A) \[C{{H}_{4}}\]
B) \[S{{F}_{4}}\]
C) \[BF_{4}^{-}\]
D) \[NH_{4}^{+}\]
Correct Answer: B
Solution :
When the number of hybrid orbitals, H is 4, the hybridization is sp3. \[H=\frac{1}{2}[V+M-C+A]\] where, V = number of valence electrons of central atom M = number of monovalent atoms C = total positive charge A = negative charge [a]\[C{{H}_{4}},\]\[H=\frac{1}{2}[4+4-0+0]=4,\]thus \[s{{p}^{3}}\]hybridization [b] For\[S{{F}_{4}},\]\[H=\frac{1}{2}[6+4-0+0]=5,\]thus \[s{{p}^{3}}d\,\,hybridization\] [c] For \[BF_{4}^{-},\]\[H=\frac{1}{2}[3+4-0+1]=4,\]thus\[s{{p}^{3}}\]hybridization [d] For\[NH_{4}^{+},\]\[H=\frac{1}{2}[5+4-1+0]=4,\]thus\[s{{p}^{3}}\]hybridization Thus, only in \[S{{F}_{4}},\] the central atom does not have sp3 hybridization.
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