A complex $PtCl X_{4} \cdot n NH X_{3}$ has Pt coordination number 6 and furnishes four ions per formula unit in water. Let $m$ be the moles of $AgCl$ obtained per mole of complex with excess $AgNO X_{3}$ . The value of $n + m$ is

Answer: A

Four ions mean one complex cation and 3 free

Cl X^{-}

, so

m = 3

and one chloride is coordinated. Coordination number

6 = n + 1

gives

n = 5

, i.e.

[Pt (NH X_{3}) X_{5} Cl] Cl X_{3}

. Then

n + m = 5 + 3 = 8