Question

What is the oxidation number of the central metal atom in the coordination compound $\left[Pt\right(N{H}_3{)}_{3}Cl]Cl?$

• A. $-1$
• B. $0$
• C. $+1$
• D. $+2$

Answer 1

The correct option is D $+2$

Firstly, the overall (addition of all) oxidation state of a neutral compound must be zero. In $\left[Pt\right(NH3\left)3Cl\right]Cl,$ the central metal atom is the Platinum($Pt$).

$\left[Pt\right(NH3\left)3Cl\right]Cl,$ ionises into $\left[Pt\right(N{H}_3{)}_{3}Cl{]}^{+}$ and $C{l}^{-}$

Let us consider the oxidation state of $Pt$ in this compound to be $x$. We know that ammonia is a neutal ligand and Chlorine is a monodentate ligand which has a unit negative charge.

$\therefore [x+(3\times 0)-1]=-1$
$x=+2$