Chemistry 03 – Classification of elements and Periodicity in properties

1) Mode of hybridization affects

(a) ‘s’ character of a bond pair electron                (b) ‘s’ & ‘p’ character of a bond pair

(c) None of these                                                      (d) Odd electron

(b) Hybridised orbital accommodates σ bond pairs.

2) The correct order of the size of the iodine species is:

(a) I > I+ > I–                (b) I > I > I+                (c) I+ > I > I                (d) I > I > I

(a) Size of an anion > size of an atom > size of a cation

3) Which of the following transitions involves maximum amount of energy?

(a) M(g) → M(g)               (b) M(g) → M+(g)               (c) M+(g) → M2+(g)               (d) M2+(g) → M3+(g)  

(a) Removal of an electron from a dipositive ion involves maximum amount of energy.

4) Which is least soluble in water

(a) CaSO4               (b) MgSO4               (c) Na2SO4               (d) BaSO4  

(d) Solvation energy > lattice energy.

5) Among NaF, NaCl, NaBr and Nal, the NaF has highest melting point because

(a) It has maximum ionic character                     (b) It has minimum ionic character              

(c) It has associated molecules                              (d) It has least molecular weight  

(a) Maximum covalencyless melting point.

6) Which possesses the largest ionic radius?

(a) Fe               (b) Fe2+               (c) Fe+              (d) Fe3+ 

(a) Fe

7) Which of the following is largest?

(a) Cl-              (b) S2-              (c) Na+              (d) F

(b) S2-

8) The low solubility of BaSO4 in water is due to

(a) Low dissociation energy              (b) Ionic bonds              (c) High value of lattice energy               (d) None of these

(c) High value of lattice energy

9) For the formation of covalent bond, the difference in the value of electronegativities should be

(a) Equal to or less than 1.7               (b) More than 1.7              (c) 1.7 or more               (d) None

(a) Equal to or less than 1.7

10) For the formation of covalent bond, the difference in the value of electronegativities should be

(a) O2               (b) O2+              (c) O22-               (d) O2-

(b) O2+

11) Amongst Cd, Hg, Ag, Au The I.P. Value order will be

(a) Ag < Cd < Au < Hg                (b) Ag < Au < Cd < Hg               (c) Ag < Hg < Au < Cd                (d) Ag = Cl < Hg < Au

(a) High effective nuclear charge makes high ionization potential.

12) In Slater’s rule, the correct arrangements of electron are according to the extent of their screening

(a) (1s, 2s), 2p                 (b) (1s), (2s, 2p)                (c) (1s), (2s), (2p)                (d) none of these

(b) (1s), (2s, 2p)

13) O–2 or S–2 formation is endothermic because

(a) More stability of O–2 or S–2                                                                                             

(b) More energy gain due to coupling                 

(c) Electrostatic repulsion outweighs the energy gain due to coupling.                 

(d) (a, b) both are correct

(c) Electrostatic repulsion is greater than the coupling energy.

14) The complexing ability of Mn+2, Fe+2, Co+2 is

(a) Mn+2 < Fe+2 < CO+2                (b) Mn+2 < CO+2 < Fe+2               (c) Co+2 < Fe+2 < Mn+2               (d) None of these

(a) Mn+2 < Fe+2 < CO+2 

15) The above phenomenon can be explained as the basis of

(a) ionic potential of metal               (b) electron affinity              (c) ionization potential               (d) None of these

(a) Ionic potential = charge/radius .

16) Highest covalent character is found in which of the following

(a) CaF2              (b) CaCl2             (c) Cal2               (d) CaBr2

(c) I is highly polarisable.

17) Which is correct in the following?

(a) Radius of Cl atom is 0.99 Å, while that of Na+ ion is 1.54 Å.              

(b) Radius of Cl atom is 0.99 Å while that of Na atom is 1.54 Å.             

(c) The radius of Cl atom is 0.95 Å while that of Cl ion is 0.81 Å.              

