- 1(current)
Q. No.| Assertion (A): | C3O2 is a linear structure. |
| Reason (R): | Each C atom in C3O2 is sp hybridized. |
| 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. | Both (A) and (R) are false. |
| Assertion (A): | On heating PCl5, it readily converts into PCl3 and Cl2 |
| Reason (R): | The axial P-Cl bond is larger than the equitorial P-Cl bond |
| 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. | Both (A) and (R) are False. |
Select the correct statement regarding among the folowing.
a. Xe is hybridised.
b. 4 bond pairs and 1 lone pair are present around Xe.
c. Shape of is square pyramidal.
d. Contains bonds.
1. a and b
2. a, and d
3. a, b and d
4. b, c and d
Match the coordination number and type of hybridization with the distribution of hybrid orbitals in space based on Valence bond theory.
| Coordination number and type of hybridisation | Distribution of hybrid orbitals in space |
| (a) 4, sp3 | (i) Trigonal bipyramidal |
| (b) 4, dsp2 | (ii) Octahedral |
| (c) 5, sp3d | (iii) Tetrahedral |
| (d) 6, d2sp3 | (iv) Square planar |
| (a) | (b) | (c) | (d) | |
| 1. | (ii) | (iii) | (iv) | (i) |
| 2. | (iii) | (iv) | (i) | (ii) |
| 3. | (iv) | (i) | (ii) | (iii) |
| 4. | (iii) | (i) | (iv) | (ii) |
Match the shape of molecules in Column I with the type of hybridization in Column II.
|
Column I |
Column II |
|
A. Tetrahedral |
1. sp2 |
|
B. Trigonal |
2. sp |
|
C. Linear |
3. sp3 |
Codes
| Options: | A | B | C |
| 1. | 3 | 1 | 2 |
| 2. | 1 | 2 | 3 |
| 3. | 5 | 4 | 3 |
| 4. | 4 | 5 | 3 |
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 |
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 |
Match the compounds of Xe in column I with the molecular structure in column II.
| Column-I | Column-II |
| (a) | (i) Square planar |
| (b) | (ii) Linear |
| (c) | (iii) Square pyramidal |
| (d) | (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) |
Match the column :
|
|
I |
|
II |
|
(P) |
(I) |
Linear; 2 bond pair and 3 lone pair |
|
|
(Q) |
(II) |
Tetrahedral, 4 bond pair and no lone pair |
|
|
(R) |
(III) |
Tetrahedral; 3 bond pair and 1 lone pair |
|
|
(S) |
(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
Match the compounds given in Column I with the hybridization and shape given in Column II and mark the correct option.
| Column I | Column II |
| A. XeF6 | 1. Distorted octahedral |
| B. XeO3 | 2. Square planar |
| C. XeOF4 | 3. Pyramidal |
| D. XeF4 | 4. Square pyramidal |
| A | B | C | D | |
| 1. | 1 | 2 | 4 | 3 |
| 2. | 4 | 3 | 1 | 2 |
| 3. | 4 | 1 | 2 | 3 |
| 4. | 1 | 3 | 4 | 2 |
- 1(current)
Q. No.
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