In the complex ion [Cr(C₂O₄)₂(H₂O)₂]^-, the oxidation number of Cr is A) +1. B) +2. C) +3. D) -2. E) -1.

Question

Quizplus · Accepted Answer

To find the oxidation number of Cr, we need to add up the oxidation numbers of all the other atoms in the complex and equate it to the overall charge of the complex ion (-1). 
The oxidation state of S is -2 (since it is part of S2–). 
The oxidation state of O is -2 (since it is usually -2 in compounds). 
The oxidation state of H is +1. 
The oxidation state of N is -3 (since it is part of CS(NH2)2, a neutral ligand). 
The oxidation state of C is +2 (since it is part of CS(NH2)2, a neutral ligand). 
We know that the complex ion has an overall charge of -1. Therefore, 
[Cr(CS(NH2)2O)(H2O)(S2)]-1: 
-2(4) + (-2)(1) + 1 + (-3)(2) + 2(2) + x - 1 = -1 
Solving for x, we get x = +3. 
Therefore, the oxidation state of Cr is +3.

In the Complex Ion [Cr(C2O4)2(H2O)2]-, the Oxidation Number of Cr

In the Complex Ion [Cr(C₂O₄)₂(H₂O)₂]^-, the Oxidation Number of Cr