Which of the following has maximum number of lone pairs associated with Xe?

We have

${XeO}_{3} :$ valence electrons $= 8 + 3 \times 6 = 26$ long pair of electrons around $Xe = \frac{1}{2} (26 - 3 \times 8) = 1$

${XeF}_{4} :$ valence electrons $8 + 4 \times 7 = 36$ long pair of electrons around $Xe = \frac{1}{2} (36 - 4 \times 8) = 3$

${XeF}_{6} :$ valence electrons $= 8 + 6 \times 7 = 50$ long pair of electrons around $Xe = \frac{1}{2} (50 - 6 \times 8) = 1$

${XeF}_{2} :$ valence electrons $= 8 + 2 \times 7 = 22$ long pair of electrons around

$Xe = \frac{1}{2} (22 - 2 \times 8) = 3$

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