Calculate the average oxidation number of element $M n$ in the compound $M n_{3} O_{4}$ .
(Report your answer upto two decimal place)

2.67

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2.66

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Solution

Let oxidation state of Mn is $x$ .
Charge of $O = - 2$

$∴ 3 x - 8 = 0$
$\Rightarrow 3 x = 8 \Rightarrow x = + \frac{8}{3}$
$∴ average oxidation number of Mn is + \frac{8}{3} = 2.66$

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