Oxidation numbers of P in PO ₄³⁻ , of S in SO ₄²⁻ and that of Cr in Cr ₂ O ₇²⁻ are respectively,

Step-by-step 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 P in PO ₄³⁻ is x . aligned & PO ₄³⁻ & x +4(-2)=-3 & x -8=-3 & x =+5 aligned Let the oxidation state of S in SO ₄²⁻ is . aligned & SO ₄²⁻ & y +4(-2)=-2 & y -8=-2 & y =+6 aligned Let the oxidation state of Cr in Cr ₂ O ₇²⁻ is . aligned & Cr ₂ O ₇²⁻ & 2 z +7(-2)=-2 & 2 z -14=-2 & z =+6 aligned Hence, oxidation state of P , S and Cr are +5 , +6 and +6 respectively.