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. 1s 2,2s 2,2p 6,3s 2,3p 6,4s 2, 3d 10,4p 6,5s 1 37Rb 1s 2,2s 2,2p 6,3s 2,3p 6,4s 1 19K 1s 2,2s 2,2p 6,3s 1 11Na 1s 2,2s 1 3Li Electronic configuration Atomic no.Element

Atomic size Covalent and van der waal’s radius: ab c

Trends of atomic size

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

Size of cation

Electrons Protons Fe 26 Fe Fe Cl – Cl 17 Protons Electrons Size of anion

Isoelectronic ions Note for isoelectronic series Na +, Mg 2+, Al 3+, N 3-, O 2-, F -, N 3- > O 2- > F - > Na + > Mg 2+ > Al 3+ 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 IE 3 > IE 2 > IE 1 IE 1 M – e M+M+ M 2+ – e M 3+ IE 2 IE 3

Factors affecting values of ionisation energy 1.Size Ionisation energy  1 Atomic size Ionisation energy KJ/mole Li Be

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 nuclear charge Ionisation energy KJ/mole Li Be

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

Factors affecting values of ionisation energy 4. Penetrating power of orbitals s>p>d>f 5. Complete octet Elements having ns 2,np 6 configuration have extremely high ionisation energy.

6. Stable Configuration  Ionisation energy 1 Stability of configuration Factors affecting values of ionisation energy Configuration Ionisation energy KJ/mole Be 2s B 2s 2 2p 1 801

Trend of ionisation energy in period and groups Exceptions (i)IE > IE II A III A ns 2 ns 2,np 1 (ii) IE > IE V A VI A ns 2,np 3 ns 2,np 4 (iii) Ionisation energy of Al > Ga Absence of d electrons in Al

In a group (column), I 1 decreases with increasing Z. valence e ’s with larger n are further from the nucleus, less tightly held Variation of I 1 with Z

Across a period (row), I 1 mainly increases with increasing Z. Because of increasing nuclear charge (Z). Variation of I 1 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: IE 1 Be > IE 1 Li  Be has stable (2s 2 ) configuration. IE 2 Li > IE 2 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

 Effective nuclear charge Li Be E.N.C EA kJ/mol Factors affecting electron affinity Electron affinity Li Na Inner shells 1 2 EA kJ/mol  1 Screening effect  1 atomic size Li Na At. size EA kJ/mol  Penetrating power s>p>d>f Li Be Config. 2s 1 2s 2 EA kJ/mol  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 F 2.Group II A have almost zero electron affinities due to stable ns 2 configuration of valence shell. 3.Group V A have very low values of electron affinities due to ns 2,np 3 configuration of valence shell.

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

Electronegativity It 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).

Periodic variation (i) In period Li Be B C N O F Valence shell configuration 2s 1 2s 2 2s 2, 2p 1 2s 2,2p 2 2s 2,2P 3 2s 2,2P 4 2s 2,2P 5 Electronegativity (ii) In groups-decreases down the group

Pauling scale of electronegativity

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.

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 1s 2,2s 2,2p 6,3s 2,3p 6,4s 2, 3d 10,4p 6,5s 1 37Rb 1s 2,2s 2,2p 6,3s 2,3p 6,4s 1 19K 1s 2,2s 2,2p 6,3s 1 11Na 1s 2,2s 1 3Li

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

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

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

Valency

Valency in period Valency Number 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 NCl 5 but phosphorous forms PCl 5.

Diagonal relationship 2 nd period 3 rd period CLi NaMg Be AlSi 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) Mg 2+ (c) Al 3+ (d) P 5+ Solution: Size of isoelectronic species decreases with increase in nuclear 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

Class Exercise - 4 Which of the following ions has smallest ionic radius? (a) Li + (b) Be 2+ (c) H – (d) All have equal radii Solution: Hence, answer is (b). More the nuclear charge on cation smaller will be the size.

Class Exercise - 5 Which one of the following is correct order of ionic size? (a) Ca 2+ > K 1+ > Cl - > S 2- (b) S 2- > Cl - > K + > Ca 2+ (c) Ca 2+ > Cl - > K 1+ > S 2- (d) S 2- > Ca 2+ > 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.

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

Class Exercise - 7 Which ion has the largest radius? (a) Ca 2+ (b) F – (c) P 3– (d) Mg 2+ Solution: Hence, answer is (c). Anions are larger in size than cation.

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 (2s 2 ) as compared to B (2s 2, 2p 1 ).

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.

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 — 1s 2, 2s 2, 2p 6, 3s 2, 3p 6, 4s 2, 3d 10, 4p 6, 5s 2, 4d 10, 5p 6, 6s 2, 4f 14, 5d 10, 6p 6, 7s 2, 5f 14, 6d 6

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 (2s 2, 2p 3 ) configuration of nitrogen.

Class Exercise - 12 The electronic configuration of an element is (n – 1)d 1, ns 2 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.

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 — 1s 2, 2s 2, 2p 6, 3s 2, 3p 6, 4s 2, 3d 5. Therefore it is d block element.

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.

Class Exercise - 15 A, B, C and D have following electronic configurations A : 1s 2, 2s 2, 2p 1 B : 1s 2, 2s 2, 2p 6, 3s 2, 3p 1 C : 1s 2, 2s 2, 2p 6, 3s 2, 3p 4 D : 1s 2, 12s 2, 2p 6, 3s 2, 3p 6, 4s 1 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

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