Question
What is the oxidation state of Xe in XeOF _4?
What is the oxidation state of Xe in XeOF _4?
D. +6
Let the oxidation state of Xe in XeOF _4 be x. The oxidation state of oxygen (O) is -2 and that of fluorine (F) is -1. Since the molecule is neutral, the sum of the oxidation states of all atoms must be zero. x + (-2) + 4(-1) = 0 x - 2 - 4 = 0 x - 6 = 0 x = +6 The oxidation state of Xe in XeOF _4 is +6. Answer: +6
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