Question

The oxidation states of transition metal atoms in $K_{2}C{r}_2{O}_7,KMn{O}_4$ and $K_{2}Fe{O}_4,$ respectively, are $x,yandz$. The sum of $x,yandz$ is

Answer 1

For $K_{2}C{r}_2{O}_7$
Let, the oxidation state of $Crbex$
$2\times 1+2x+(-2\times 7)=0$
$2+2x=14$
$x=6$
$K_2\stackrel{+6}{C{r}_2}{O}_7$
For $KMn{O}_4$
Let, the oxidation state of $Mnbey$
$+1+y+(-2\times 4)=0$
$+1+y=8$
$y=+7$
For $K_{2}Fe{O}_4$
$1\times 2+z+(-2\times 4)=0$
$2+z=8$
$z=+6$
$K_2\left[\stackrel{+6}{F}e{O}_4\right]$
Hence,
$6+7+6=19$