Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Hybridisation of the underline atom changes in

Steric Number

To determine if the hybridisation of a central atom changes during a reaction, we must rely on the concept of the Steric Number ($Z$). The steric number is the sum of the number of sigma bonds (bond pairs) and the number of lone pairs on the central atom.

The value of $Z$ directly dictates the hybridisation: $Z=2$ means $sp$, $Z=3$ means $s{p}^2$, and $Z=4$ means $s{p}^3$. Let's apply this master key to each option.

The Electron-Deficient Aluminum

Consider aluminum hydride, ${\text{AlH}}_3$. Aluminum is a Group 13 element, meaning it has 3 valence electrons. In ${\text{AlH}}_3$, it uses all 3 electrons to form single bonds with hydrogen. There are no lone pairs left.

A steric number of 3 corresponds to $s{p}^2$ hybridisation, giving the molecule a flat, trigonal planar geometry.

When ${\text{AlH}}_3$ reacts with a hydride ion (${\text{H}}^{-}$), the aluminum atom accepts the electron pair from the hydride. The central aluminum now effectively has 4 valence electrons (due to the negative charge) and forms 4 sigma bonds with zero lone pairs.

The steric number has increased to 4! This means the hybridisation has changed from $s{p}^2$ to $s{p}^3$, and the geometry has shifted to tetrahedral. We have found our answer, but let's verify the others.

The Generous Donors

In option B, water (${\text{H}}_2\text{O}$) reacts with a proton (${\text{H}}^{+}$) to form the hydronium ion (${\text{H}}_3{\text{O}}^{+}$). Oxygen in water has 2 bond pairs and 2 lone pairs ($Z=2+2=4⟹s{p}^3$). When it donates one of its lone pairs to the proton, that lone pair simply becomes a bond pair. The new configuration is 3 bond pairs and 1 lone pair. The sum is still 4 ($Z=3+1=4$). The hybridisation remains $s{p}^3$.

Similarly, in option C, ammonia (${\text{NH}}_3$) has 3 bond pairs and 1 lone pair ($Z=3+1=4⟹s{p}^3$). Upon reacting with a proton to form ammonium (${\text{NH}}_4^{+}$), the lone pair is converted into a bond pair. The new configuration is 4 bond pairs and 0 lone pairs. The sum is still 4 ($Z=4+0=4$). The hybridisation remains $s{p}^3$.

In both water and ammonia, the central atom acts as a Lewis base, donating a lone pair to form a coordinate bond. This action changes the shape of the molecule but preserves the total steric number, keeping the hybridisation constant. Only in the case of ${\text{AlH}}_3$ does the central atom accept a new electron pair without sacrificing an existing one, thereby increasing its steric number and changing its hybridisation. Therefore, Option (a) is the correct answer.