At 143k the reaction of XeF₄ with O₂F₂ produces a xenon compound y. The total number of lone pairs electrons present on the whole molecule of y is _____________

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answer is 19.

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Detailed Solution

${XeF}_{4} + O_{2} F_{2} \to {XeF}_{6} + O_{2}$
XeF₄ Contains 3 L.P on each F and 1 L.P on Xe .Therefore total number of lone pairs in molecule y=3× 6+1=19

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