TOPIC WISE REVIEW CPP-II -CB

SINGLE CORRECT CHOICE TYPE

1. Which of the following statements is incorrect for the dipole moment measurement of the compound.
(A) It helps to predict the percentage ionic character in a bond.
(B) It helps to predict the shape of the molecule.
(C) It helps to predict the particular cis trans isomer.
(D) It help to predict the bond energies of all bonds within the molecule.
2. The hybridisation of the central atom of anionic part and cationic part of solid N2O5 are _____ and
____ respectively.
(A) sp and sp2 (B) sp2 and sp3 (C) sp2 and sp2 (D) sp2 and sp
3. Calculate the electronegativity of Cl in Pauling scale
[Given : E.A. (Cl) = 4.0 eV / atom and I.E. (Cl) = 13.0 eV / atom]
(A) 2 (B) 3.03 (C) 3.5 (D) 4.0
4. Which of the following molecule is having shortest bond length of C–O bond.
(A) CH3OH (B) H2CO (C) CO (D) Na2CO3
5. Which of the following Lewis structure of N2O is least adopted to describe the structure of N2O.
    2
 
(A) N = N = O (B) N  N – O (C) N – N  O (D) None of these
6. Molecules with See-Saw shape is
(A) CH2F2 (B) XeOF4 (C) XeO2F2 (D) HgCl2
7. Which of the following compounds are the common product/s obtained in the hydrolysis of XeF6 and
XeF4.
(A) XeO2F2 (B) HF (C) XeO3 (D) Both (B) & (C)
8. Which of the following statement is incorrect for CO molecule.
(A) Intramolecular Lewis acid - base interaction is present
(B) Charge separation is present.
(C) -bond, -bond and back-bond all are present together.
(D) Direction of dipole moment is from 'C' to 'O'
9. Which of the following species is/are isoelectronic to each other?

(I) BH3  C  N (II) CH3 – C  CH (III) N2O (IV) CN 22
(A) I and II only (B) II and III only (C) I, II and III only (D) All four

10 Which of the set of isomers of C6H4Cl2 is having equal dipole moment with C6H5Cl and C6H6
respectively
(A) ortho and meta (B) meta and para
(C) ortho and para (D) para and ortho

11. The compound of Vanadium has magnetic moment of 1.73 BM. The vanadium chloride has the formula:
(A) VCl2 (B) VCl3 (C) VCl4 (D) VCl5

12. In the hypothetical molecule AX2Ln (where A : Central atom, X is Surrounding atom, L is lone pair,
n is the number of lone pair) for which possible value of ‘n’ the dipole moment of the molecule will
be minimum.
(A) Zero (B) 1 (C) 2 (D*) 3
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REASONING TYPE
13. Statement-1 : In sp3d hybridisation for TBP geometry only dz orbital is the most suitable among
2

all other d orbitals.

Statement-2 : All d orbitals are degenerate.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for
statement 1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
MORE THAN ONE CORRECT OPTION
14. Which of the following relationship is correct for the following figure.
O
x y
C
H z F

(A) x = y = z (B) x = y > z (C) x > 120° (D*) z < 120°

15. Which of following statements are correct for HCN molecule.
(A) Two -bonds are perpendicular to each other.
(B) Lone pair lying in the orbital which is perpendicular to both -bonds.
(C) The dipole moment value is non zero.
(D) The direction of dipole moment is from N to H atom.
16. The type of bonds present in CuSO4 . 5H2O.
(A) Covalent bond (B) Co-ordinate bond
(C) Electrovalent bond (D) H-bond

17. Which of the following ionisation energy order is/are correct :

(A) F > Cl– (B) F > Cl (C) Cl > F– (D) Cl– > F–
18. d x 2  y 2 orbital is involved in which of the following hybridisation.
(A) sp3d (TBP) (B) sp3d2 (C) sp3d3 (D) None of these
COMPREHENSION-1
Ionisation energy of the elements increases along the period and decreases along the group. Ionisation
energy of an element is numerically identical with electron affinity of the respective univalent cation.
19. The first ionisation potential (in eV) of N, O atoms are:
(A) 14.6, 13.6 (B) 13.6, 14.6 (C) 13.6, 13.6 (D) 14.6, 14.6
20. For the process
X (g) + e¯  X¯ (g) , H = x
and X¯ (g)  X (g) + e¯, H = y
Select correct alternate:
(A) Ionisation energy of X¯ (g) is y (B) Electron affinity of X(g) is x
(C) Electron affinity of X(g) is –y (D) All are correct statements
21. Sodium forms Na+ ion but it does not form Na2+ because:
(A) Very low value of (IE)1 and (IE)2 (B) Very high value of (IE)1 and (IE)2
(C) Low value of (IE)1 and low value of (IE)2 (D) Low value of (IE)1 and high value of (IE)2

