The ion that has sp^3d^2 hybridization for th | vedclass

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(B) To determine the hybridization,we use the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons of the central at

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$[ICl_4]^-$

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(B) To determine the hybridization,we use the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.For $[IF_6]^-$: $H = \frac{1}{2}(7 + 6 + 1) = 7$,which corresponds to $sp^3d^3$ hybridization.For $[ICl_4]^-$: $H = \frac{1}{2}(7 + 4 + 1) = 6$,which corresponds to $sp^3d^2$ hybridization.For $[ICl_2]^-$: $H = \frac{1}{2}(7 + 2 + 1) = 5$,which corresponds to $sp^3d$ hybridization.For $[BrF_2]^-$: $H = \frac{1}{2}(7 + 2 + 1) = 5$,which corresponds to $sp^3d$ hybridization.Therefore,the ion with $sp^3d^2$ hybridization is $[ICl_4]^-$,which corresponds to option $B$.

List-$I$	List-$II$
$(1)$ $SF_4$	$(A)$ $sp^3d^2$
$(2)$ $IF_5$	$(B)$ $d^2sp^3$
$(3)$ $NO_2^+$	$(C)$ $sp^3d$
$(4)$ $NH_4^+$	$(D)$ $sp^3$
	$(E)$ $sp$

The ion that has $sp^3d^2$ hybridization for the central atom is

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

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Match the species in List-$I$ with their correct hybrid orbitals in List-$II$.

List-$I$ List-$II$

$(1)$ $SF_4$ $(A)$ $sp^3d^2$

$(2)$ $IF_5$ $(B)$ $d^2sp^3$

$(3)$ $NO_2^+$ $(C)$ $sp^3d$

$(4)$ $NH_4^+$ $(D)$ $sp^3$

$(E)$ $sp$

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