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(D) To determine the hybridisation of the central atom,we use the formula: $	ext{Steric Number} = \frac{1}{2} [V + M - C + A]$,where $V$ is the numbe

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$BF_3$

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(D) To determine the hybridisation of the central atom,we use the formula: $	ext{Steric Number} = \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 $NH_3$: $	ext{Steric Number} = \frac{1}{2} [5 + 3] = 4$,which corresponds to $sp^3$ hybridisation.$B$) For $H_2O$: $	ext{Steric Number} = \frac{1}{2} [6 + 2] = 4$,which corresponds to $sp^3$ hybridisation.$C$) For $CH_4$: $	ext{Steric Number} = \frac{1}{2} [4 + 4] = 4$,which corresponds to $sp^3$ hybridisation.$D$) For $BF_3$: $	ext{Steric Number} = \frac{1}{2} [3 + 3] = 3$,which corresponds to $sp^2$ hybridisation.Thus,the central atom in $BF_3$ does not undergo $sp^3$ hybridisation.

The molecule in which the central atom does $NOT$ undergo $sp^3$ hybridisation is

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