Question

$N{H}_4^{+}$ ion is isoelectronic and isostructural with :

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

Answer 1

The correct option is C $C{H}_4$

$H=\frac{1}{2}[V+M-C+A]$
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.
$N{H}_{4+}$
$H=\frac{1}{2}[5+4-1]=4$.
$\Rightarrow s{p}^3$ hybridized state.
Tetradhedral in shape with 10 electrons.
Consider the hybridization of the options:-
A) $H_{3}O+$
$H=\frac{1}{2}[5+4-1]=4$.
$\Rightarrow s{p}^3$ hybridized state.
Trigonal Pyramidal in shape with 10 electrons.
B)$C{H}_3^{-}$
$H=\frac{1}{2}[4+3+1]=4$.
$\Rightarrow s{p}^3$ hybridized state.
Trigonal Pyramidal in shape with 10 electrons.
C) $C{H}_4$
$H=\frac{1}{2}[4+4]=4$.
$\Rightarrow s{p}^3$ hybridized state.
Tetradhedral in shape with 10 electrons.
D)$C{H}_3^{+}$
$H=\frac{1}{2}[4+3-1]=3$.
$\Rightarrow s{p}^2$ hybridized state.
Trigonal Planar in shape with 10 electrons.
Hence option C is the right answer.