Question

When excess of AgNO\(_3\) is added to a complex, one mole of AgCl is precipitated. The formula of complex is _______.

• A. \( \text{[CoCl}_2\text{(NH}_3)_4\text{]Cl} \)
• B. \( \text{[CoCl(NH}_3)_5\text{]Cl}_2 \)
• C. \( \text{[CoCl}_3\text{(NH}_3)_3\text{]} \)
• D. \( \text{[Co(NH}_3)_6\text{]Cl}_3 \)

Hint:

Think of the square brackets in a coordination compound as a protective box. Reagents like AgNO\(_3\) can only react with the ions that are \textit{outside} this box. The number of moles of AgCl formed directly tells you the number of Cl\(^-\) ions outside the brackets.

Answer 1

The Correct Option is A

Step 1: Understand precipitation in coordination compounds. When silver nitrate (AgNO\(_3\)) is added to a solution of a coordination complex, it only reacts with the halide ions that are outside the coordination sphere (the square brackets). These are known as counter-ions. Ions inside the coordination sphere are covalently bonded to the central metal atom and do not dissociate.
Step 2: Relate the amount of precipitate to the number of counter-ions. The problem states that one mole of silver chloride (AgCl) is precipitated per mole of the complex. This means there must be exactly one chloride ion (\(\text{Cl}^-\)) available as a counter-ion in the formula.
Step 3: Examine the given formulas.
- \(\text{[CoCl}_2\text{(NH}_3)_4\text{]Cl}\): Has one \(\text{Cl}^-\) outside the brackets. Will precipitate 1 mole of AgCl.
- \(\text{[CoCl(NH}_3)_5\text{]Cl}_2\): Has two \(\text{Cl}^-\) outside the brackets. Will precipitate 2 moles of AgCl.
- \(\text{[CoCl}_3\text{(NH}_3)_3\text{]}\): Has zero \(\text{Cl}^-\) outside the brackets. Will precipitate 0 moles of AgCl.
- \(\text{[Co(NH}_3)_6\text{]Cl}_3\): Has three \(\text{Cl}^-\) outside the brackets. Will precipitate 3 moles of AgCl.
Based on this analysis, formula is the correct answer.