The oxidation states of Cr in K_2Cr_2O_7 and | vedclass

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Question

(C) For $K_2Cr_2O_7$: Let the oxidation state of $Cr$ be $x$. $2(+1) + 2x + 7(-2) = 0$ $2 + 2x - 14 = 0$ $2x = 12$ $x = +6$ For $CrO_5$: This compound

Vedclass · Accepted Answer

$+6, +6$

Vedclass · Answer

(C) For $K_2Cr_2O_7$: Let the oxidation state of $Cr$ be $x$. $2(+1) + 2x + 7(-2) = 0$ $2 + 2x - 14 = 0$ $2x = 12$ $x = +6$ For $CrO_5$: This compound has a butterfly structure containing four peroxy oxygen atoms (each with $-1$ charge) and one oxo oxygen atom (with $-2$ charge). Let the oxidation state of $Cr$ be $x$. $x + 4(-1) + 1(-2) = 0$ $x - 4 - 2 = 0$ $x - 6 = 0$ $x = +6$ Thus,the oxidation states of $Cr$ in both compounds are $+6$.

The oxidation states of $Cr$ in $K_2Cr_2O_7$ and $CrO_5$ are,respectively:

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