Question

The hybridizations of atomic orbitals of nitrogen in $NO _{2}^{+}, NO _{3}^{-}$ and $NH _{4}^{+}$ respectively are

• A. $sp , sp ^{3}$ and $sp ^{2}$
• B. $sp ^{2}, sp ^{3}$ and $sp$
• C. $s p, s p^{2}$ and $s p^{3}$
• D. $sp ^{2}, sp$ and $sp ^{3}$

Answer 1

The Correct Option is C

The hybridization of atomic orbitals of nitrogen in the following species can be determined based on their structure and bonding: 

• NO₂⁺ (Nitronium ion): This ion has a linear structure with two bonds to oxygen atoms. In this case, the nitrogen atom undergoes sp hybridization to form two equivalent bonds with the oxygen atoms. The bond angles are 180°, which is characteristic of an sp hybridized atom. Hence, the hybridization of nitrogen in NO₂⁺ is sp.
• NO₃⁻ (Nitrate ion): The nitrate ion has a trigonal planar structure with three bonds to oxygen atoms. The nitrogen atom in this case undergoes sp² hybridization, with the remaining electron in the p orbital. The bond angles are 120°, which is typical of sp² hybridization. Therefore, the hybridization of nitrogen in NO₃⁻ is sp².
• NH₄⁺ (Ammonium ion): The ammonium ion has a tetrahedral structure, with four hydrogen atoms symmetrically surrounding the nitrogen atom. This requires sp³ hybridization for nitrogen, as it forms four equivalent bonds with the hydrogen atoms. Hence, the hybridization of nitrogen in NH₄⁺ is sp³.

Thus, the hybridizations of nitrogen in these species are:

NO₂⁺: sp, NO₃⁻: sp², NH₄⁺: sp³.

So, the correct answer is:

sp, sp², and sp³.