Chemistry

Classification of elements-II

Session Objectives

Session Opener

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.

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

Causes of periodicity

Repetition of similar valence shell configuration after regular interval.

1s2,2s2,2p6,3s2,3p6,4s2,3d10,4p6,5s1

37Rb

1s2,2s2,2p6,3s2,3p6,4s119K

1s2,2s2,2p6,3s111Na

1s2,2s13Li

Electronic configuration

Atomic no.Element

Atomic sizeCovalent and van der waal’s radius:

a b

c

Distance between a and bCovalent radius

2

Distance between b and cvan der Waal's radius=

2

Trends of atomic size

Ask your self

Which element has highest covalent radius?

Cs

Solution:

Size of cation

Size of cation

Electrons

ProtonsFe

26

26

Fe2+

26

24

Fe3+

23

26

Cl–

17

18

Cl

17

17Protons

Electrons

Size of anion

Isoelectronic ions4 3 2C N O rF

r r r

no.of electrons 10 10 10 10

nuclear charge 6 7 8 9

2 3Na Mg Alr 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

Ionisation energy

+ h

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

Successive ionisation energies

IE3 > IE2 > IE1

IE1

M – eM+

– eM2+ – e M3+

IE2IE3

Factors affecting values of ionisation energy

1. Size

Ionisation energy 1

Atomic size

Atomic size

Ionisation energy KJ/mole

Li1.23

520

Be0.89

899

Factors affecting values of ionisation energy

Ionization energy Effective nuclear charge

2. Effective nuclear charge

Is net nuclear attraction towards the valence shell electrons .

Effective nuclearcharge

Ionisation energy KJ/mole

Li+3

520

Be+4

899

Factors affecting values of ionisation energy

3. Screening effect orshielding effect

Combined effect of attractive and repulsive forces between electron and proton.

Ionisation energy1

Number of inner shells

No. of inner shells

Ionisation energy KJ/mole

Li1

520

Na2

496

Factors affecting values of ionisation energy

4. Penetrating power of orbitals

s>p>d>f

5. Complete octet

Elements having ns2,np6 configuration have extremelyhigh ionisation energy.

6. Stable Configuration

Ionisation energy1

Stability of configuration

Factors affecting values of ionisation energy

Configuration

Ionisation energy KJ/mole

Be2s2

899

B2s22p1

801

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

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

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

Variation of I1 with Z

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 Bee ––

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.

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

Effective nuclear

charge

Li BeE.N.C 1.23 0.89EA kJ/mol -57 66

Factors affecting electron affinity

Electron affinity

Li NaInner shells 1 2EA kJ/mol -57 -21

1

Screening effect

1

atomic size

Li NaAt. size 1.23 1.57EA kJ/mol -57 -21

Penetrating power

s>p>d>f

Li BeConfig. 2s1 2s2

EA kJ/mol -57 66

1

Stable configuration

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 F2. 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.

Do you know?

More the value of electron affinitygreater is the oxidising power.

ElectronegativityIt is the relative tendency to attract shared pair of electrons towards itself.

Factors effecting electronegativity

1. Electronegativity 1

Atomic size

2. Electronegativity is higher for nearly filled configuration e.g. O(3.5) and F(4.0).

H H.. H Cl

.x x x

x x xx

Periodic variation

(i) In period

Li Be B C N O F

Valence shellconfiguration

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

Pauling scale of electronegativity

AB AA BB AB

2AB 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.

Mulliken’s scale of electronegativityElectronegativity 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 EAX

2Relation with Pauling's electronegativity

X = 0.336(X - 0.615)

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

• 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

1s2,2s2,2p6,3s2,3p6,4s119K

1s2,2s2,2p6,3s111Na

1s2,2s13Li

• Valency of s block elements issame 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

Valency

•Valency of p block elements isequal to number of electrons invalence shell.

e.g.- Al has 3 electrons in valence shell.Therefore, its valency is 3. Or8-number of electrons in valence shell.e.g- valency of oxygen is 8 – 6 =2

Valency

• Valency of d and f block elements variable.

Iron shows the valence 2 and 3

Valency

Valency in period

01234321Valency

87654321Number of electrons in valence shell

NeFONCBBeLiElement

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:

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.

Diagonal relationship

2nd period

3rd period

CLi

Na Mg

Be

Al Si

B

Causes of diagonal relationship

• Similarity in size.

• Similarity in ionisation energy.

• Similarity in electron affinity.

Class Test

Class Exercise - 1

Which has the smallest size?

(a) Na+ (b) Mg2+

(c) Al3+ (d) P5+

Solution:

Size of isoelectronic species decreases with increase innuclear charge.

Hence, answer is (d).

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.

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 2.5F O N C H

Class Exercise - 4

Which of the following ions hassmallest ionic radius?

(a) Li+ (b) Be2+

(c) H– (d) All have equal radii

Solution:

Hence, answer is (b).

More the nuclear charge on cation smallerwill be the size.

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 increasein nuclear charge, more interelectronic repulsion inS and Cl is the reason of their increased size.

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 isN < O < S < Cl

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.

Class Exercise - 8

In which of the following pairs thereis an exception in the periodic trendfor 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).

Class Exercise - 9

The first three successive ionisationenergies 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 ionisationenergies is very high, it belongs to group 1.

Class Exercise - 10

Atomic number of element is 108.This element is placed in ____ blockof 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

Class Exercise - 11

Which of the following values inelectron volt per atom representthe 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 oxygenbecause of stable (2s2, 2p3) configuration of nitrogen.

Class Exercise - 12

The electronic configuration of anelement is (n – 1)d1, ns2 wheren = 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 principalquantum number.

Class Exercise - 13

An element havingatomic 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.

Class Exercise - 14

In its structure an element has 4shells. 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.

Class Exercise - 15

A, B, C and D have following electronicconfigurationsA : 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 periodElement B — 3rd periodElement C — 3rd periodElement D — 4th period

Thank you