The correct order of magnetic moment (spin only values in B.M.) is:
I. `[MnBr_4]^(2-)` II. `[Fe(CN)_6]^(4-)` III. `[CoBr_4]^(2-)`

Options

A. `Iigt III gt I`
B. `I gt II gt III`
C. `II gt I gt III`
D. `I gt III gt II`

Answer

To determine the correct order of magnetic moments for the given complexes, we will first calculate the number of unpaired electrons in each complex. The magnetic moment (spin-only) can be calculated using the formula:

\[ \mu = \sqrt{n(n+2)} \text{ B.M.} \]

where \( n \) is the number of unpaired electrons.

### Step 1: Analyze the complexes

1. **For \([MnBr_4]^{2-}\)**:
- Manganese (Mn) has an atomic number of 25. In its elemental state, the electron configuration is \([Ar] 3d^5 4s^2\).
- In the \([MnBr_4]^{2-}\) complex, Mn is in the +2 oxidation state, which means it loses 2 electrons from the 4s orbital and 1 from the 3d orbital.
- Thus, the configuration becomes \(3d^5\).
- All 5 electrons in the 3d subshell are unpaired.
- Therefore, \(n = 5\).

\[ \mu_{MnBr_4^{2-}} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ B.M.} \]

2. **For \([Fe(CN)_6]^{4-}\)**:
- Iron (Fe) has an atomic number of 26. Its electron configuration is \([Ar] 3d^6 4s^2\).
- In the \([Fe(CN)_6]^{4-}\) complex, Fe is in the +2 oxidation state (losing 2 electrons).
- Thus, the configuration becomes \(3d^6\).
- Since CN⁻ is a strong field ligand, it causes pairing of electrons. Therefore, the configuration will be \(3d^6\) with 4 paired and 2 unpaired electrons.
- Thus, \(n = 4\).

\[ \mu_{Fe(CN)_6^{4-}} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ B.M.} \]

3. **For \([CoBr_4]^{2-}\)**:
- Cobalt (Co) has an atomic number of 27. Its electron configuration is \([Ar] 3d^7 4s^2\).
- In the \([CoBr_4]^{2-}\) complex, Co is in the +2 oxidation state.
- Thus, the configuration becomes \(3d^7\).
- Br⁻ is a weak field ligand, so it does not cause pairing. Therefore, there are 3 unpaired electrons.
- Thus, \(n = 3\).

\[ \mu_{CoBr_4^{2-}} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ B.M.} \]

### Step 2: Compare the magnetic moments

Now we have the magnetic moments for each complex:

- \([MnBr_4]^{2-}\): \(5.92 \text{ B.M.}\)
- \([Fe(CN)_6]^{4-}\): \(4.90 \text{ B.M.}\)
- \([CoBr_4]^{2-}\): \(3.87 \text{ B.M.}\)

### Final Order of Magnetic Moments

The correct order of magnetic moments from highest to lowest is:

\[ [MnBr_4]^{2-} > [Fe(CN)_6]^{4-} > [CoBr_4]^{2-} \]