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Coordination Compounds question

2023 · 10 Apr · Shift 2 · Q14
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Coordination Compounds question

2023 · 10 Apr · Shift 2 · Q14

JEE MainChemistryCoordination CompoundsNumerical+4 / −1
For a metal ion, the calculated magnetic moment is 4.90 BM4.90 ~\mathrm{BM}4.90 BM. This metal ion has ‾\underline{\hspace{2cm}}​ number of unpaired electrons.
Numerical answer
View written solutionFree

Correct answer: 4

  1. For a transition metal ion, the spin-only magnetic moment is given by

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

where nnn is the number of unpaired electrons.

  1. Given:

μ=4.90  BM\mu = 4.90\;\text{BM}μ=4.90BM

So,

n(n+2)=4.90\sqrt{n(n+2)} = 4.90n(n+2)​=4.90

  1. Squaring both sides,

n(n+2)=(4.90)2=24.01n(n+2) = (4.90)^2 = 24.01n(n+2)=(4.90)2=24.01

n2+2n−24.01=0n^2 + 2n - 24.01 = 0n2+2n−24.01=0

  1. Now test integer values of nnn:
  • For n=4n=4n=4, μ=4(4+2)=24=4.90  BM (approximately)\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90\;\text{BM (approximately)}μ=4(4+2)​=24​=4.90BM (approximately)

This matches the given value.

  1. Therefore, the number of unpaired electrons is

4\boxed{4}4​

Comparison with stored answer

Stored correct answer = 4

My derived answer = 4

They agree.

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