Which of the following species uses both axia | vedclass

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Question

(C) The hybridization of the central atom determines the $d-$ orbitals involved in bonding.For $PBr^{+}_4$,the central atom $P$ has $4$ bonding pairs

Vedclass · Accepted Answer

$ICl^{-}_4$

Vedclass · Answer

(C) The hybridization of the central atom determines the $d-$ orbitals involved in bonding.For $PBr^{+}_4$,the central atom $P$ has $4$ bonding pairs and $0$ lone pairs,resulting in $sp^3$ hybridization.For $PCl^{-}_4$,the central atom $P$ has $4$ bonding pairs and $1$ lone pair,resulting in $sp^3d$ hybridization.For $ICl^{-}_4$,the central atom $I$ has $4$ bonding pairs and $2$ lone pairs,resulting in $sp^3d^2$ hybridization.In $sp^3d^2$ hybridization,the $d-$ orbitals involved are $d_{z^2}$ and $d_{x^2-y^2}$,which are the axial set of $d-$ orbitals.Therefore,$ICl^{-}_4$ is the correct species.

Molecule	Hybridisation of central atom	Shape
$(a)$ $PCl_5$	$dsp^3$	square pyramidal
$(b)$ $[Ni(CN)_4]^{2-}$	$sp^3$	tetrahedral
$(c)$ $SF_6$	$sp^3d^2$	octahedral
$(d)$ $IF_3$	$dsp^3$	pyramidal

Which of the following species uses both axial set of $d-$ orbitals in the hybridization of the central atom?

Similar Questions

What is the correct mode of hybridization of the central atom in the following compounds: $NO_2^+$,$SF_4$,$PF_6^-$?

$C-H$ bond length is least in

Identify the correct set from the following table:
Molecule Hybridisation of central atom Shape
$(a)$ $PCl_5$ $dsp^3$ square pyramidal
$(b)$ $[Ni(CN)_4]^{2-}$ $sp^3$ tetrahedral
$(c)$ $SF_6$ $sp^3d^2$ octahedral
$(d)$ $IF_3$ $dsp^3$ pyramidal

In $PCl_5$, phosphorus is in $sp^3d$ hybridised state but all its five bonds are not equivalent. Justify your answer with reason.

Which one of the following pairs is isostructural ($i.e.$,having the same shape and hybridization)?

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