JEE Main 2023 · 30 January · evening · Single correct · +4 / −1
1 L, 0.02 M solution of [Co(NH₃)₅ SO₄] Br is mixed with 1 L, 0.02 M solution of [Co(NH₃)₅ Br] SO₄. The resulting solution is divided into two equal parts (X) and treated with excess of AgNO₃ solution and BaCl₂ solution respectively as shown below:
1 L Solution (X)+AgNO₃ solution (excess) → Y
1 L Solution (X)+BaCl₂ solution (excess) → Z
The number of moles of Y and Z respectively are
- (a) 0 .01,0.01
- (b) 0.01,0.02
- (c) 0.02,0.01
- (d) 0.02,0.02
Answer and solution
Answer (a) 0 .01,0.01
Solution
Chemistry ma’am · 3 steps
Only ions outside the square bracket are free to precipitate, so count those in one half of the mix.
- Moles in each half. Each solution has 1 × 0.02 = 0.02 mol of complex. After mixing and splitting into two equal 1 L parts, each part holds 0.01 mol of [Co(NH₃)₅SO₄]Br and 0.01 mol of [Co(NH₃)₅Br]SO₄.
- Free versus bound ions. Ions inside the square bracket are bound to cobalt and do not react. Only the counter-ions outside the bracket are free in solution to form a precipitate.
Complex (0.01 mol each) Free ion Bound ion Precipitate it can give [Co(NH₃)₅SO₄]Br Br⁻, 0.01 mol SO₄²⁻ 0.01 mol AgBr with AgNO₃ [Co(NH₃)₅Br]SO₄ SO₄²⁻, 0.01 mol Br⁻ 0.01 mol BaSO₄ with BaCl₂ - So. Y is AgBr (0.01 mol) and Z is BaSO₄ (0.01 mol), which is option (a).
The idea to keep. In a coordination compound, only the ions outside the square bracket show up in precipitation tests. Ligands inside the bracket stay attached to the metal.
Answer from the NTA final answer key. Solution by PrepWiser.