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COMPREHENSION-2
One hydrogen bonding represented as X – H ..... D denotes the interaction between the donor species
D and the acceptor species X – H through the hydrogen end. H atom should remain between the
atoms (i.e. X and D) of more electronegative non metallic elements.
22. B–H–B bridges (in boro hydrides) and W–H–W links in (OC)6 W H W (CO)6 will be considered as
(A) Vander Waals force (B) Hydrogen bond
(C) Ionic Bond (D) None of these
23. In which of the following compound does not have hydrogen bonding.
(A) K2HPO4 (B) K2HPO3 (C) Chloralhydrate (D) H2O
24. The strength of H-bonding order is
(A) H2O > H2O2 > H–F (B) H2O2 > H2O > HF
(C) HF > H2O > H2O2 (D) HF = H2O = H2O2
COMPREHENSION-3
Hybridisation is the chemists attempt to explain the observed molecular shape by constructing hybridised
atomic orbitals with the appropriate inter orbital angles. The molecule for which deviation from normal
bond angle is observed, VSEPR theory suggest electron pair repulsive interaction (p – p > p – bp
> bp – bp ). While from hybridisation point of view that is departure from normal hybridisation because
the angle between any equivalent hybrid orbitals determine the fraction of s and p character of the
hybrid and vice - versa.
25. An element 'A' has outer shell configuration of 5s25p6. If A forms covalent compound AF2 with fluorine.
The orbitals used by 'A' for bonding are
(A) d-orbitals (B) p-orbitals (C) sp-hybridised orbital (D) sp3d hybrid orbital
26. In which species number of lone pair on iodine and number of d - orbitals used in hybridisation by
iodine are same
(A) ICl 2 (B) ICl 2 (C) IF7 (D) ICl 4
27. Which line of properties A, B, C or D correctly listed, properties of the given three compounds.
XeF4 CCl4 Zn2+
(A) Square planar Non-planar µ = 8 BM
(B) µ=0 Tetrahedral Diamagnetic
(C) Non-planar Polar µ = Zero
(D) Polar Square planar Total number of 12 electrons having m
(Magnetic quantum number) = Zero
COMPREHENSION-4
I2 has less solubility in water and it’s solubility increases on adding KI solution. When KI and I2 reacts
then a species 'X' is formed by which solubility of I2 increases.
28. Hybridisation of anionic part of 'X' is
(A) sp2 (B) sp3 (C) sp3d (D) sp3d2
29. Geometry of anionic part of 'X' is
(A) linear (B) T shape (C) pyramidal (D) See-saw
30. Which of the following is/are the correct characteristics of anionic part of 'X' .[Polar and Non polar
nature to be considered on the basis of dipole moment]
(I) Planar (II) Non planar (III) Polar (IV) Nonpolar
(A) I and III (B) I and IV (C) II and III (D) II and IV
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MATRIX MATCH TYPE QUESTIONS
31. Column I Column II
(Number of electron pair (Probable geometry)
around the central atom)
(A) 2 (P) Bent
(B) 3 (Q) Linear
(C) 4 (R) Sea Saw
(D) 5 (S) Trigonal Pyramidal
32. Column I Column II
(A) Ionic compound (P) HCl (vap)
(B) Non-polar covalent compound (Q) SiO2
(C) Polar covalent compound having (R) KNO 3
sigma bond only (S) NO
SUBJECTIVE TYPE
33. Write the total number of FIF angles in IF7 having the bond angle value of 90°.
34. Count the total number of X–O bonds are having equal length in HSO4– and S3O62– respectively. (If
the answer is 5 and 3 then represent as 53)
35. Count the total number of X–O bonds are having equal length in HSO4– and S3O62– respectively. (If
the answer is 5 and 3 then represent as 53)
36. The amount of energy released when one million atoms of iodine in vapour state are converted to I–
ions is 4.9 × 10–13 J. What is the electron affinity of Iodine in eV per atom.
37. Given
Bond energy of F – F bond = 38 Kcal / mol–1
Bond energy of Cl – Cl bond = 58 Kcal / mol–1
Bond energy of Cl – F bond = 61 Kcal / mol–1
Electronegativity of fluorine = 4 eV
Calculate the electronegativity of chlorine atom.
38. The number of lone pair(s) on each F and B atom in BF4 species are ____and ___ respectively..
[If the answer is 1 and 2 then represent as 12]
^
39. The numbers of F I F angles equal to 90° and less than 90° in IF7 are ______ and _____ respectively..
[If the answer is 15 and 9 then represent as 159 ]

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 TOPIC WISE REVIEW CPP-II -CB
ANSWER, HINT AND SOLUTION
SINGLE CORRECT CHOICE TYPE
1. D 2. D 3. B 4. C 5. C 6. C
7. D 8. D 9. D 10. B 11. C 12. A
13. B 14. C, D 15. A, B, C 16. A, B, C, D 17. A, B, C, D
18. B, C 19. A 20. A 21. D 22. D 23. B
24. C 25. D 26. D 27. B 28. C 29. A 30. B
31. (A) Q (B) P, Q (C) P,Q,S (D) P,Q,R 32. (A) R (B) Q (C) P]
SUBJECTIVE TYPE
33. 0010 34. 36 35. 36 36. 3.06 37. 3.22 eV38. 30 39.
105

