Question

Find the oxidation number of$\mathrm{Cr}$ in ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$.

Answer 1

Step-1 Calculating the charge on ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$

• ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$is a neutral Inorganic compound in nature.
• Therefore the overall charge on ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ = $0$

Step-2 Calculating the oxidation number of $\mathrm{Cr}$ in${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$-:

1. As per standard, the oxidation number of Potassium$\left(K\right)$= $+1$
2. The oxidation number on Oxygen$\left(\mathrm{O}\right)$= $-2$
3. The oxidation number on Chromium $\left(\mathrm{Cr}\right)$ = $x$
   $FormalChargeon{K}_{2}C{r}_2{O}_7=2\times OxidationnumberofK+7\times OxidationnumberofO+2\times OxidationnumberofCr$

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

$0=2x+2-14$

$$\begin{gathered} 2x=12 \\ x=6 \end{gathered}$$

Hence, the oxidation number of Chromium in ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is $6$.