Oxidation numbers of $\mathrm{P}$ in $\mathrm{PO}_4^{3-}$, of $\mathrm{S}$ in $\mathrm{SO}_4^{2-}$ and that…
Oxidation numbers of $\mathrm{P}$ in $\mathrm{PO}_4^{3-}$, of $\mathrm{S}$ in $\mathrm{SO}_4^{2-}$ and that of $\mathrm{Cr}$ in $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ are respectively,
$+5,+6$ and +6
$+3,+6$ and +5
$+5,+3$ and +6
$-3,+6$ and +6
Solution
Key Idea (i) Sum of oxidation states of all atoms charge of ion.
(ii) Oxidation number of oxygen $=-2$.
Let the oxidation state of $\mathrm{P}$ in $\mathrm{PO}_4^{3-}$ is $x$.
$\therefore \quad \begin{aligned}
& \mathrm{PO}_4^{3-} \\
& \quad \therefore \quad \mathrm{x}+4(-2)=-3 \\
& \mathrm{x}-8=-3 \\
& \mathrm{x}=+5
\end{aligned}$
Let the oxidation state of $\mathrm{S}$ in $\mathrm{SO}_4^{2-}$ is .
$\begin{aligned}
& \mathrm{SO}_4^{2-} \\
& \therefore \quad \mathrm{y}+4(-2)=-2 \\
& \mathrm{y}-8=-2 \\
& \mathrm{y}=+6
\end{aligned}$
Let the oxidation state of $\mathrm{Cr}$ in $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ is.
$\therefore \quad \begin{aligned}
& \mathrm{Cr}_2 \mathrm{O}_7^{2-} \\
& 2 \times \mathrm{z}+7(-2)=-2 \\
& 2 \mathrm{z}-14=-2 \\
& \mathrm{z}=+6
\end{aligned}$
Hence, oxidation state of $\mathrm{P}, \mathrm{S}$ and $\mathrm{Cr}$ are +5 , +6 and +6 respectively.