The oxidation state of Cr in $K_{2} C r_{2} O_{7}$ is:

+ 7

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+ 6

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+ 3

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+ 2

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Solution

The correct option is C $+ 6$
Let the oxidation number of $C r$ in $K_{2} {C r}_{2} O_{7}$ be $x$ . The oxidation states of $K$ and $O$ are +1 and -2 respectively. Since the atom as whole is neutral, following equation holds true:
$2 \times (+ 1) + 2 \times (x) + 7 \times (- 2) = 0 ⟹ 2 + 2 x - 14 = 0 ⟹ 2 x = 14 - 2 ⟹ 2 x = 12 ⟹ x = \frac{12}{2} = 6$

Hence, answer is $B$ .

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