Question

What is the oxidation number of Ru in [Ru(NH\(_3\))\(_5\)H\(_2\)O]Cl\(_2\)?

• A. +6
• B. +5
• C. +1
• D. +2

Hint:

To determine the oxidation state of a metal in a complex, balance the charges from the ligands and counterions with the overall charge of the complex.

Answer 1

The Correct Option is D

Step 1: Understanding oxidation states.
The complex \([Ru(NH_3)_5H_2O]Cl_2\) consists of the \([Ru(NH_3)_5H_2O]^2+\) cation and two chloride anions (Cl\(^-\)). The ammonia (NH\(_3\)) and water (H\(_2\)O) molecules are neutral ligands. The oxidation number of chloride is \(-1\).

Step 2: Calculating the oxidation number of Ru.
The overall charge of the complex is \(+2\), and since the chloride ions are each \(-1\), we have: \[ x + 5(0) + 1(0) - 2 = 0, \] where \(x\) is the oxidation state of Ru. Solving for \(x\), we get \(x = +2\).
Step 3: Conclusion.
Thus, the oxidation number of Ru in the complex is \(\mathbf{+2}\).