Question

$N{H}_2^{-}$ is isoelectronic and isostructural with:

• A. $B{H}_3$
• B. $C{H}_4$
• C. $H_{2}O$
• D. $H_3{O}^{+}$

Answer 1

Answer: (C)

$H_{2}O$

$H=\frac{1}{2}[V+M-C+A]$
where,
H= Number of orbitals involved in hybridization.
V=Valence electrons of central atom.
M- Number of monovalent atoms linked to central atom.
C= Charge of cation.
A= Charge of anion.
$N{H}_2^{-}$
$H=\frac{1}{2}[5+2+1]=4$.
$\Rightarrow s{p}^3$ hybridized state.
Bent V shape in shape with 10 electrons.
Consider the hybridization of the options:-
A) $B{H}_3$
$H=\frac{1}{2}[3+3]=3$.
$\Rightarrow s{p}^2$ hybridized state.
Trigonal Planar in shape with 8 electrons.
B)$C{H}_4$
$H=\frac{1}{2}[4+4]=4$.
$\Rightarrow s{p}^3$ hybridized state.
Tetradhedral in shape with 10 electrons.
C) $H_{2}O$
$H=\frac{1}{2}[6+2]=4$.
$\Rightarrow s{p}^3$ hybridized state.
Bent V shape in shape with 10 electrons.
D)$H_3{O}^{+}$
$H=\frac{1}{2}[6+3-1]=4$.
$\Rightarrow s{p}^3$ hybridized state.
Trigonal Pyramidal in shape with 10 electrons.
Hence option C is the right answer.