Arrange the following in increasing order of oxidation states of $M n$ ,
$K_{2} M n O_{4}, M n O_{2}, K M n O_{4}$

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Solution

$K^{+} = H, O^{- 2} = - 2$
$K_{2} M n O_{4}$
$(+ 1 \times 2) + x + (- 2 \times 4) = 0$
$2 + x - 8 = 0$
$x = + 6$
$M n O_{2}$
$↓$
$x + (- 2 \times 2) = 0$
$x = + 4$
$K M n O_{4}$
$↓$
$(+ 1) + x + (- 2 \times 4) = 0$
$x = 8 - 1 = + 7$

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