Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$2$

$\left[Pt{\left(N{H}_3\right)}_{3}Cl\right]Cl$

It can be written as, ${\left[Pt{\left(N{H}_3\right)}_{3}Cl\right]}^{+}Cl$

The central atom in ${\left[Pt{\left(N{H}_3\right)}_{3}Cl\right]}^{+}$ is $Pt$.

The oxidation state of ${NH}_3=0$ and $Cl=-1$

In general,

Let oxidation state of $Pt=x$

Applying formula, $\text{Sum of total oxidation state of all atoms}=\text{Overall charge on the compound}.x+3\times 0+1\times (-1)=+1,x=2$

$\therefore$ Oxidation state of $Pt=+2$