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Monodentate Ligands

Dichromate io...

Question

Dichromate ion is treated with base, the oxidation number of Cr in the product formed is
JEE MAINS 2021

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Solution

$C r_{2} O_{7}^{2 -} O H^{-} ⟶ C r O_{4}^{2 -}$
$For C r O_{4}^{2 -}$
$x - 8 = - 2 x = + 6$
$∴$ Oxidation state of Cr in $C r O_{4}^{2 -}$ is +6.

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