What is the metal oxidation state in $dichromate ion$ $"dichromate ion"$ , $C r_{2} O_{7}^{2 -}$ $Cr_2O_7^(2-)$ ?

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Answer 1 · 2017-05-28T08:33:59

In $dichromate$ $"dichromate"$ , $C r_{2} O_{7}^{2 -}$ $Cr_2O_7^(2-)$ ? We got $C r (V I +)$ $Cr(VI+)$ .

The SUM of the individual oxidation states of chromium and the oxygen atoms in dichromate anion is equal to the charge on the ion.

And thus $2 \times C r_{oxidation number} + 7 \times O_{oxidation number} = - 2$ $2xxCr_"oxidation number"+7xxO_"oxidation number"=-2$ .

Now oxygen generally has an oxidation of $- 2$ $-2$ in its compounds, and it does so here, so.............

$2 \times C r_{oxidation number} + 7 \times (- 2) = - 2$ $2xxCr_"oxidation number"+7xx(-2)=-2$ .

Add +14 to both sides..............

$2 \times C r_{oxidation number} = + 12$ $2xxCr_"oxidation number"=+12$ .

$C r_{oxidation number} = + \frac{12}{2} = + V I$ $Cr_"oxidation number"=+12/2=+VI$ , as required................

What is the metal oxidation state in dichromate ion"dichromate ion", Cr2O2−7Cr_2O_7^(2-)?