Question

What is oxidation number of [Cr(H2O)2(NH3)4]????

Answer 1

First, the charge of complex anion is 2-. NH3 is a neutral ligand, whereas CN is 1-, O is 2- and the PEROXIDO ligand (O2) has an oxidation number of -2: so 2x(-1)+2x(-2)+1x(-2)=-8. Therefore, the formal oxidation number for chromium ion must be +6, a d0 metal center. 1k Views · View Upvoters.

Answer 2

The oxidation state of Cr in Cr(H₂O)₆³⁺ is +3.
Let's recall some of the rules for calculating oxidation states.
• The oxidation state of O is almost always -2.
• The oxidation state of H is almost always +1.
• The sum of all the oxidation states equals the charge on the ion.
If H = +1, then 2H = +2. If O = -2, then 2H + O = +2 - 2 = 0.
Thus, the H₂O units do not contribute to the charge on the ion. The only atom that contributes is the Cr atom.
Since the charge on the ion is 3+, the oxidation state of Cr is +3.