The compound in which the number of $d\pi -p\pi$ bonds are equal to those present in ${\mathrm{ClO}}_4^{-}$

• ${InClO}_4^{-}$, we have 4 bond pairs and 0 lone pair of electrons and so, it has $s^3$ hybridization. Here, we have $3\mathrm{Cl}=\mathrm{O}$ bonds (chlorine is doubly bonded with the oxygen) and $1\mathrm{Cl}-{\mathrm{O}}^{-}$bond. As, both $s$ and $p$ orbitals are occupied so, only d-orbital is available to participate in $\mathrm{\pi }$ - bonding. So, here the number of $\mathrm{d\pi }-\mathrm{p\pi }$ bonds are 3.

Now, from the given options, ${\mathrm{XeF}}_4$ and ${\mathrm{XeF}}_6$ do not form $\mathrm{d\pi }-\mathrm{p\pi }$ bonds (as the fluorine is linked to xenon by single bonds).

• Considering ${\mathrm{XeO}}_3$ and ${\mathrm{XeO}}_4,{\mathrm{XeO}}_3$ has 3 bond pairs and 1 lone pair and ${\mathrm{XeO}}_4$ has 4 bond pair and 0 lone pairs thus, resulting in $s{p}^3$ hybridization.

But, ${\mathrm{XeO}}_3$ has $3\mathrm{Xe}=\mathrm{O}$ bonds along with 1 lone pair of electrons and the number of $d\pi -p\pi$ bonds are 3 which is same as that of ${\mathrm{ClO}}_4^{-}$.

Hence, the compound in which the number of $\mathrm{d\pi }-\mathrm{p\pi }$ bonds are equal to those present in ${\mathrm{ClO}}_4^{-}$is ${\mathrm{XeO}}_3$.