Question

The types of hybrid orbitals of nitrogen in $$N{O}_2^{+}$$, $$N{O}_3^{-}$$ and $$N{H}_4^{+}$$ respectively are expected to be:
A. $$sp,s{p}^{3}ands{p}^2$$
B. $$sp,s{p}^{2}ands{p}^3$$
C. $$s{p}^2\text{, }spands{p}^3$$
D. $$s{p}^2,s{p}^{3}andsp$$

Answer 1

Hint: To find the types of hybrid orbitals involved in hybridization there is a formula. It is as follows.
$$H={\displaystyle \frac{1}{2}}[VC+A]$$
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$$N{O}_2^{+}$$, $$N{O}_3^{-}$$ and $$N{H}_4^{+}$$.
- The structure of $$N{O}_2^{+}$$ 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
$$\begin{array}{rl}& H={\displaystyle \frac{1}{2}}[V+M-C+A]\\ & H={\displaystyle \frac{1}{2}}[5+0-1+0]\\ & H=2\end{array}$$
- Number of hybrid orbital is equal to two means the hybridization is $$sp$$.

- The structure of $$N{O}_3^{-}$$ 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
$$\begin{array}{rl}& H={\displaystyle \frac{1}{2}}[V+M-C+A]\\ & H={\displaystyle \frac{1}{2}}[5+0-0+1]\\ & H=3\end{array}$$
-Number of hybrid orbital is equal to three means the hybridization is $$s{p}^2$$.

-The structure of $$N{H}_4^{+}$$ 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
$$\begin{array}{rl}& H={\displaystyle \frac{1}{2}}[V+M-C+A]\\ & H={\displaystyle \frac{1}{2}}[5+4-1+0]\\ & H=4\end{array}$$
-Number of hybrid orbital is equal to four means the hybridization is $$s{p}^3$$.
-The hybrid orbitals of nitrogen in $$N{O}_2^{+}$$, $$N{O}_3^{-}$$ and $$N{H}_4^{+}$$ are $$sp,s{p}^{2}ands{p}^3$$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 $$s{p}^3{d}^3$$.