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JEE Main
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Chemistry
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Practical Chemistry

JEE Main · Chemistry · Practical Chemistry

Two unknown 3d transition metal chlorides, Salt A and Salt B, are...

Two unknown 3d3d transition metal chlorides, Salt A and Salt B, are tested with aqueous potassium ferrocyanide, K4[Fe(CN)6]\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]. Salt A produces a chocolate brown precipitate, while Salt B produces a Prussian blue precipitate. The spin-only magnetic moments (in BM) of the metal cations present in Salt A and Salt B, respectively, are:

Options

  1. A5.925.92 and 1.731.73
  2. B1.731.73 and 5.925.92
  3. C1.731.73 and 4.904.90
  4. D5.925.92 and 4.904.90

Correct answer

B. 1.731.73 and 5.925.92

Step-by-step solution

The reaction of metal cations with potassium ferrocyanide, K4[Fe(CN)6]\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6], gives characteristic precipitates.

Salt A forms a chocolate brown precipitate, which indicates the presence of the Cu2+\mathrm{Cu}^{2+} ion. The precipitate is cupric ferrocyanide, Cu2[Fe(CN)6]\mathrm{Cu}_2[\mathrm{Fe}(\mathrm{CN})_6].
The electronic configuration of Cu2+\mathrm{Cu}^{2+} is [Ar]3d9[\mathrm{Ar}] 3d^9. It has 11 unpaired electron (n=1n = 1).
Spin-only magnetic moment μ=n(n+2)=1(1+2)=3≈1.73 BM\mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ BM}.

Salt B forms a Prussian blue precipitate, which indicates the presence of the Fe3+\mathrm{Fe}^{3+} ion. The precipitate is ferric ferrocyanide, Fe4[Fe(CN)6]3\mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3.
The electronic configuration of Fe3+\mathrm{Fe}^{3+} is [Ar]3d5[\mathrm{Ar}] 3d^5. It has 55 unpaired electrons (n=5n = 5).
Spin-only magnetic moment μ=n(n+2)=5(5+2)=35≈5.92 BM\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM}.

Therefore, the spin-only magnetic moments are 1.731.73 and 5.925.92 respectively.

Answer: 1.731.73 and 5.925.92
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