Question

The oxidation number of Pt in $\mathrm{[Pt(en)_2Cl_2]^{2+}}$ is :

• A. + 3
• B. + 4
• C. + 5
• D. + 6

Hint:

'en' is neutral (0) and each Cl is −1; balance against the +2 charge to get Pt = +4.

Answer 1

The Correct Option is B

Concept: The charge written on top of a complex is the total of the metal's charge plus all the ligand charges. So if we know every ligand's charge, we can work backwards and find the metal's oxidation number.

Step 1: Look at each ligand and its charge
In $\mathrm{[Pt(en)_2Cl_2]^{2+}}$ there are two ligands. 'en' is ethylenediamine, a neutral molecule, so it carries no charge, that is $0$. Chloride is $\mathrm{Cl^-}$ and carries $-1$, and there are two of them, so together they give $2\times(-1)=-2$.

Step 2: Write a simple balance
Let the oxidation number of Pt be $x$. The whole thing has a charge of $+2$ on it. So the metal plus all the ligands must add up to $+2$: \[ x + 2(0) + 2(-1) = +2 \]

Step 3: Solve it
\[ x - 2 = +2 \;\Rightarrow\; x = +4 \] So platinum is in the $+4$ state here.

Answer: Option (B), Pt is $+4$.