Question

When $K_{2}C{r}_2{O}_7$ is converted into $K_{2}Cr{O}_4$ , the change in oxidation number of chromium is :

• A. $0$
• B. $5$
• C. $7$
• D. $9$

Answer 1

Answer: (A)

The sum of oxidation state of all elements in a compound is always zero.

Let, oxidation state of $Cr$ in $K_{2}C{r}_2{O}_7=x$

$(+1\times 2)+2x+(-2\times 7)=0$

or $+2+2x-14=0$

$\therefore x=+6$

Let oxidation sate of $Cr$ in $K_{2}Cr{O}_4=x$

$\therefore +1\times 2+x+(-2\times 4)=0$

$2+x-8=0$

$\therefore x=6$

$\therefore$ Change in oxidation state of $Cr$ is zero when it changes from $K_{2}C{r}_2{O}_7$ to $K_{2}Cr{O}_4$