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. 

${\mathrm{AlCl}}_3$ $+$ ${\mathrm{Cl}}^{-}$ $\to$ ${\mathrm{AlCl}}_4^{-}$

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. \(\operatorname{sp}^{2} \text { to } d s p^{2}\)

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

$1.$ $1\mathrm{s}$ $\mathrm{and}$ $2\mathrm{s}$
$2.$ $1\mathrm{s}$ $\mathrm{and}$ $2{\mathrm{p}}_{\mathrm{x}}$
$3.$ $2{\mathrm{p}}_{\mathrm{y}}$ $\mathrm{and}$ $2{\mathrm{p}}_{\mathrm{y}}$
$4.$ $1\mathrm{s}$ $\mathrm{and}$ $1\mathrm{s}$

CH₄ does not exhibit square planar geometry because:

The shapes of sp, ${\mathrm{sp}}^2$, ${\mathrm{sp}}^3$ 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

${\mathrm{BF}}_3+{\mathrm{NH}}_3\to {\mathrm{F}}_3\mathrm{B}.{\mathrm{NH}}_3$

The correct statement about the above reaction is :

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.

The axial bonds are longer as compared to equatorial bonds in PCl₅ because:

The incorrect statement among the following is: