Arrange the hydrides $\mathrm{NH}_3, \mathrm{HF}, \mathrm{H}_2 \mathrm{O}, \mathrm{HCl}$ in the increasing…

Arrange the hydrides $\mathrm{NH}_3, \mathrm{HF}, \mathrm{H}_2 \mathrm{O}, \mathrm{HCl}$ in the increasing order of their boiling points
  1. $\mathrm{HF} < \mathrm{NH}_3 < \mathrm{HCl} < \mathrm{H}_2 \mathrm{O}$
  2. $\mathrm{H}_2 \mathrm{O} < \mathrm{HF} < \mathrm{HCl} < \mathrm{NH}_3$
  3. $\mathrm{NH}_3 < \mathrm{HCl} < \mathrm{H}_2 \mathrm{O} < \mathrm{HF}$
  4. $\mathrm{HCl} < \mathrm{NH}_3 < \mathrm{HF} < \mathrm{H}_2 \mathrm{O}$

Solution

(d$\mathrm{H}_2 \mathrm{O}$, HF and $\mathrm{NH}_3$ have hydrogen bonding as the intermolecular forces while $\mathrm{HCl}$ has weaker dipoledipole interaction. Thus, $\mathrm{HCl}$ has the lowest boiling point. Among $\mathrm{H}_2 \mathrm{O}$, HF, and $\mathrm{NH}_3 \mathrm{H}_2 \mathrm{O}$ has the highest boiling point due to extensive network of hydrogen bonds. Among $\mathrm{NH}_3$ and $\mathrm{HF}$, $\mathrm{HF}$ forms stronger hydrogen bonds due to very high electronegativity of fluorine. Thus, the order of boiling point will be :- $ \mathrm{HCl} < \mathrm{NH}_3 < \mathrm{HF} < \mathrm{H}_2 \mathrm{O} $

Asked in: AP EAMCET 2023 (18 May Shift 2)

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