Question

The hybridization of atomic orbitals of N in $NO_{2}^{+}$, $NO_{3}^{-}$ and $NH_{4}^{+}$ are respectively.

(A) \[sp\], \[s{{p}^{2}}\], \[s{{p}^{3}}\]
(B) \[sp\],\[s{{p}^{3}}\], \[s{{p}^{2}}\]
(C) \[s{{p}^{2}}\], \[sp\], \[s{{p}^{3}}\]
(D) \[s{{p}^{2}}\], \[s{{p}^{3}}\], \[sp\]

Answer 1

Hint: Hybridization of a molecule is determined using the postulates of the Valence band theory and it is basically the phenomenon of mixing of orbitals.

Complete step by step solution:
Hybridization (mixing of orbitals) when two or more than two types of orbitals having less energy difference are combined and form the same number of new orbitals having similar shape and energy but different orientation.
Formula of hybridization; -
$H=\dfrac{(V+M-C+A)}{2}$
Here,
H= hybridization
V = number of valence shell electron of central atom
M= number of monovalent atom surrounding the atom
C= charge on cation
A= charge on anion

-First of all, we have to find out which is the central Atom here so the nitrogen is the central Atom here.

-Therefore, the electronic configuration of nitrogen atom is 1\[{{s}^{2}}\] 2\[{{s}^{2}}\] 2\[{{p}^{3}}\].

-According to the electronic configuration the first (inner) shell will have two electrons and the outermost shell will have two electrons and the outermost shell will have 5 electrons.

-Number of outermost shell electrons is the number of valence shell electrons of Atom.

So, now we are solving for $NO_{2}^{+}$;

By formula of hybridization

 $H=\dfrac{(V+M-C+A)}{2}$
where,

V= 5; C= 1
M= 0; A= 0
Put these values in formula

H =$\dfrac{\left( 5\text{ }+\text{ }0\text{ }-1\text{ }+\text{ }0\text{ } \right)}{2}$
H= $\dfrac{4}{2}$= 2
So, the hybridization = \[sp\]

 Shape = linear

Now solving for $NO_{3}^{-}$;
Again, by formula

V = 5; M= 0
C= 0; A = 1

Put these values in formula
$H=\dfrac{(V+M-C+A)}{2}$
H= \[\dfrac{\left( 5\text{ }+\text{ }0\text{ }-\text{ }0\text{ }+\text{ }1 \right)}{2}\]
H= $\dfrac{6}{2}$ = 3
So, the hybridization = \[s{{p}^{2}}\]

Shape = trigonal planar

Now solving for $NH_{4}^{+}$;
Again, by formula
$H=\dfrac{(V+M-C+A)}{2}$
V= 5; M = 4
C = 1; A = 0
Put these values in formula
H= $\dfrac{\left( 5\text{ }+\text{ }4\text{ }-\text{ }1\text{ }+\text{ }0 \right)}{2}$
H=$\dfrac{8}{2}$= 4
So, the hybridization is =\[s{{p}^{3}}\]

Shape = tetrahedral

So, by solving these the hybridization of atomic orbitals of N in $NO_{2}^{+}$, $NO_{3}^{-}$ and$NH_{4}^{+}$ are respectively \[sp\], \[s{{p}^{2}}\], \[s{{p}^{3}}\].
Therefore, the correct option is option (A) \[sp\], \[s{{p}^{2}}\], \[s{{p}^{3}}\].


Note:
-Check properly the number of valence shell electrons of the central Atom.
-Check properly the number of monovalent atoms.
-Check properly the charge of cation and anions