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In which of the following molecules/ions $BF_3$, $NO_2^-$, $NH_2^-$ and $H_2O$, the central atom is $sp^2$ hybridized?
A
$BF_3$ and $NO_2^-$
B
$NO_2^-$ and $NH_2^-$
C
$NH_2^-$ and $H_2O$
D
$NO_2^-$ and $H_2O$
Detailed Solution
Steric number = σ bonds + lone pairs on the central atom; 3 means $sp^2$ and 4 means $sp^3$.
$BF_3$: 3 σ bonds + 0 lone pairs = 3, so $sp^2$.
$NO_2^-$: 2 σ bonds + 1 lone pair = 3, so $sp^2$.
$NH_2^-$: 2 σ bonds + 2 lone pairs = 4, so $sp^3$.
$H_2O$: 2 σ bonds + 2 lone pairs = 4, so $sp^3$.
Hence the central atom is $sp^2$ hybridised in $BF_3$ and $NO_2^-$.
$BF_3$: 3 σ bonds + 0 lone pairs = 3, so $sp^2$.
$NO_2^-$: 2 σ bonds + 1 lone pair = 3, so $sp^2$.
$NH_2^-$: 2 σ bonds + 2 lone pairs = 4, so $sp^3$.
$H_2O$: 2 σ bonds + 2 lone pairs = 4, so $sp^3$.
Hence the central atom is $sp^2$ hybridised in $BF_3$ and $NO_2^-$.