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The hybridizations of atomic orbitals of nitrogen in $NO_2^+$, $NO_3^-$ and $NH_4^+$ respectively are
A
sp, $sp^2$ and $sp^3$
B
$sp^2$, sp and $sp^3$
C
sp, $sp^3$ and $sp^2$
D
$sp^2$, $sp^3$ and sp
Explanation
Linear, trigonal planar and tetrahedral.
Detailed Solution
$NO_2^+$ = sp (linear)
$NO_3^-$ = $sp^2$ (trigonal planar)
$NH_4^+$ = $sp^3$ (tetrahedral)
$NO_3^-$ = $sp^2$ (trigonal planar)
$NH_4^+$ = $sp^3$ (tetrahedral)
