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.
How does the measurement of bond angles in methane (\(\text{CH}_4\)) rule out a square planar structure?
| 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 d2sp3 hybridization. |
The shapes of sp, , 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
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. \(\operatorname{sp}^{2} \text { to } d s p^{2}\)
The 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'. |
The total number of sigma and pi bonds in C2H4 is:
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.
Considering the X-axis as the internuclear axis, a sigma bond will not be formed in:
Hybridizations of C1, and C2 in the structure given below are
4. None of the above