Q.

Partial hydrolysis of XeF6 forms oxo fluorides (X) and (Y). Oxidation state of Xe in (X) and (Y) is respectively

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a

4,4

b

2,6

c

6,4 

d

6,6

answer is D.

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Detailed Solution

XeF6 + H2XeOF4 (X) + 2HF

XeF6 + 2H2XeO2F2 (Y) + 4HF

In X and Y  oxidation state of Xe = 6

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