Question

(Cr(C2O4)2Cl) raised to -4 Find oxidation number of Cr.​

Answer 1

oxidation no

[ Cr (C2O 4)2 cl ] raised to -4

let x be thie oxidation no of Cr

then ,

by oxidation no method

x -2× 2 -1 = -4

x = -4+5

x = 1

therefore oxidation no Cr in given complex is 1

Answer 2

Oxidation Number of Cr = +1