Question

The types of hybrid orbitals of nitrogen in NO\(_2\)\(^+\), NO\(_3\)\(^-\), and NH\(_4\)\(^+\) respectively are

• A. sp, sp\(^2\) and sp\(^3\)
• B. sp, sp\(^3\) and sp\(^2\)
• C. sp\(^2\), sp and sp\(^3\)
• D. sp\(^2\), sp\(^3\) and sp

Hint:

An alternative to the formula is to draw the Lewis structure.

• NO\(_2\)\(^+\): [O=N=O]\(^+\). N has 2 sigma bonds and 0 lone pairs. Steric number = 2 \(\rightarrow\) sp.
• NO\(_3\)\(^-\): Resonance structures with one N=O and two N-O bonds. N has 3 sigma bonds and 0 lone pairs. Steric number = 3 \(\rightarrow\) sp\(^2\).
• NH\(_4\)\(^+\): N is bonded to four H atoms. N has 4 sigma bonds and 0 lone pairs. Steric number = 4 \(\rightarrow\) sp\(^3\).

Answer 1

The Correct Option is A

Step 1: Key Formula for Hybridization.
We can determine the hybridization of the central atom using the steric number method. The steric number (H) is calculated as:
\[ H = \frac{1}{2} [V + M - C + A] \] Where: V = number of valence electrons of the central atom, M = number of monovalent atoms attached, C = magnitude of positive charge, A = magnitude of negative charge.
The hybridization is then determined from H: H=2 \(\rightarrow\) sp, H=3 \(\rightarrow\) sp\(^2\), H=4 \(\rightarrow\) sp\(^3\).
Step 2: Calculating Hybridization for each ion.
For NO\(_2\)\(^+\) (Nitronium ion):
Central atom is Nitrogen (N).
V = 5, M = 0 (Oxygen is divalent), C = 1, A = 0.
\[ H = \frac{1}{2} [5 + 0 - 1 + 0] = \frac{4}{2} = 2 \] H = 2 corresponds to sp hybridization.
For NO\(_3\)\(^-\) (Nitrate ion):
Central atom is Nitrogen (N).
V = 5, M = 0, C = 0, A = 1.
\[ H = \frac{1}{2} [5 + 0 - 0 + 1] = \frac{6}{2} = 3 \] H = 3 corresponds to sp\(^2\) hybridization.
For NH\(_4\)\(^+\) (Ammonium ion):
Central atom is Nitrogen (N).
V = 5, M = 4 (four monovalent H atoms), C = 1, A = 0.
\[ H = \frac{1}{2} [5 + 4 - 1 + 0] = \frac{8}{2} = 4 \] H = 4 corresponds to sp\(^3\) hybridization.
Step 3: Final Answer.
The respective hybridizations for NO\(_2\)\(^+\), NO\(_3\)\(^-\), and NH\(_4\)\(^+\) are sp, sp\(^2\), and sp\(^3\).