Total number of moles of AgCl precipitated on addition of excess of $\mathrm{AgNO}_{3}$ to one mole each of the following complexes $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}$, $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2},\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$ and $\left[\mathrm{Pd}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}$ is

JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Coordination Compounds

Total number of moles of AgCl precipitated on addition of excess of

AgNO_{3}

to one mole each of the following complexes

[Co (NH_{3})_{4} Cl_{2}] Cl

[Ni (H_{2} O)_{6}] Cl_{2}, [Pt (NH_{3})_{2} Cl_{2}]

and

[Pd (NH_{3})_{4}] Cl_{2}

Answer: 5

Solution

Only the chloride ions lying outside the coordination sphere precipitate as silver chloride. In

[Co (N H_{3})_{4} C l_{2}] Cl

a single chloride is ionisable, giving one mole; in

[Ni (H_{2} O)_{6}] C l_{2}

two chlorides lie outside, giving two moles;

[Pt (N H_{3})_{2} C l_{2}]

is a neutral complex with both chlorides bound, giving none; and

[Pd (N H_{3})_{4}] C l_{2}

has two ionisable chlorides, giving two moles. Adding these, the total silver chloride precipitated is five moles.

JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Integer answer

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