Co Ordination Compounds Question 201
Question: Magnetic moment of $ {{[Cu{{(NH_3)}_4}]}^{2+}} $ ion is
[RPET 2003]
Options:
A) 1.414
B) 1.73
C) 2.23
D) 2.38
Show Answer
Answer:
Correct Answer: B
Solution:
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First, we need to understand what determines the magnetic moment of a complex ion.
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The magnetic moment (μ) is related to the number of unpaired electrons in the ion. It can be calculated using the spin-only formula:
μ = √[n(n+2)] BM (Bohr Magnetons)
where n is the number of unpaired electrons.
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For the $[Cu(NH_3)_4]^{2+}$ ion:
- Cu is in the +2 oxidation state
- Cu²⁺ has an electron configuration of [Ar] 3d⁹
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In a tetrahedral complex (which $[Cu(NH_3)_4]^{2+}$ is), the d-orbitals split into e and t₂ sets, but the splitting is small.
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Cu²⁺ has 9 d-electrons. In this case, there will be one unpaired electron.
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So, n = 1
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Plugging this into our formula:
μ = √[1(1+2)] = √3 BM
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√3 ≈ 1.73 BM
Therefore, the magnetic moment of $[Cu(NH_3)_4]^{2+}$ ion is approximately 1.73 BM.
The correct answer is B) 1.73.
BETA
