JEE Chemistry 2023 Question with Solution

Given: One mole each of $\mathrm{[Co(NH_3)_4Cl_2]Cl}$, $\mathrm{[Ni(H_2O)_6]Cl_2}$, $\mathrm{[Pt(NH_3)_2Cl_2]}$ and $\mathrm{[Pd(NH_3)_4]Cl_2}$ is treated with excess $\mathrm{AgNO_3}$.

Find: Total moles of AgCl precipitated.

Only the chloride ions present outside the coordination sphere are ionizable and precipitate with $\mathrm{Ag^+}$ as AgCl.

For $\mathrm{[Co(NH_3)_4Cl_2]Cl}$, there is 1 chloride ion outside the coordination sphere, so it gives 1 mole of AgCl.

For $\mathrm{[Ni(H_2O)_6]Cl_2}$, there are 2 chloride ions outside the coordination sphere, so it gives 2 moles of AgCl.

For $\mathrm{[Pt(NH_3)_2Cl_2]}$, there is no chloride ion outside the coordination sphere, so it gives 0 mole of AgCl.

For $\mathrm{[Pd(NH_3)_4]Cl_2}$, there are 2 chloride ions outside the coordination sphere, so it gives 2 moles of AgCl.

Therefore,

$$\text{Total moles of AgCl} = 1 + 2 + 0 + 2 = 5$$

Therefore, the total number of moles of AgCl precipitated is $5$.

Given: Excess $\mathrm{AgNO_3}$ is added to one mole each complex.

Find: The total moles of AgCl formed.

Use the idea that only counter ions outside square brackets react with $\mathrm{AgNO_3}$.

$$\mathrm{[Co(NH_3)_4Cl_2]Cl} \rightarrow \text{gives 1 mole AgCl}$$$$\mathrm{[Ni(H_2O)_6]Cl_2} \rightarrow \text{gives 2 moles AgCl}$$$$\mathrm{[Pt(NH_3)_2Cl_2]} \rightarrow \text{gives no AgCl}$$$$\mathrm{[Pd(NH_3)_4]Cl_2} \rightarrow \text{gives 2 moles AgCl}$$

Adding them,

$$1 + 2 + 0 + 2 = 5$$

Hence, the required numerical value is $5$.