The green complex PtCl4.3NH3 (Pt coordination number 6) gives 1 mol of AgCl per mole with excess AgNO3. Let a be the number of ions per formula unit in water and b the number of stereoisomers of the complex cation. The value of a + b is

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The green complex $PtCl X_{4} \cdot 3 NH X_{3}$ (Pt coordination number 6) gives 1 mol of $AgCl$ per mole with excess $AgNO X_{3}$ . Let $a$ be the number of ions per formula unit in water and $b$ the number of stereoisomers of the complex cation. The value of $a + b$ is

One free chloride means the formula is

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

, which gives 2 ions, so

a = 2

. The cation is

Ma X_{3} b X_{3}

with fac and mer forms, both achiral, so

b = 2

. Hence

a + b = 4