Question

Arrange the following complexes in the increasing order of their spin only magnetic moment (in B.M)
I. \([Fe(CN)₆]⁴⁻\)
II. \([MnCl₄]²⁻\)
III. \([Mn(CN)₆]³⁻\)
IV. \([Cr(NH₃)₆]³⁺\)

• A. II<IV<I<III
• B. III<II<I<IV
• C. I<IV<II<III
• D. I<III<IV<II

Hint:

To solve these problems quickly, remember the spectrochemical series to identify strong-field (CN⁻, CO) and weak-field (halides, H₂O) ligands. Strong ligands cause pairing for \(d^4-d^7\) configurations in octahedral complexes. For tetrahedral complexes, pairing rarely occurs. The magnetic moment increases with the number of unpaired electrons, so you only need to compare \(n\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The spin-only magnetic moment (\(\mu\)) of a transition metal complex is determined by the number of unpaired electrons (\(n\)). It is calculated using the formula:
\[ \mu = \sqrt{n(n+2)} \text{ B.M. (Bohr Magnetons)} \] A larger number of unpaired electrons results in a larger magnetic moment. We need to determine \(n\) for each complex by considering the metal's oxidation state, its d-electron count, and whether the ligand is strong-field (causing pairing) or weak-field (not causing pairing).
Step 2: Analyzing Each Complex:

- I. \([Fe(CN)₆]⁴⁻\):
- Oxidation state of Fe: \(x + 6(-1) = -4 \implies x = +2\). Fe²⁺.
- Electronic configuration of Fe²⁺: \([Ar] 3d^6\).
- Ligand: CN⁻ is a strong-field ligand, causing electron pairing.
- In an octahedral field, the six \(d\) electrons will fill the lower \(t_{2g}\) orbitals: \(t_{2g}^6 e_g^0\).
- Number of unpaired electrons, \(n = 0\).
- Magnetic moment, \(\mu = \sqrt{0(0+2)} = 0\) B.M.

- II. \([MnCl₄]²⁻\):
- Oxidation state of Mn: \(x + 4(-1) = -2 \implies x = +2\). Mn²⁺.
- Electronic configuration of Mn²⁺: \([Ar] 3d^5\).
- Ligand: Cl⁻ is a weak-field ligand.
- The complex is tetrahedral. The five \(d\) electrons will occupy the orbitals singly: \(e^2 t_2^3\).
- Number of unpaired electrons, \(n = 5\).
- Magnetic moment, \(\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\) B.M.

- III. \([Mn(CN)₆]³⁻\):
- Oxidation state of Mn: \(x + 6(-1) = -3 \implies x = +3\). Mn³⁺.
- Electronic configuration of Mn³⁺: \([Ar] 3d^4\).
- Ligand: CN⁻ is a strong-field ligand, causing pairing.
- In an octahedral field, the four \(d\) electrons will occupy the lower \(t_{2g}\) orbitals: \(t_{2g}^4 e_g^0\).
- Number of unpaired electrons, \(n = 2\).
- Magnetic moment, \(\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83\) B.M.

- IV. \([Cr(NH₃)₆]³⁺\):
- Oxidation state of Cr: \(x + 6(0) = +3 \implies x = +3\). Cr³⁺.
- Electronic configuration of Cr³⁺: \([Ar] 3d^3\).
- Ligand: NH₃ is a moderately strong ligand, but with only three \(d\) electrons, pairing is not an issue.
- In an octahedral field, the three \(d\) electrons will occupy the lower \(t_{2g}\) orbitals singly: \(t_{2g}^3 e_g^0\).
- Number of unpaired electrons, \(n = 3\).
- Magnetic moment, \(\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\) B.M.
Step 3: Ordering the Magnetic Moments:
Let's order the complexes by their number of unpaired electrons (\(n\)), which corresponds to the order of their magnetic moments.

- I (\(n=0\))
- III (\(n=2\))
- IV (\(n=3\))
- II (\(n=5\))
The increasing order of magnetic moment is I<III<IV<II.

Step 4: Final Answer:
The correct increasing order of spin only magnetic moment is I<III<IV<II. This corresponds to option (D).