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Q. No.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
Column I
Column II
H3O+
Linear
HCCH
Angular
ClO2-
Tetrahedral
NH4+
Trigonal bipyramidal
Pyramidal
Subtopic: Hybridisation | V.S.E.P.R & V.B.T |
89%
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Match the species in Column I with the bond order in Column II.
|
Column I |
Column II |
||
| A. |
NO |
1. |
1.5 |
| B. |
CO |
2. |
2.0 |
| C. |
\(O_2^-\) |
3. |
2.5 |
| D. |
O2 |
4. |
3.0 |
Choose the correct option:
| Options: | A | B | C | D |
| 1. | 3 | 4 | 1 | 2 |
| 2. | 1 | 2 | 3 | 5 |
| 3. | 5 | 4 | 3 | 2 |
| 4. | 4 | 5 | 3 | 2 |
Subtopic: M.O.T |
87%
From NCERT
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Match the items given in Column I with examples given in Column II.
|
Column I |
Column II |
||
| A. |
Hydrogen bond |
1. |
C |
| B. |
Resonance |
2. |
LiF |
| C. |
Ionic solid |
3. |
H2 |
| D. |
Covalent solid |
4. |
HF |
| 5. |
O3 |
Codes:
| Options: | A | B | C | D |
| 1. | 2 | 3 | 4 | 1 |
| 2. | 1 | 2 | 3 | 5 |
| 3. | 5 | 4 | 3 | 2 |
| 4. | 4 | 5 | 2 | 1 |
Subtopic: Resonance & Nature of Compounds |
92%
From NCERT
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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 |
Subtopic: Hybridisation |
94%
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Given below are two statements:
| Assertion (A): | Sodium chloride, formed by the action of chlorine gas on sodium metal, is a stable compound. |
| Reason (R): | Sodium and chloride ions acquire octets in sodium chloride formation. |
| 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: Ionic Bond |
84%
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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 |
88%
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Given below are two statements:
| Assertion (A): | Among the two O-H bonds in H2O molecule, the energy required to break the first O-H bond and the other O-H bond is the same. |
| Reason (R): | The electronic environment around oxygen is the same even after the breakage of one O-H 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. |
Subtopic: Lattice & Hydration Energy |
56%
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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 |
89%
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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: V.S.E.P.R & V.B.T |
78%
From NCERT
NEET - 2020
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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) |
Subtopic: Hybridisation |
83%
From NCERT
NEET - 2020
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