Q5 / 225 · JEE Main 2004 · Chemistry

The correct order of bond angles (smallest first) in $\mathrm{H}_2 \mathrm{~S}, \mathrm{NH}_3, \mathrm{BF}_3$ and $\mathrm{SiH}_4$ is

  • A
    $\mathrm{H}_2 \mathrm{~S} < \mathrm{SiH}_4 < \mathrm{NH}_3 < \mathrm{BF}_3$
  • B
    $\mathrm{H}_2 \mathrm{~S} < \mathrm{NH}_3 < \mathrm{BF}_3 < \mathrm{SiH}_4$
  • Correct answer
    C
    $\mathrm{H}_2 \mathrm{~S} < \mathrm{NH}_3 < \mathrm{SiH}_4 < \mathrm{BF}_3$
  • D
    $\mathrm{NH}_3 < \mathrm{H}_2 \mathrm{~S} < \mathrm{SiH}_4 < \mathrm{BF}_3$

Solution

$\mathrm{H}_2 \mathrm{~S} \longrightarrow \mathrm{sp}^3$ $\mathrm{NH}_3 \longrightarrow \mathrm{sp}^3$ $\mathrm{BF}_3 \longrightarrow \mathrm{sp}^2$ $\mathrm{SiH}_4 \longrightarrow \mathrm{sp}^3$

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

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