Concept:Determine the hybridization of the central atom to see which one utilizes \( \text{sp} \) hybrid orbitals to bond with the unhybridized p-orbitals of a halogen.
Formula:$$ \text{Steric Number (SN)} = 2 \implies \text{sp hybridization} $$
Solution:- Beryllium in \( \text{BeCl}_2 \) forms 2 single bonds with Chlorine and has 0 lone pairs.
- A steric number of 2 means Be is \( \text{sp} \) hybridized.
- Chlorine uses its unhybridized 3p orbital to bond. Thus, the bond is a direct \( \text{sp - p} \) overlap.
Why other options are incorrect:- Option B: Boron in \( \text{BF}_3 \) is \( \text{sp}^2 \) hybridized.
- Option C & Option D: Oxygen in water and Nitrogen in ammonia are \( \text{sp}^3 \) hybridized (and hydrogen uses s-orbitals, not p-orbitals).
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