Question

What is the oxidation number of chromium in `K_2Cr_2O_7`?

Answer 1

We has `Cr(VI+)`........

Explanation:

The ion is `Cr_2O_7^(2-)`. The sum of the oxidation numbers of the individual elements must equal the charge on the ion. And thus `2xxCr_"oxidation number"+7xxO_"oxidation number"+=-2`.

The oxidation number of oxygen in its compounds is usually `-II`, and it is here..........

So `Cr_"oxidation number"=(-2+14)/2=+6`, i.e. `Cr^(VI+)`.