(d) Radius of Na atom is 0.95 Å, while that of Na+ ion is 1.54 Å.

(b) Radius of Cl atom is 0.99 Å while that of Na atom is 1.54 Å. 

18) In the transition of Cu to Cu2+, there is a decrease in

(a) Atomic number              (b) Atomic mass             (c) Equivalent weight               (d) Number of valency

(d) No. of valency electrons will reduced.

19) The IP1, IP2, IP3, IP4 and IP5 of an element are 7.1, 14.3, 34.5, 46.8, 162.2 eV respectively. The element is likely to be

(a) Na              (b) Si             (c) F               (d) Ca

(b) Characteristic of silicon.

20) The correct order of second ionization potential of carbon, nitrogen, oxygen and fluorine is

(a) C > N > O > F               (b) O > N > F > C            (c) O > F > N > C               (d) F > O > N > C

(c) The ionization potential increases across the period but the second ionization potential of oxygen is highest among them because after the removal of 1e the 2e is to be removed from half-filled orbital which is difficult.

21) Electron affinity is positive when

(a) O is formed from O                                                                (b) O2– is formed from O             

(c) electron affinity in always a positive value               (d) electron affinity is always a negative value

(b) O2– is formed from O

22) Which of the order for ionisation energy is correct

(a) Be > B > C > N > O               (b) B < Be < C < O < N           (c) B < Be < C < N < O               (d) B < Be < N < C < O

(b) Ionisation energy increases across the period but due to stable half filled configuration of VA group, its I.E. is more than VI-A group.

23) Melting points of NaCl, NaBr, NaI and NaF will be in order

(a) NaI > NaBr > NaCl > NaF                    (b) NaF > NaCl > NaBr > NaI            

(c) NaBr > NaF > NaCl > NaI                     (d) NaCl > NaI > NaF > NaBr

(b) NaF > NaCl > NaBr > NaI

24) For a p – block element, its 3d, 3s, 3p and 4s orbitals are completely filled and the differentiating electron goes to the 4p orbital. The element should have its atomic number in the range

(a) 13 – 18               (b) 21 – 26           (c) 31 – 36               (d) 49 – 54

(c) 31– 36 ⇒ Ga to Kr.

25) Calculate the dipole moment of HBr molecule if the bond length is 1.2476

μ = q.r. = (1.6023 × 10–19-C) (1.2476 × 1010 m) = 1.99 × 10–29 cm

= (1.99 × 10-29)/(3.336 × 10-30)

= 5.99 D

26) HBr molecule has internuclear distance of 1.27 × 10–10 m. The electronic charge is 4.8 × 10–10 esu. Observed dipole moment is 1.03 D. find % ionic character of the bond.

% ionic character = [μ(observed)(Theo)] × 100

= (1.03×100)/(1.27×10-8×4.8×10-18)

= (1.03/6.096) × 100

= 16.80

27) The electronegativity of H, N are 2.1, 3.0 respectively. Calculate percentage ionic character of H – N bond.

% Ionic character = 16 (xA – xB) + 3.5 (xA – xB)2

xA = 3.0, xB = 2.1

∴ % IC = 16 × 0.9 + 3.5 × 0.81

= 17.24

28) Calculate effective nuclear charge on a valence electron in a nitrogen atom N = 1s2 2s2 2p3

= 4 × 0.35 + 2 × 0.85 = 3.1

= Z* = Z – σ

= 7 – 3.1 = 3.9

29) Calculate AlredRochow electro negativity in F using Slater rule, rF = 0.7065 Å.

 X = (0.359/r2) Z* + 0.744

= 0.359 × [(0.35×6+0.85×2)/(0.7065)2] + 0.744

= 4.23

30) IP1 and IP2 of Mg are 178 and 348 kcal mole–1. The energy required for the reaction Mg → Mg2+ + 2e– is

I.P. = IP1 + IP2

=178 + 348

= +526 Kcal

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