Question

The hybridization of $NO_{3}^{-}$ is the same as the hybridization of _________.
(A) C in $C{{H}_{4}}$
(B) C in ${{C}_{2}}{{H}_{2}}$
(C) C in ${{C}_{2}}{{H}_{4}}$
(D) C in ${{C}_{3}}{{H}_{8}}$

Answer 1

Hint: Draw the structure of $NO_{3}^{-}$ and find out its hybridization. In the options, some hydrocarbons are given. We know the kind of hybridization in alkanes, alkenes and alkynes. Match the corresponding hybridization to get the answer.

Complete answer:
- Let’s start by writing the electronic configuration of nitrogen and oxygen.
\[{}^{7}N=1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}\]
\[{}^{8}O=1{{s}^{2}}2{{s}^{2}}2{{p}^{4}}\]
- The number of valence electrons in nitrogen is 5 and in oxygen is 6. In nitrate, $NO_{3}^{-}$, there is one nitrogen and three oxygen atoms and a -1 charge.
- Therefore, the total number of electrons present in nitrate is $5+\left( 3\times 6 \right)+1=24$
- Therefore, the total number of electrons present in nitrate is $\frac{24}{2}=12$
- So, the structure of nitrate is

- From the structure, we can conclude that nitrogen forms three $\text{ }\!\!\sigma\!\!\text{ -bonds}$ and one $\text{ }\!\!\pi\!\!\text{ -bond}$.
- Therefore, nitrate is $s{{p}^{2}}$ hybridized due to the presence of one double bond which indicates lateral overlapping of 2p orbitals of nitrogen and oxygen.
- Now, let’s have a look at the options.
(A) C in $C{{H}_{4}}$- In methane, carbon is $s{{p}^{3}}$ hybridized.
(B) C in ${{C}_{2}}{{H}_{2}}$- In acetylene, carbon is sp hybridized.
(C) C in ${{C}_{2}}{{H}_{4}}$- In ethene, carbon is $s{{p}^{2}}$hybridized.
(D) C in ${{C}_{3}}{{H}_{8}}$- In propane, carbon is $s{{p}^{3}}$ hybridized.
- Therefore, the hybridization of $NO_{3}^{-}$ is same as the hybridization of C in ethene, \[{{C}_{2}}{{H}_{2}}\].

Therefore, option (C) is the correct answer.

Note:
Remember the presence of double bond indicates lateral overlapping of unhybridized orbitals. Also, for such problems, derive the structure of the compound or molecule or ion first and then go for hybridization.