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