An element A in a compound ABD has oxidation number (A^n- ). It is oxidised by (Cr2O7)^2- in acidic medium. In the experiment 1.68×10^-3 moles of K2Cr2O7 were used for 3.16×10^-3 moles of ABD. The new oxidation number of A after oxidation will be?

Equivalent of Cr2O72− = Equivalent of ABD or A−nmole×valency factor=mole×valency factorin acidic medium Cr2O72− changes to Cr3+ hence n−factor or valency factro is 6 (change in oxi. no is 3 per atom)so n−factor for A can be calculated1.68×10−3×6=3.26×10−3×n−factorn−factor=3so A−n will be oxidized so oxidation no. increase from n to 3−n.Happy to help