Question: 73: Amongst the following which one will have maximum ‘lone pair – lone pair’ electron repulsions?
(1) \mathrm{SF}_{4}
(2) \mathrm{XeF}_{2}
(3) \mathrm{ClF}_{3}
(4) \mathrm{IF}_{5}
Answer: Option (2)
Explanation:
According to VSEPR theory, the strength of electron pair repulsions follows the order lone pair–lone pair > lone pair–bond pair > bond pair–bond pair.
The molecule \mathrm{XeF}_{2} has a total of five electron pairs around xenon, out of which three are lone pairs and two are bond pairs.
Thus, \mathrm{XeF}_{2} has three lone pairs on the central atom, leading to significant lone pair–lone pair repulsions.
In \mathrm{SF}_{4}, there is only one lone pair on sulfur.
In \mathrm{ClF}_{3}, there are two lone pairs on chlorine.
In \mathrm{IF}_{5}, there is only one lone pair on iodine.
Since \mathrm{XeF}_{2} has the maximum number of lone pairs on the central atom,
it exhibits the maximum lone pair–lone pair electron repulsions.