The incorrect statement about NH3 and H2O is -

1. The bond angle in NH3 is less than in H2O.

2. Both have distorted tetrahedral geometries.

3. The bond angle in H2O is less than in NH3.

4. Both are sp3 hybridized. 

Subtopic:  Hybridisation |
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AlCl3 + Cl-  AlCl4-

The change in the hybridization of the Al atom in the above reaction is-
1. \(s p^{2} \text { to } s p^{3}\)
2. \(\mathrm{sp}^{3} \text { to } s p^{2}\)
3. \(\mathrm{sp}^{3} \text { to } \mathrm{dsp}{ }^{2}\)
4. \(\operatorname{sp}^{2} \text { to } d s p^{2}\)

Subtopic:  Hybridisation |
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Considering the X-axis as the internuclear axis, a sigma bond will not be formed in:

1. 1s and 2s2. 1s and 2px3. 2py and 2py4. 1s and 1s

Subtopic:  Hybridisation |
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CH4 does not exhibit square planar geometry because-

1. For square planar geometry, 5 bonds are required.

2. Carbon does not have d-orbitals to undergo dsp2 hybridization.

3. Due to steric hindrance CH4 does not exhibit square planar geometry.

4. Carbon does not have d-orbitals to undergo d2sphybridization.

Subtopic:  Hybridisation |
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The shapes of sp, sp2, sp3 orbitals formed due to hybridization of atomic orbitals would be respectively-

1. Trigonal planar, linear, tetrahedral

2. Linear, trigonal planar, tetrahedral

3. Linear, tetrahedral, trigonal planar

4. Tetrahedral, trigonal planar, linear

Subtopic:  Hybridisation |
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BF3+NH3F3B.NH3

Correct statement about the above reaction is -

1. Hybridization of ‘B’ changes to sp2 from sp3 while there is no change in the hybridization of 'N'

2. Hybridization of ‘N’ changes to sp3 from sp2 while there is no change in the hybridization of 'B'

3. Hybridization of ‘N’ changes to sp2 from sp3 while there is no change in the hybridization of 'B'

4. Hybridization of ‘B’ changes to sp3 from sp2 while there is no change in the hybridization of 'N'.

Subtopic:  Hybridisation |
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The total number of sigma and pi bonds in C2His

1. 6 sigma bonds and 1 pi-bond.

2. 3 sigma bonds and 3 pi-bonds.

3. 5 sigma bonds and 1 pi-bond.

4. 2 sigma bonds and 2 pi-bonds.

Subtopic:  Hybridisation |
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The axial bonds are longer as compared to equatorial bonds in PCl5 because -

1. Axial bond pairs suffer more repulsion from the equatorial bond pairs

2. Equatorial bond pairs suffer more repulsion from the axial bond pairs

3. Both 1, and 2

4. None of the above

Subtopic:  Hybridisation |
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The incorrect statement from the following is

1. The sigma bond forms via head-on overlap and pie bond forms via sidewise overlapping of orbitals.

2. s and p orbitals are combined to form a sigma bond as well as a pie bond.

3. Hybrid orbitals only form sigma bonds.

4. Sigma bonds are stronger than pie bonds.

 

 

Subtopic:  Hybridisation |
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Hybridization of C1, C2 , and C3 in the structure given below are
 
 
1. C1 = sp2 , C2 = sp2 , C3 = sp2
 
2. C1 = sp3 , C2 = sp3 , C3 = sp2
 
3. C1 = sp3 , C2 = sp2 , C3 = sp2
 
4. C1 = sp3 , C2 = sp2 , C3 = sp3
Subtopic:  Hybridisation |
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