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ANSWER, HINT AND SOLUTION
SINGLE CORRECT CHOICE TYPE
1. D
[Sol. With the knowledge of dipole moment we can’t tell any thing about the bond energies of all bonds
within the molecule. ]
2. D
[Sol. NO+2 NO3 ]
sp sp2
3. B
4. C
5. C
6. C

F
O
Xe
[Sol. O]
F
See-saw

7. D
8. D
9. D
10. B

Cl Cl Cl

[Sol. Same as
Cl
net = C–Cl Cl
net = 0 net = 0

net m-dichlorobenzene =  2C– Cl   2CCl  2 C2 Cl cos120 = C–Cl ]

11. C
[Sol. n (n  2) = 1.73 V [Ar] 3d3 4s2
for n = 1, µ = 1.73 and n is number of unpaired electron  V4+ must be formed thus VCl4.]
12. A
X

A A A
[Sol. X —A— X
X X X X
X
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µ=0 µ0 µ0 µ=0 ]
REASONING TYPE
13. B
[Sol. In sp3d hybridisation dz2 is involved
All d-orbitals are degenerate this is also true. ]

MORE THAN ONE CORRECT OPTION

14. C, D
15. A, B, C

16. A, B, C, D

H2 O H2 O O O
S
Cu2+ H2 O
[Sol. ]
O
H2 O H2 O O

17. A, B, C, D

18. B, C
COMPREHENSION-2
19. A
[Sol. IE1 N > O half filled configuration ]

20. A
[Sol. Heg = –E.A. ]
21. D
Na IE1
Na  I
E2
[Sol.    Na 2
I E2 >> I E1  Sodium gets ionised only once after which it attains inert gas configuration.
Na+  1s2 2s2 2p6.
COMPREHENSION-3
22. D

(W H B Covalent bridge bond)

[Sol. (OC)6 W–H ]
W(Co)6
23. B

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 H
O O
Cl O
+ +
[Sol. H–P–O¯ K K O¯—P—O—H Cl—C—C—H ]
+ + It will show
O¯ K O¯ K H-bonding Cl O
H
Since it is not connected to
Chloralhydrate
electronegtaive atom & hence
it will not show H-bonding

24. C
COMPREHENSION-4
25. D
[Sol. A = Xe F
Compound = XeF2 Geometry 
Xe sp3d, Linear
(3 p – 2 bp) = 5 ]
F
26. D

I
[Sol. ICl2+ sp3
Cl Cl
Cl

I
ICl2– sp3 d
Cl
F F
F
I F
IF7 sp3d3
F
F F
Cl Cl

I
ICl4– sp3d2 ]
Cl Cl

27. B

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 F F

Xe
[Sol. XeF4 Planar µ = 0 (Non polar)
F F

CCl4 sp3 Tetrahedral (Non planar)

Zn+2  [Ar]3d10 4s0 Diamagnetic U.P.E. = Zero.]
COMPREHENSION-5
28. C
29. A
30. B
MATRIX MATCH TYPE QUESTIONS

31. (A) Q (B) P, Q (C) P,Q,S (D) P,Q,R

32. (A) R (B) Q (C) P]

[Sol. (P) + – (Polar covalent compound )
H Cl
O

(Q) O Si O Si O

O O

O Si O Si O

3-D network like structure each Si is sp3 hybridised  non-polar compound.

O
+
(R) K N ionic compound with  and  bonds.
O O

(S) N O. ]

SUBJECTIVE TYPE
33. 0010
[Sol. IF7  sp3d3 PbP

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 F axial
F
F F
I
Five 90° angle of each axial fluorine atom. ]
F F
F axial

34. 36

35. 36
[Sol. Count the total X–O bonds
O OH O O
S O S—S S O

O O¯ ¯O O¯
Ans. 3 & 6 Due to Resonance ]

36. 3.06
[Sol. I(g)  I– (g) Ereleased = 4.9 × 10–13 J for one million atoms.

4.9 10 13

For one atom, Energy released = = 4.9 × 10–3 × 10–6 = 4.9 × 10–19 Joule
106

4.9 10 19

Ereleased = eV/atm = 3.606 eV / atom. Ans. (1 eV = 1.6 × 10–19 Joule) ]
1.6 1019
37. 3.22 eV

[Sol. Electronegativity of F – Electronegativity of Cl = 0.208 61  38  58

4eV – Electronegativity of Cl = 0.779 eV
Electronegativity of Cl = (4 – 0.779) eV = 3.22 eV Ans.]
38. 30
39. 105
2 4
Z = 16 [Ne] 3s 3p  Two unpaired electron p-block element

 less negative Heg than the element to its right side element in same
period.
2
Z = 82 element is Pb [Xe] 6s2 6p  Two unpaired electrons, p-block element I.E. is more

than above element (exception.) ]

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