Question

K2[Cr(CN)2(O)2(O2)(NH3)] in the given compound oxidation number of Cr is

Answer 1

O2 shows -1 oxidation state
there fore
os of Cr=-6+2=+4

Answer 2

Answer : The oxidation number of Cr is (+6).

Explanation :

The given compound is $K_2[Cr(CN)_2(O)_2(O_2)(NH_3)]$

$K_2[Cr(CN)_2(O)_2(O_2)(NH_3)]=K^{+2}+[Cr(CN)_2(O)_2(O_2)(NH_3)]^{-2}$

The oxidation state of given ligands are :

CN (cyno) = -1

O (oxo) = -2

$O_2$ (peroxo) = -2

$NH_3$ is a neutral ligand.

Now calculating the oxidation number of negative part of the compound.

Let the oxidation number of Cr (chromium) be 'x'

$x+2(-1)+2(-2)+(-2)=-2$

x = +6

Therefore, the oxidation number of Cr is (+6).