A coordination compound is made of $\mathrm{Co}^{3+}$, $\mathrm{NH}_3$, and $\mathrm{Cl}^{-}, 0.1…

A coordination compound is made of $\mathrm{Co}^{3+}$, $\mathrm{NH}_3$, and $\mathrm{Cl}^{-}, 0.1 \mathrm{M}$ solution of this complex when treated with excess silver nitrate gave no precipitate. The formula of the complex and secondary valency of metal are respectively.
  1. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right], 6$
  2. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}_3 \mathrm{Cl}_2, 6\right.$
  3. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right], 3$
  4. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}, 6$

Solution

Since, no precipitate of $\mathrm{AgCl}$ is obtained on treating the compound with silver nitrate that means no $\mathrm{Cl}^{-}$ions are available outside the coordination sphere. So, correct formula of coordination compound is $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right.$. Since, central metal $\mathrm{Co}$ forms six coordination bonds thus the secondary valency is 6 .

Asked in: AP EAMCET 2018 (23 Apr Shift 2)

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