) Na+
(b) Mg2+
(c) Al3+
(d) P5+
Solution:
Size of isoelectronic species decreases with increase in
nuclear charge.
Hence, answer is (d).

46. Class Exercise - 2
If the first ionization energy of helium is 2.37 kJ/mole,
the first ionization energy of neon in kJ/mole is:
(a) 0.11 (b) 2.37
(c) 2.68 (d) 2.08
Solution:
Hence, answer is (d).
Ionization energy decreases down the group.

47. Class Exercise - 3
The relative electronegativities of F, O, N, C and H are
(a) F > C > H > N > O (b) F > O > N > C > H
(c) F > N > C > H > O (d) F > N > H > C > O
Solution:
Hence, answer is (b).
The correct order of electronegativities is
> > > >
4.0 3.0 2.13.5 2.5
F O N C H

48. Class Exercise - 4
Which of the following ions has
smallest ionic radius?
(a) Li+
(b) Be2+
(c) H–
(d) All have equal radii
Solution:
Hence, answer is (b).
More the nuclear charge on cation smaller
will be the size.

49. Class Exercise - 5
Which one of the following is correct order of ionic size?
(a) Ca2+
> K1+
> Cl-
> S2-
(b) S2-
> Cl-
> K+
> Ca2+
(c) Ca2+
> Cl-
> K1+
> S2-
(d) S2-
> Ca2+
> Cl-
> K+
Solution:
Hence, answer is (b).
Size of iso electronic species decreases with increase
in nuclear charge, more interelectronic repulsion in
S and Cl is the reason of their increased size.

50. Class Exercise - 6
The electron affinities of N,O, S and Cl are
(a) N < O < S < Cl
(b) O < N < Cl < S
(c) O = Cl < N = S
(d) O < S < Cl < N
Solution:
Hence, answer is (a).
The correct order of electron affirmities is
N < O < S < Cl

51. Class Exercise - 7
Which ion has the largest radius?
(a) Ca2+
(b) F–
(c) P3–
(d) Mg2+
Solution:
Hence, answer is (c).
Anions are larger in size than cation.

52. Class Exercise - 8
In which of the following pairs there
is an exception in the periodic trend
for the ionization energy?
(a) Fe – Ni (b) C – N
(c) Be – B (d) O – F
Solution:
Hence, answer is (c).
Since Be has stable configuration (2s2
) as compared to B
(2s2
, 2p1
).

53. Class Exercise - 9
The first three successive ionisation
energies of an element Z are 520,
7297 and 9810 kJ mol–1
respectively.
The element Z belongs to
(a) group 2 (b) group 1
(c) group 15 (d) group 16
Solution:
Hence, answer is (b).
Since the difference in first and second ionisation
energies is very high, it belongs to group 1.

54. Class Exercise - 10
Atomic number of element is 108.
This element is placed in ____ block
of periodic table.
(a) s (b) p (c) d (d) f
Solution:
Hence, answer is (c).
Atomic number — 108
Configuration — 1s2
, 2s2
, 2p6
, 3s2
, 3p6
, 4s2
, 3d10
,
4p6
, 5s2
, 4d10
, 5p6
, 6s2
, 4f14
, 5d10
, 6p6
, 7s2
, 5f14
, 6d6

55. Class Exercise - 11
Which of the following values in
electron volt per atom represent
the first ionisation energies of
oxygen and nitrogen atom respectively
(a) 14.6, 13.6 (b) 13.6, 14.6
(c) 13.6, 13.6 (d) 14.6, 14.6
Solution:
Hence, answer is (d).
First ionisation energy of nitrogen is more than oxygen
because of stable (2s2
, 2p3
) configuration of nitrogen.

56. Class Exercise - 12
The electronic configuration of an
element is (n – 1)d1
, ns2
where
n = 4. The element belongs to ____
period of periodic table.
(a) 3 (b) 2 (c) 5 (d) 4
Solution:
Hence, answer is (d).
The period number is same as maximum value of principal
quantum number.

57. Class Exercise - 13
An element having
atomic number 25 belongs to
(a) s (b) p (c) f (d) d
Solution:
Hence, answer is (d).
Electronic configuration — 1s2
, 2s2
, 2p6
, 3s2
, 3p6
,
4s2
, 3d5
.
Therefore it is d block element.

58. Class Exercise - 14
In its structure an element has 4
shells. Therefore it belongs to
(a) 3rd period (b) 4th period
(c) 2nd period (d) None of these
Solution:
Hence, answer is (b).
Group 2 is present in s block and for them group
number = number of electrons in valence shell.

59. Class Exercise - 15
A, B, C and D have following electronic
configurations
A : 1s2
, 2s2
, 2p1
B : 1s2
, 2s2
, 2p6
, 3s2
, 3p1
C : 1s2
, 2s2
, 2p6
, 3s2
, 3p4
D : 1s2
, 12s2
, 2p6
, 3s2
, 3p6
, 4s1
Find out the periods of A, B, C and D.
Solution:
Hence, answer is (b).
Period number is equal to maximum
value of principal quantum number.
Element A — 2nd period
Element B — 3rd period
Element C — 3rd period
Element D — 4th period

60. Thank you