The species, having bond angles of 120° is :
1. PH3
2. ClF3
3. NCl3
4. BCl3
| 1. | \(\mathrm{{SF}_6}\) | 2. | \(\mathrm{NCl_3}\) |
| 3. | \(\mathrm{BCl_3}\) | 4. | \(\mathrm{PH_3}\) |
In which of the following molecules, all atoms are coplanar ?

The pair of electrons in the given carbanion, CH3C≡C-, is present in which of the following orbitals?
| 1. | sp3 | 2. | sp2 |
| 3. | sp | 4. | 2p |
The hybridisations of atomic orbitals of nitrogen in \(\mathrm{NO}^{+}\), \(N O_3^{-}\) and NH3 respectively are:
1.
2.
3.
4.
Which of the following pairs are isoelectronic and isostructural?
1.
2.
3.
4.
In which of the following molecules, all atoms are coplanar?
| 1. | 2. | ||
| 3. | 4. |
In which orbital is the pair of electrons located in the provided carbanion, CH3C≡C– ?
1. sp3
2. sp2
3. sp
4. 2p
| 1. | The H-O-H bond angle in H2O is larger than the H-C-H bond angle in CH4 |
| 2. | The H-C-H bond angle in CH4 is larger than the H-N-H bond angle in NH3 |
| 3. | The H-C-H bond angle in CH4, the H-N-H bond angle in NH3 and the H-0-H bond angle in H2O are all greater than 90o |
| 4. | The H-O-H bond angle in H2O is smaller than the H-N-H bond angle in NH3 |
The correct geometry and hybridization for XeF4 are :
1. Octahedral, sp3d2
2. Trigonal bipyramidal, sp3d3
3. Planar triangle, sp3d3
4. Square planar, sp3d2