Question

The heating of $(NH_4 )_2Cr_2O_7$ 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 $(NH_4)_2Cr_2O_7$ produces $Cr_2O_3$ compound along with $N_2.$
$(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} N_{2}+Cr_{2}O_{3}+4H_{2}O$
Let the oxidation state of $Cr$ be $x$
In $\left(NH_{4}\right)_{2}Cr_{2}O_{7},$
$=2\left(+1\right)+2\left(x\right)+7\left(-2\right)=0$
$2+2x-14=0$
$2x=12\Rightarrow x=+6$
In $Cr_2O_3$,
$2\left(x\right)+3\left(-2\right)=0$
$2x+\left(-6\right)=0$
$2x=+6 \Rightarrow x=+3$
$\therefore $ The change of the oxidation state of $Cr$ in the reaction is from + $6$ to +$3$.