Oxidation numbers of P in PO_4^3-, of S in SO_4^2- and that of Cr in Cr_2 O_7^2- are respectively,

Key Idea (i) Sum of oxidation states of all atoms charge of ion. (ii) Oxidation number of oxygen $=-2$=−2. Let the oxidation state of $\mathrm{P}$P in $\mathrm{PO}_4^{3-}$PO43−​ is $x$x. $$\therefore \quad \begin{aligned} & \mathrm{PO}_4^{3-} \\ & \quad \therefore \quad \mathrm{x}+4(-2)=-3 \\ & \mathrm{x}-8=-3 \\ & \mathrm{x}=+5 \end{aligned}$$∴​PO43−​∴x+4(−2)=−3x−8=−3x=+5​ Let the oxidation state of $\mathrm{S}$S in $\mathrm{SO}_4^{2-}$SO42−​ is . $$\begin{aligned} & \mathrm{SO}_4^{2-} \\ & \therefore \quad \mathrm{y}+4(-2)=-2 \\ & \mathrm{y}-8=-2 \\ & \mathrm{y}=+6 \end{aligned}$$​SO42−​∴y+4(−2)=−2y−8=−2y=+6​ Let the oxidation state of $\mathrm{Cr}$Cr in $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$Cr2​O72−​ is . $$\therefore \quad \begin{aligned} & \mathrm{Cr}_2 \mathrm{O}_7^{2-} \\ & 2 \times \mathrm{z}+7(-2)=-2 \\ & 2 \mathrm{z}-14=-2 \\ & \mathrm{z}=+6 \end{aligned}$$∴​Cr2​O72−​2×z+7(−2)=−22z−14=−2z=+6​ Hence, oxidation state of $\mathrm{P}, \mathrm{S}$P,S and $\mathrm{Cr}$Cr are +5 , +6 and +6 respectively.