- 1(current)
Q. No.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) |
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. |
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 |
Among the following, the correct statements about
| (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) |
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
- 1(current)
Q. No.
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