Question

The species in which the $N$ atom is in a state of $sp$ hybridization is :

• A. $NO^{+}_2$
• B. $NO^{-}_2$
• C. $NO^{-}_3$
• D. $NO_2$

Answer 1

The Correct Option is A

To determine the hybridization of the nitrogen atom in different species, we need to understand the underlying principles of hybridization. Hybridization involves the mixing of atomic orbitals to form new hybrid orbitals, which explain the molecular geometry of covalent compounds.

1. For the NO_2^+ species:
   • The total number of valence electrons for nitrogen is 5 and for each oxygen is 6. The positive charge indicates a loss of one electron.
   • Total electrons = \(5 + 2\times6 - 1 = 16\).
   • Nitrogen forms two sigma bonds with oxygen atoms and carries a positive charge without any lone pair, making the number of regions of electron density around the nitrogen atom 2.
   • This corresponds to sp hybridization, indicating a linear shape.
2. Ruling out other options:
   • For NO_2^-\: The nitrogen carries a lone pair along with sigma bonds, leading to sp^2 hybridization.
   • For NO_3^-\: The nitrogen atom in nitrate forms three equivalent bonds with oxygens and it has sp^2 hybridization due to resonance and three regions of electron density.
   • For NO_2: The nitrogen in nitrite has one lone pair and two bonds, which gives rise to sp^2 hybridization.

Thus, the nitrogen atom in NO_2^+ exhibits sp hybridization, making this the correct answer.