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Exercise · Q22

Q.CoCl3.6NH3\text{CoCl}_3.6\text{NH}_3, CoCl3.5NH3\text{CoCl}_3.5\text{NH}_3, and CoCl3.4NH3\text{CoCl}_3.4\text{NH}_3 give 3, 2, and 1 mole of AgCl\text{AgCl} respectively on treatment with excess AgNO3\text{AgNO}_3 solution. Using Werner's theory, write the correct formula for each compound and explain the trend.

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Cobalt's secondary valence (coordination number) is fixed at 6 in all three compounds, regardless of exactly which groups fill those six positions. In CoCl3.6NH3\text{CoCl}_3.6\text{NH}_3, all six secondary-valence positions are filled by the six NH3\text{NH}_3 molecules, leaving all three Cl−\text{Cl}^- free (satisfying the primary valence): [Co(NH3)6]Cl3[\text{Co}(\text{NH}_3)_6]\text{Cl}_3, giving 3 mol AgCl\text{AgCl}. In CoCl3.5NH3\text{CoCl}_3.5\text{NH}_3, only five NH3\text{NH}_3 molecules are available, so one Cl−\text{Cl}^- must move inside the coordination sphere to complete the fixed secondary valence of 6; this leaves only two Cl−\text{Cl}^- free: [Co(NH3)5Cl]Cl2[\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2, giving 2 mol AgCl\text{AgCl} (the coordinated chloride, bonded directly to cobalt, does not react with AgNO3\text{AgNO}_3 under these conditions). In CoCl3.4NH3\text{CoCl}_3.4\text{NH}_3, only four NH3\text{NH}_3 molecules are available, so two Cl−\text{Cl}^- must move inside to complete the secondary valence of 6, leaving just one Cl−\text{Cl}^- free: [Co(NH3)4Cl2]Cl[\text{Co}(\text{NH}_3)_4\text{Cl}_2]\text{Cl}, giving 1 mol AgCl\text{AgCl}. The clear trend is that the metal's coordination number never changes, but as fewer neutral NH3\text{NH}_3 ligands are available to fill it, progressively more chloride ions are pulled from bein …

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