Match the compounds of Xe in Column I with the molecular structure in Column II.

Column-I Column-II
(a) XeF2 (i) Square planar
(b) XeF4 (ii) Linear
(c) XeO3 (iii) Square pyramidal
(d) XeOF4 (iv) Pyramidal
 
(a) (b) (c) (d)
1. (ii) (i) (iii) (iv)
2. (ii) (iv) (iii) (i)
3. (ii) (iii) (i) (iv)
4. (ii) (i) (iv) (iii)
Subtopic:  V.S.E.P.R & V.B.T |
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Given below are two statements: 

Assertion (A): Though the central atom of both NH3 and H2O molecules are sp3 hybridised, yet H-N-H bond angle is greater than that of H-O-H.
Reason (R): This is because nitrogen atom has one lone pair and oxygen atom has two lone pairs.

1. Both (A) and (R) are true and (R) is the correct explanation of (A).
2. Both (A) and (R) are true but (R) is not the correct explanation of (A).
3. (A) is true but (R) is false.
4. (A) is false but (R) is true.
Subtopic:  V.S.E.P.R & V.B.T |
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The incorrect statements among the following are-

(a) NaCl being an ionic compound is a good conductor of electricity in the solid-state
(b) In canonical structure, there is a difference in the arrangement of atoms
(c) Hybrid orbitals form stronger bonds than pure orbitals
(d) VSEPR theory can explain the square planar geometry of XeF4
 
1. (a) and (b) only 2. (b) and (c) only
3. (c) and (d) only 4. (b) and (d) only
Subtopic:  Resonance & Nature of Compounds | V.S.E.P.R & V.B.T |
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Among the following, the correct statements about CO32

(a)

The hybridisation of the central atom is sp3

(b)

Its resonance structure has one C-O single bond and two C=O double bonds

(c)

The average formal charge on each oxygen atom is 0.67 units

(d)

All C-O bond lengths are equal

Choose the correct option :

1. (a), (b)

2. (b), (c)

3. (c), (d)

4. (b), (d)

Subtopic:  Octate, Isoelectronic Species, Lewis Structure & Formal Charge | Resonance & Nature of Compounds | V.S.E.P.R & V.B.T |
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Match the species in Column I with the shape in Column II.

Column I

Column II

A. H3O+

1. Linear

B. HCCH

2. Angular

C. ClO2-

3. Tetrahedral

D. NH4+

4. Trigonal bipyramidal

5. Pyramidal

Codes

Options:  A   B   C   D 
1. 5  1  2  3
2. 1  2  3  5
3. 5  4  3  2
4. 4  5  3  2

Subtopic:  Hybridisation | V.S.E.P.R & V.B.T |
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Match the species in Column I with the type of hybrid orbitals in Column II.

Column I

Column II

A. SF4

1. sp3d2

B. IF5

2. d2sp3

C. NO2+

3. sp3d

D. NH4+

4. sp3

5. sp

Codes:

Options:  A   B   C   D 
1. 3 1 5 4
2. 1 2 3 5
3. 5 4 3 2
4. 4 5 3 2

       

Subtopic:  Hybridisation | V.S.E.P.R & V.B.T |
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Match the compounds of Xe in column I with the molecular structure in column II.

Column-I Column-II
(a) XeF2 (i) Square planar
(b) XeF4 (ii) Linear
(c) XeO3 (iii) Square pyramidal
(d) XeOF4 (iv) Pyramidal
 
(a) (b) (c) (d)
1. (ii) (i) (iii) (iv)
2. (ii) (iv) (iii) (i)
3. (ii) (iii) (i) (iv)
4. (ii) (i) (iv) (iii)
Subtopic:  Hybridisation | V.S.E.P.R & V.B.T |
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Match the column :

 

I

 

II

(P)

BH4-

(I)

Linear; 2 bond pair and 3 lone pair

(Q)

ICl2+

(II)

Tetrahedral, 4 bond pair and no lone pair

(R)

ICl2-

(III)

Tetrahedral; 3 bond pair and 1 lone pair

(S)

ICl4-

(IV)

V shaped; 2 bond pair and 2 lone pair

 

 

(V)

Square planar; 4 bond pair and 2 lone pair

 

 

(VI)

Square planar; 4 bond pair and no lone pair

1. P = II, Q = IV, R = III, S = VI

2.  P = II, Q = IV, R = I, S = V

3.  P = II, Q = I, R = V, S = VI  

4.  P = II, Q = I, R = III, S = IV

Subtopic:  Hybridisation | V.S.E.P.R & V.B.T |
 82%
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