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Electronic Configuration and Oxidation States

n factor for ...

Question

$n$ factor for $H_{2} S$ during its oxidation to $S O_{2}$ is:

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Solution

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$H_{2} S \to S O_{2}$
-2 +4
So, n-factor of $H_{2} S = 4 - (- 2) = 6$ .

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n

F e C_{2} O_{4}

during its oxidation by acidified

K M n O_{4}

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