Identify the pair of molecules in which the c | vedclass

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(A) To determine the hybridization,we calculate the steric number $(SN)$ using the formula: $SN = \frac{1}{2} (V + M - C + A)$,where $V$ is the number

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$H_2O, NH_3$

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(A) To determine the hybridization,we calculate the steric number $(SN)$ using the formula: $SN = \frac{1}{2} (V + M - C + A)$,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.$A$: For $H_2O$,$SN = \frac{1}{2} (6 + 2) = 4$ $(sp^3)$. For $NH_3$,$SN = \frac{1}{2} (5 + 3) = 4$ $(sp^3)$. Both have $sp^3$ hybridization.$B$: For $ClF_3$,$SN = \frac{1}{2} (7 + 3) = 5$ $(sp^3d)$. For $NH_3$,$SN = 4$ $(sp^3)$.$C$: For $XeF_2$,$SN = \frac{1}{2} (8 + 2) = 5$ $(sp^3d)$. For $ClF_5$,$SN = \frac{1}{2} (7 + 5) = 6$ $(sp^3d^2)$.$D$: For $SF_4$,$SN = \frac{1}{2} (6 + 4) = 5$ $(sp^3d)$. For $CF_4$,$SN = \frac{1}{2} (4 + 4) = 4$ $(sp^3)$.Thus,the correct pair is $H_2O$ and $NH_3$.

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

Identify the pair of molecules in which the central atom has the same hybridization.

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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

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