The compound in which the number of dpi - ppi | vedclass

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(B) In the perchlorate ion $(ClO_4^{-})$,the central chlorine atom is $sp^3$ hybridized. It forms four $Cl-O$ bonds,out of which three are double bond

Vedclass · Accepted Answer

$XeO_3$

Vedclass · Answer

(B) In the perchlorate ion $(ClO_4^{-})$,the central chlorine atom is $sp^3$ hybridized. It forms four $Cl-O$ bonds,out of which three are double bonds involving $d\pi - p\pi$ back bonding between the filled $2p$ orbitals of oxygen and empty $3d$ orbitals of chlorine. Thus,$ClO_4^{-}$ contains $3$ $d\pi - p\pi$ bonds.Now,let us analyze the given options:$1$. $XeF_4$: It has no $d\pi - p\pi$ bonds.$2$. $XeO_3$: The xenon atom is $sp^3$ hybridized with one lone pair. It forms three $Xe=O$ double bonds,each involving $d\pi - p\pi$ bonding. Thus,it contains $3$ $d\pi - p\pi$ bonds.$3$. $XeO_4$: The xenon atom is $sp^3$ hybridized. It forms four $Xe=O$ double bonds,each involving $d\pi - p\pi$ bonding. Thus,it contains $4$ $d\pi - p\pi$ bonds.$4$. $XeF_6$: It has no $d\pi - p\pi$ bonds.Therefore,the compound with the same number of $d\pi - p\pi$ bonds as $ClO_4^{-}$ is $XeO_3$.

List-$I$ (Molecule / Species)	List-$II$
$(P)$ $NO_2$ (Unpaired electron in)	$(1)$ Vacant $p$-orbital involved in hybridization
$(Q)$ $B_2H_6$	$(2)$ $sp^2$-orbital
$(R)$ $ClO_3$ (Unpaired electron in)	$(3)$ $sp^3$-orbital
$(S)$ $CH_3^+$	$(4)$ $120^\circ$ Bond angle

The compound in which the number of $d\pi - p\pi$ bonds are equal to those present in $ClO_4^{-}$ is:

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