Which of the following arrangements represent the increasing oxidation number of the central atom?

$C r O_{2}^{-}, C l O_{3}^{-}, C r O_{4}^{2 -}, M n O_{4}^{-}$

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$C r O_{2}^{-}, C l O_{3}^{-}, M n O_{4}^{-}, C r O_{4}^{2 -}$

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$C l O_{3}^{-}, C r O_{4}^{2 -}, M n O_{4}^{-}, C r O_{2}^{-}$

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$C r O_{4}^{2 -}, M n O_{4}^{-}, C r O_{2}^{-}, C l O_{3}^{-}$

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Solution

The correct option is A
$C r O_{2}^{-}, C l O_{3}^{-}, C r O_{4}^{2 -}, M n O_{4}^{-}$

$C r O_{2}^{-}$

Let oxidation number of $C r$ be $a$

$a + 2 \times (- 2) = - 1$

$a - 4 = - 1$

Oxidation number of $C r$ is $+ 3$

$C l O_{3}^{-}$

Let oxidation number of $C l$ be $b$

$b + 3 \times (- 2) = - 1$

$b - 6 = - 1$

$b = + 5$

Oxidation number of $C l$ is $+ 5.$

$C r O_{4}^{2 -}$

Let oxidation number of $C r$ be $c$

$c + 4 \times (- 2) = - 2$

$c - 8 = - 2$

$c = + 6$

Oxidation number of $C r$ is $+ 6.$

$M n O_{4}^{-}$

Let oxidation number of $M n$ be $d$

$d + 4 \times (- 2) = - 1$

$d - 8 = - 1$

$d = + 7$

Oxidation number of $M n$ is $+ 7$

$C r O_{2}^{-} (C r = + 3), C l O_{3}^{-} (C l = + 5),$

$C r O_{4}^{2 -} (C r = + 6), M n O_{4}^{-} (M n = + 7)$

Thus, order is:

$C r O_{2}^{-}, C l O_{3}^{-}, C r O_{4}^{2 -}, M n O_{4}^{-}$

Hence, correct answer is option $(A) .$

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