Question

What is oxidation number of $Cr$ in $(N{H}_4{)}_{2}C{r}_2{O}_7$?

Answer 1

$(N{H}_4{)}_{2}C{r}_2{O}_7$

$2N{H}_4^{+}+C{r}_2{O}_7^{2-}$

Let oxidation state of $Cr$ be $x$

$2x+2-2\times 7=0$

$x=+6$