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ExamsJEE AdvancedChemistry

Choose the correct order from the following:

  1. Thermal stability: LiOH < KOH < RbOH
  2. Bond strength: 1s-2s < 1s-2p < 1s-1s < 2p-2p
  3. Bond length: HF < HCl < HBr < HI
  4. Bond angle: NH4(+) < NH3 < NH2(-)

Correct answer: Bond length: HF < HCl < HBr < HI

Solution

Option A: Thermal stability of alkali metal hydroxides increases down the group (cation polarization decreases). LiOH < KOH < RbOH is correct in direction. True. Option B: Bond strength: 1s-1s (H-H) is the strongest sigma bond formed by pure s orbitals. 1s-2s < 1s-2p is reasonable because 2p orbitals are more directional and allow better overlap with 1s. Then 1s-1s should be stronger than 1s-2p (H-H bond ~436 kJ/mol vs C-H ~413 kJ/mol). So order should be 1s-2s < 1s-2p < 1s-1s, and then 2p-2p (like C-C sigma ~347 kJ/mol) < 1s-1s. So the order 1s-2s < 1s-2p < 1s-1s < 2p-2p is WRONG (2p-2p is weaker than 1s-1s). False. Option C: Bond length in HF < HCl < HBr < HI (increasing as halogen size increases). This is correct. Option D: Bond angle order: In NH4(+) all 4 bonds are bonding pairs; angle = 109.5 deg. In NH3: 3 bonds + 1 lone pair; angle = 107 deg. In NH2(-): 2 bonds + 2 lone pairs; angle = 105 deg approx. So NH2(-) < NH3 < NH4(+). The stated order NH4(+) < NH3 < NH2(-) is WRONG. Option C is the only correct statement.

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