Question

Dichromate ion is treated with base, the oxidation number of $\mathrm{Cr}$ in the product formed in the following reaction is:

${\mathrm{Cr}}_2{\mathrm{O}}_7^{-2}+2{\mathrm{OH}}^{-}\rightleftharpoons 2{\mathrm{CrO}}_4^{-2}+{\mathrm{H}}_2\mathrm{O}$

Answer 1

Step 1: The given reaction is

${\mathrm{Cr}}_2{\mathrm{O}}_7^{-2}+2{\mathrm{OH}}^{-}\rightleftharpoons 2{\mathrm{CrO}}_4^{-2}+{\mathrm{H}}_2\mathrm{O}$

Step 2: Determine the oxidation number of $\mathbf{Cr}$ in ${\mathbf{CrO}}_{\mathbf{4}}^{\mathbf{-}\mathbf{2}}$:

Formula to find the oxidation number of an atom in a compound is

$Totalcharge=[\left(NumberofChrominumatoms\right)(OxidationnumberofChromium\left)\right]+[\left(NumberofOxygenatoms\right)(OxidationnumberofOxygen\left)\right]$

Let $\text{'}x\text{'}$ be the oxidation number of $\mathrm{Cr}$,

$$\begin{gathered} \Rightarrow -2=1\left(x\right)+4\left(-2\right) \\ \Rightarrow x=-2+8 \\ \Rightarrow x=+6 \\ \Rightarrow x=+6 \end{gathered}$$

$\therefore$ Oxidation number of $\mathrm{Cr}$ in ${\mathrm{CrO}}_4^{-2}$ is $+6$.