Q28 / 90 · JEE Main 2021 (26 Feb Shift 1) · Chemistry

Dichromate ion is treated with base, the oxidation number of $Cr$ in the product formed is

Correct answer: 6

Solution

${Cr}_{2} O_{7}^{2 -} + {OH}^{-} ⟶ {CrO}_{4}^{2 -}$

Oxidation state of $Cr$ in ${CrO}_{4}^{2 -}$ is $+ 6$

Hence, the correct answer is 6.

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