Question

The heating of $(N{H}_4{)}_{2}C{r}_2{O}_7$ produces another chromium compound along with $N_2$ gas. The change of the oxidation state of $Cr$ in the reaction is

• A. +6 to +2
• B. +7 to +4
• C. +8 to +4
• D. +6 to +3

Answer 1

Answer: (D)

The heating of $(N{H}_4{)}_{2}C{r}_2{O}_7$ produces $C{r}_2{O}_3$ compound along with $N_2.$
$(N{H}_4{)}_{2}C{r}_2{O}_7\stackrel{\Delta }{\to }{N}_2+C{r}_2{O}_3+4H_{2}O$
Let the oxidation state of $Cr$ be $x$
In ${\left(N{H}_4\right)}_{2}C{r}_2{O}_7,$
$=2\left(+1\right)+2\left(x\right)+7\left(-2\right)=0$
$2+2x-14=0$
$2x=12\Rightarrow x=+6$
In $C{r}_2{O}_3$,
$2\left(x\right)+3\left(-2\right)=0$
$2x+\left(-6\right)=0$
$2x=+6\Rightarrow x=+3$
$\therefore$ The change of the oxidation state of $Cr$ in the reaction is from + $6$ to +$3$.