$\underline{X}$ and $\underline{Y}$ are the two covalent molecules in which the hybridisation of the central…
$\underline{X}$ and $\underline{Y}$ are the two covalent molecules in which the hybridisation of the central atoms is same, but shapes are different. $\underline{X}$ and $\underline{Y}$ are
$\mathrm{XeF}_4, \mathrm{NH}_3$
$\mathrm{XeF}_2, \mathrm{PF}_5$
$\mathrm{BF}_3, \mathrm{H}_2 \mathrm{O}$
$\mathrm{CH}_4, \mathrm{BeCl}_2$
Solution
The hybridisation in $\mathrm{XeF}_2$ and $\mathrm{PF}_5$ both is $s p^3 d$ thus they must have trigonal bi-pyramidal geometry. But, due to the presence of 3 lone pair and 2 bond pair shape of $\mathrm{XeF}_2$ get distorted and the molecule appears linear. But in case of $\mathrm{PCl}_5$ this is not so.