Looking for classes? Ksquare Career Institute, Bengaluru →
Which of the two ions from the list given below have the geometry that is explained by the same hybridization of orbitals: $NO_2^-$, $NO_3^-$, $NH_2^-$, $NH_4^+$, $SCN^-$?
A
$NO_2^-$ and $NH_2^-$
B
$NO_2^-$ and $NO_3^-$
C
$NH_4^+$ and $NO_3^-$
D
$SCN^-$ and $NH_2^-$
Detailed Solution
Find the steric number (σ bonds + lone pairs) of the central atom in each ion.
$NO_2^-$: N has 2 σ bonds + 1 lone pair = 3, so $sp^2$.
$NO_3^-$: N has 3 σ bonds + 0 lone pairs = 3, so $sp^2$.
$NH_2^-$: N has 2 σ bonds + 2 lone pairs = 4, so $sp^3$.
$NH_4^+$: N has 4 σ bonds + 0 lone pairs = 4, so $sp^3$.
$SCN^-$: C has 2 σ bonds + 0 lone pairs = 2, so sp.
Among the given pairs, only $NO_2^-$ and $NO_3^-$ have the same hybridisation ($sp^2$).
$NO_2^-$: N has 2 σ bonds + 1 lone pair = 3, so $sp^2$.
$NO_3^-$: N has 3 σ bonds + 0 lone pairs = 3, so $sp^2$.
$NH_2^-$: N has 2 σ bonds + 2 lone pairs = 4, so $sp^3$.
$NH_4^+$: N has 4 σ bonds + 0 lone pairs = 4, so $sp^3$.
$SCN^-$: C has 2 σ bonds + 0 lone pairs = 2, so sp.
Among the given pairs, only $NO_2^-$ and $NO_3^-$ have the same hybridisation ($sp^2$).