The strong fluoride acceptor SbF5 converts XeF2, XeF4 and XeF6 into XeF+, XeF3+ and XeF5+ respectively. The total number of hybrid orbitals used by Xe in these three cations is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ph154

The strong fluoride acceptor $SbF X_{5}$ converts $XeF X_{2}$ , $XeF X_{4}$ and $XeF X_{6}$ into $XeF X^{+}$ , $XeF X_{3} X^{+}$ and $XeF X_{5} X^{+}$ respectively. The total number of hybrid orbitals used by Xe in these three cations is

XeF X^{+}

: 1 bond pair + 3 lone pairs = 4 pairs,

s p^{3}

(4 orbitals).

XeF X_{3} X^{+}

: 3 + 2 = 5 pairs,

s p^{3} d

(5).

XeF X_{5} X^{+}

: 5 + 1 = 6 pairs,

s p^{3} d^{2}

(6). Total =

4 + 5 + 6 = 15