Q41 / 150 · MHT CET 2025 (25 Apr Shift 2) · Chemistry

Which of the following pair of compounds consists equal number of lone pair of electrons in the valence shell of central atom?

  • Correct answer
    A
    $\mathrm{BrF}_5$ and $\mathrm{XeF}_6$
  • B
    ICl and $\mathrm{H}_2 \mathrm{~S}$
  • C
    $\mathrm{ClF}_3$ and $\mathrm{XeF}_2$
  • D
    $\mathrm{IF}_7$ and $\mathrm{XeF}_4$

Solution

Determining lone pairs on central atoms requires analyzing valence electrons and bonding structure. Compounds with analogous electron domains exhibit equal lone pair counts when their bonding patterns yield identical remaining electron pairs.

$\mathrm{BrF}_5$ and $\mathrm{XeF}_6$: Bromine (Group 17) has 7 valence electrons; bonding to 5 fluorine atoms uses 5 electrons, leaving 2 (1 lone pair). Xenon (Group 18) has 8 valence electrons; bonding to 6 fluorine atoms uses 6 electrons, leaving 2 (1 lone pair). Both thus have one lone pair.

$\mathrm{ICl}$ has 3 lone pairs on iodine (7 valence electrons minus 1 bonding electron), while $\mathrm{H}_2\mathrm{S}$ has 2 lone pairs on sulfur (6 valence electrons minus 2 bonding electrons).

$\mathrm{ClF}_3$ has 2 lone pairs on chlorine; $\mathrm{XeF}_2$ has 3 lone pairs on xenon.

$\mathrm{IF}_7$ has 0 lone pairs; $\mathrm{XeF}_4$ has 2 lone pairs.

Only $\mathrm{BrF}_5$ and $\mathrm{XeF}_6$ share an identical lone pair count.

Final answer: $\boxed{\text{A}}$

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

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