The type of hybridisation and number of lone pair(s) of electrons of Xe in $\mathrm{XeOF}_{4}$ respectively are

(A) $\mathrm{sp}^{3} \mathrm{~d}^{2}$ and 1

(B) $\mathrm{sp}^{3} \mathrm{~d}$ and 2

(C) $\mathrm{sp}^{3} \mathrm{~d}^{2}$ and 2

(D) $\mathrm{sp}^{3} \mathrm{~d}$ and 1

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Chemical Bonding and Molecular Structure

The type of hybridisation and number of lone pair(s) of electrons of Xe in

XeOF_{4}

respectively are

(A) $sp^{3} d^{2}$ and 1✓ Correct
(B) $sp^{3} d$ and 2
(C) $sp^{3} d^{2}$ and 2
(D) $sp^{3} d$ and 1

Answer: (A)

Solution

XeOF X_{4}

the xenon is surrounded by five bonding regions, four to fluorine and one to oxygen, plus a single lone pair, so the steric number is six. Six electron domains call for

sp X_{3} X_{2}^{23} d X^{2}

hybridisation, and with one position taken by the lone pair the shape is square pyramidal. So the hybridisation is

sp X_{3} X_{2}^{23} d X^{2}

and there is one lone pair on xenon.

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option

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