If the central atom is of third row or below this in the periodic table, then lone pair will occupy a stereochemically inactive s-orbital and bonding will be through almost pure p-orbitals and bond an

Question

If the central atom is of third row or below this in the periodic table, then 
lone pair will occupy a stereochemically inactive s-orbital and bonding will be 
through almost pure p-orbitals and bond angles are nearly 90°, if the 
substituent's electronegativity value is $\le$ 2.5.

Prashant Kotian · Accepted Answer

Correct answer: D. Step 1: Identify the nature of intermolecular forces in HF, NH3, and H2S. HF and NH3 exhibit strong intermolecular hydrogen bonding, whereas H2S only exhibits weaker van der Waals forces. Step 2: Compare the hydrogen bond strengths of HF and NH3. Since fluorine is more electronegative than nitrogen, HF forms much stronger hydrogen bonds than NH3, giving HF a higher boiling point. Step 3: This results in the boiling point order: HF > NH3 > H2S, corresponding to option (a).

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