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JEE Advanced levelSingle correct optionDifficulty 3 of 3ph153

$XeF X_{4}$ accepts a fluoride ion from a strong fluoride donor to form a planar anion. The $F - Xe - F$ angle between adjacent fluorines in this anion is

A $10 8^{\circ}$
B $7 2^{\circ}$
C $9 0^{\circ}$
D $12 0^{\circ}$

Answer: B

XeF X_{4} + F X^{-} XeF X_{5} X^{-}

, which has 5 bond pairs and 2 lone pairs (

s p^{3} d^{3}

). The lone pairs are axial, so five F atoms lie in a regular pentagon around Xe. The angle between adjacent F atoms =

36 0^{\circ} /5 = 7 2^{\circ}

.

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