When K_2Cr_2O_7 is converted to K_2CrO_4,the | vedclass

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(A) In $K_2Cr_2O_7$,the oxidation state of $Cr$ is calculated as: $2(+1) + 2(x) + 7(-2) = 0 \implies 2 + 2x - 14 = 0 \implies 2x = 12 \implies x = +6$

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(A) In $K_2Cr_2O_7$,the oxidation state of $Cr$ is calculated as: $2(+1) + 2(x) + 7(-2) = 0 \implies 2 + 2x - 14 = 0 \implies 2x = 12 \implies x = +6$.In $K_2CrO_4$,the oxidation state of $Cr$ is calculated as: $2(+1) + x + 4(-2) = 0 \implies 2 + x - 8 = 0 \implies x = +6$.Since the oxidation state of $Cr$ remains $+6$ in both compounds,the change in the oxidation state is $6 - 6 = 0$.

When $K_2Cr_2O_7$ is converted to $K_2CrO_4$,the change in the oxidation state of chromium is

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