Question

What is the value of $x$ in $PtC{l}_4\cdot xN{H}_3$ to form $\left[Pt\right(N{H}_3{)}_6]C{l}_4$?

Answer 1

The value of $x$ in $PtC{l}_4\cdot xN{H}_3$ to form $\left[Pt\right(N{H}_3{)}_6]C{l}_4$ is $6$.

$Pt$ exhibits in $+2$ oxidation state so in the complex $6$ moles of $N{H}_3$ are required.

The correct answer is $6$.