What is the type of hybridization of the cent | vedclass

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(A) The hybridization is determined by the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of

Vedclass · Accepted Answer

$NH_3: sp^3, PCl_5: sp^3d, SF_6: sp^3d^2, CCl_4: sp^3$

Vedclass · Answer

(A) The hybridization is determined by the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$1. NH_3: H = \frac{1}{2}(5 + 3) = 4 \rightarrow sp^3$ hybridization.$2. PCl_5: H = \frac{1}{2}(5 + 5) = 5 \rightarrow sp^3d$ hybridization.$3. SF_6: H = \frac{1}{2}(6 + 6) = 6 \rightarrow sp^3d^2$ hybridization.$4. CCl_4: H = \frac{1}{2}(4 + 4) = 4 \rightarrow sp^3$ hybridization.Therefore,the correct hybridization sequence is $NH_3: sp^3, PCl_5: sp^3d, SF_6: sp^3d^2, CCl_4: sp^3$.

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

What is the type of hybridization of the central atom in the following molecules: $NH_3, PCl_5, SF_6, CCl_4$?

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Molecule Hybridisation of central atom Shape
$(a)$ $PCl_5$ $dsp^3$ square pyramidal
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$(c)$ $SF_6$ $sp^3d^2$ octahedral
$(d)$ $IF_3$ $dsp^3$ pyramidal

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