Question

Which of the following set of species have the same hybridisation state but different shapes?

• A. $N{O}_2^{+},N{O}_2,N{O}_2^{-}$
• B. $Cl{O}_4^{-},S{F}_4,Xe{F}_4$
• C. $N{H}_4^{+},H_3{O}^{+},O{F}_2$
• D. $S{O}_4^{2-},P{O}_4^{3-},Cl{O}_4^{-}$

Answer 1

The correct option is C $N{H}_4^{+},H_3{O}^{+},O{F}_2$

Steric factor of the central atom of a molecule/ion
$=\frac{1}{2}\text{(Valence electrons on the central atom}\text{+ number of atoms attached to the central atom by single bonds}\text{+ magnitude of negative charge on the ion}\text{- magnitude of positive charge on the ion)}$

So, for $N{H}_4^{+}=\frac{1}{2}[5+4+0-1]=4$
Hybridisation = $s{p}^3$
Shape is tetrahedral.


For $H_3{O}^{+}=\frac{1}{2}[6+3+0-1]=4$
Hybridisation = $s{p}^3$
Shape is trigonal pyramidal.


For $O{F}_2=\frac{1}{2}[6+2+0-0]$ = 4
Hybridisation = $s{p}^3$
Shape is bent.

So $N{H}_4^{+}$, $H_3{O}^{+}$ and $O{F}_2$ have the same hybridisation state, but different shapes.

In options (a) and (b), the compounds have different hybridisations and different shapes while in option (d), the compounds have the same hybridisation as well as shape.