Total number of moles of AgCl precipitated on addition of excess of $\mathrm{AgNO}_3$ to one mole each of the following complexes $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]_{21} \mathrm{Cl},\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_3\right]_2 \mathrm{Cl}_2,\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$ and $\left[\mathrm{Pd}\left(\mathrm{NH}_3\right)_4\right]_{\mathrm{Cl}}$ is
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Solution
Sol. $\quad\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl} \Rightarrow$ Gives 1 mole AgCl
$\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right){ }_8\right] \mathrm{Cl}_2 \Rightarrow$ Gives 2 moles AgCl
$\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right] \Rightarrow$ Gives No AgCl
$\left[\mathrm{Pd}\left(\mathrm{NH}_3\right)_4\right] \mathrm{Cl}_2 \Rightarrow$ Gives 2 moles AgCl
Total number of moles of AgCl = 5 mole.
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