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D and F Block Elements question

2021 · 17 Mar · Shift 1 · Q2
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D and F Block Elements question

2021 · 17 Mar · Shift 1 · Q2

JEE MainChemistryD and F Block ElementsMCQ+4 / −1
What is the spin-only magnetic moment value (BM) of a divalent metal ion with atomic number 25, in it's aqueous solution?
  1. A
    zero
  2. B
    5.26
  3. C
    5.0
  4. D
    5.92
View written solutionFree

Correct answer: D

  1. Identify the element

Atomic number 252525 corresponds to manganese (Mn).

Its electronic configuration is:

Mn:[Ar] 3d54s2\mathrm{Mn}: [Ar] \, 3d^5 4s^2Mn:[Ar]3d54s2
  1. Find the divalent ion

For the divalent ion, remove two electrons from the 4s4s4s orbital first:

Mn2+:[Ar] 3d5\mathrm{Mn^{2+}}: [Ar] \, 3d^5Mn2+:[Ar]3d5
  1. Consider aqueous solution

In aqueous solution, Mn2+\mathrm{Mn^{2+}}Mn2+ forms a weak-field complex like [Mn(H2O)6]2+[\mathrm{Mn(H_2O)_6}]^{2+}[Mn(H2​O)6​]2+. Since H2O\mathrm{H_2O}H2​O is a weak-field ligand, the ion remains high spin.

For 3d53d^53d5 high spin, the number of unpaired electrons is:

n=5n = 5n=5
  1. Use the spin-only magnetic moment formula

The spin-only magnetic moment is:

μ=n(n+2) BM\mu = \sqrt{n(n+2)} \text{ BM}μ=n(n+2)​ BM

Substitute n=5n=5n=5:

μ=5(5+2)=35\mu = \sqrt{5(5+2)} = \sqrt{35}μ=5(5+2)​=35​ μ≈5.92 BM\mu \approx 5.92 \text{ BM}μ≈5.92 BM
  1. Check options
  • A: zero →\to→ incorrect
  • B: 5.265.265.26 →\to→ incorrect
  • C: 5.05.05.0 →\to→ incorrect
  • D: 5.925.925.92 →\to→ correct

Hence, the correct answer is D.

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