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Q.

${XeF}_{6}$ on partial hydrolysis gives

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a

${XeOF}_{4}$ only

b

${XeO}_{2} F_{2}$ only

c

Both ${XeOF}_{4}$ and ${XeO}_{2} F_{2}$

d

${XeOF}_{4}$ or ${XeO}_{2} F_{2}$

answer is D.

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Detailed Solution

${XeF}_{6} + H_{2} O \to {XeOF}_{4} + 2 HF .$

${XeF}_{6} + H_{2} O \to {XeO}_{2} F_{2} + 4 HF .$

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