Q13 / 150 · MHT CET 2025 (20 Apr Shift 2) · Chemistry

Identify the correct decreasing order of boiling point for hydrides of halogens.

  • A
    $\mathrm{HF}>\mathrm{HCl}>\mathrm{HBr}>\mathrm{HI}$
  • B
    $\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}>\mathrm{HF}$
  • Correct answer
    C
    $\mathrm{HF}>\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}$
  • D
    $\mathrm{HI}>\mathrm{HF}>\mathrm{HBr}>\mathrm{HCl}$

Solution

The boiling points of hydrogen halides are determined by their intermolecular forces.

Hydrogen fluoride (HF) exhibits strong hydrogen bonding due to fluorine's high electronegativity and small size, resulting in a significantly elevated boiling point compared to other hydrogen halides.

Hydrogen chloride (HCl), hydrogen bromide (HBr), and hydrogen iodide (HI) lack hydrogen bonding capabilities. Their boiling points depend on London dispersion forces, which strengthen with increasing molecular size and electron count.

$\text{HI} > \text{HBr} > \text{HCl}$ due to HI's largest molecular size and highest electron count, producing the strongest London dispersion forces. HBr exhibits intermediate strength, while HCl has the weakest forces.

HF's hydrogen bonding gives it the highest boiling point, exceeding even HI's. The complete ordering is $\text{HF} > \text{HI} > \text{HBr} > \text{HCl}$.

This corresponds to option C.

Hence, the correct answer is option: 'C'.

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