Concept:The hybridization of an atom is dictated by its steric number (the sum of sigma bonds and lone pairs). For nitrogen forming consecutive double bonds, the geometry is forced to be linear.
Formula:$$ \text{Steric Number} = (\sigma \text{ bonds}) + (\text{lone pairs}) $$
Solution:- In the given cumulated double-bond structure (similar to an isocyanate group, \( -\text{N}=\text{C}=\text{O} \)), the Nitrogen atom is double-bonded to the adjacent Carbon atom and double-bonded to the aromatic ring (or another group acting as a cation/anion depending on resonance).
- Based on the explanatory notes provided in the source key, this specific Nitrogen atom is forming exactly 2 sigma (\( \sigma \)) bonds and 2 pi (\( \pi \)) bonds.
- Because it has only 2 sigma bonding domains (and the pi bonds utilize the unhybridized p-orbitals), its steric number is 2.
- A steric number of 2 corresponds to \( \text{sp} \) hybridization, giving that segment of the molecule a linear geometry.
Why other options are incorrect:- Option B: Requires 3 sigma domains (e.g., a standard double bond and a lone pair, as in a typical imine).
- Option C: Requires 4 sigma domains (e.g., standard single bonds as in ammonia).
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