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13/11/2021 12:11 pm
Topic starter
A compound of Xe and F is found to have 53.5% of Xe. What is oxidation number of Xe in this compound?
(a) –4
(b) 0
(c) +4
(d) +6
1 Answer
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13/11/2021 12:14 pm
Correct answer: (d) +6
Explanation:
Xe = 53.5 % ∴ F = 46.5%
Relative number of atoms Xe
= \frac{53.5}{131.2} = 0.4 and
F = \frac{46.5}{19} = 2.4
Simple ratio Xe = 1 and F = 6 ; Molecular formula is XeF6
O.N. of Xe is +6