The heating of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ produces another chromium compound along with ${\mathrm{N}}_2$ gas. The change of the oxidation state of $\mathrm{Cr}$ in the reaction is

The heating of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ produces ${\mathrm{Cr}}_2{\mathrm{O}}_3$ compound along with ${\mathrm{N}}_2.$

${\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\stackrel{\mathrm{\Delta }}{⟶}{\mathrm{N}}_2+{\mathrm{Cr}}_2{\mathrm{O}}_3+4{\mathrm{H}}_2\mathrm{O}$

Let the oxidation state of $\mathrm{Cr}$ be $\mathrm{x}$

In ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$

$=2\left(+1\right)+2\left(\mathrm{x}\right)+7\left(-2\right)=0$

$2+2\mathrm{x}-14=0$

$\mathrm{x}=+6$

In ${\mathrm{Cr}}_2{\mathrm{O}}_3$

$2\left(\mathrm{x}\right)+3\left(-2\right)==$

$2\mathrm{x}-6=0$

$\mathrm{x}=+3$

$\therefore$ The change of the oxidation state of $\mathrm{Cr}$ in the reaction is from $+6$ to $+3$