Question

The types of hybrid orbitals of nitrogen in $NO_{2}^{+}$, $NO_{3}^{-}$ and $NH_{4}^{+}$ respectively are expected to be:
A. $sp,\text{ }s{{p}^{3}}\text{ }and\text{ }s{{p}^{2}}$
B. $sp,\text{ }s{{p}^{2}}\text{ }and\text{ }s{{p}^{3}}$
C. $s{{p}^{2}}\text{, }sp\text{ }and\text{ }s{{p}^{3}}$
D. $s{{p}^{2}},\text{ }s{{p}^{3}}\text{ }and\text{ }sp$

Answer 1

To find the types of hybrid orbitals involved in hybridization there is a formula. It is as follows.
$H=\dfrac{1}{2}[VC+A]$
Where H is the number of orbitals involved in hybridization
V is the number of valence electrons of central atom
M is the number of monovalent atoms attached to central atom
C = charge on the cation
A = charge on the anion

Complete step by step answer:
- The given molecules are$NO_{2}^{+}$, $NO_{3}^{-}$ and $NH_{4}^{+}$.
- The structure of $NO_{2}^{+}$ is as follows.

-From the above structure we can say that
V= number of valence electrons of central atom = 5
M = number of monovalent atoms attached to central atom = 0
C = charge on the cation = 1
A = charge on the anion = 0
- Therefore

& H=\dfrac{1}{2}[V+M-C+A] \\\ & H=\dfrac{1}{2}[5+0-1+0] \\\ & H=2 \\\ \end{aligned}$$ \- Number of hybrid orbital is equal to two means the hybridization is $$sp$$. \- The structure of $$NO_{3}^{-}$$ is as follows.  -From the above structure we can say that V= number of valence electrons of central atom = 5 M = number of monovalent atoms attached to central atom = 0 C = charge on the cation = 0 A = charge on the anion = 1 \- Therefore $$\begin{aligned} & H=\dfrac{1}{2}[V+M-C+A] \\\ & H=\dfrac{1}{2}[5+0-0+1] \\\ & H=3 \\\ \end{aligned}$$ -Number of hybrid orbital is equal to three means the hybridization is $$s{{p}^{2}}$$. -The structure of $$NH_{4}^{+}$$ is as follows.  -From the above structure we can say that V= number of valence electrons of central atom = 5 M = number of monovalent atoms attached to central atom = 4 C = charge on the cation = 1 A = charge on the anion = 0 \- Therefore $$\begin{aligned} & H=\dfrac{1}{2}[V+M-C+A] \\\ & H=\dfrac{1}{2}[5+4-1+0] \\\ & H=4 \\\ \end{aligned}$$ -Number of hybrid orbital is equal to four means the hybridization is $$s{{p}^{3}}$$. -The hybrid orbitals of nitrogen in $$NO_{2}^{+}$$, $$NO_{3}^{-}$$ and $$NH_{4}^{+}$$ are $$sp,\text{ }s{{p}^{2}}\text{ }and\text{ }s{{p}^{3}}$$respectively. **So, the correct answer is “Option B”.** **Note:** If the number of orbitals involved in hybridization are 4 then the hybridization of the central atom is $$s{{p}^{3}}d$$. If the number of orbitals involved in hybridization are 5 then the hybridization of the central atom is $$s{{p}^{3}}{{d}^{2}}$$. If the number of orbitals involved in hybridization are 6 then the hybridization of the central atom is $$s{{p}^{3}}{{d}^{3}}$$.