The correct order of the spin - only magnetic moments of the following complexes is
(I) `[Cr(H_(2)O)_(5)]Br_(2)`
(II) `Na_(4)[Fe(CN)_(6)]`
(III) `Na_(3)[Fe(C_(2)O_(4))_(3)](Delta_(0) gtP)`
(IV) `(Et_(4)N)_(4)[CoCl_(4)]`

Options

A. (I) gt (IV) gt (III) gt (II)
B. (II) `~~` (I) gt (IV) gt (III)
C. (III) gt (I) gt (IV) gt (III)
D. (III) gt (I) gt (II) gt (IV)

Answer

To determine the correct order of the spin-only magnetic moments of the given complexes, we need to analyze each complex step by step.

### Step 1: Determine the oxidation states and electron configurations

1. **Complex I: \([Cr(H_2O)_5]Br_2\)**
- Bromide (Br) has a charge of -1, and there are 2 bromides, giving a total charge of -2.
- The complex is neutral, so the oxidation state of Cr must be +2.
- The electron configuration of Cr in the +2 state is:
- Cr: \([Ar] 3d^4\) (since it loses 2 electrons from the 4s and 3d orbitals).

2. **Complex II: \(Na_4[Fe(CN)_6]\)**
- CN is a strong field ligand and is neutral.
- The overall charge of the complex is 0, and since there are 4 Na+ (which contribute +4), the oxidation state of Fe must be +2.
- The electron configuration of Fe in the +2 state is:
- Fe: \([Ar] 3d^6\).

3. **Complex III: \(Na_3[Fe(C_2O_4)_3](\Delta_0 > P)\)**
- Oxalate (C2O4) is a bidentate ligand and is also neutral.
- The overall charge of the complex is 0, and since there are 3 Na+ (which contribute +3), the oxidation state of Fe must be +3.
- The electron configuration of Fe in the +3 state is:
- Fe: \([Ar] 3d^5\).

4. **Complex IV: \((Et_4N)_4[CoCl_4]\)**
- Chloride (Cl) has a charge of -1, and there are 4 chlorides, giving a total charge of -4.
- The overall charge of the complex is 0, so the oxidation state of Co must be +4.
- The electron configuration of Co in the +4 state is:
- Co: \([Ar] 3d^6\).

### Step 2: Determine the spin-only magnetic moment

The spin-only magnetic moment (\(\mu\)) can be calculated using the formula:
\[ \mu = \sqrt{n(n+2)} \]
where \(n\) is the number of unpaired electrons.

1. **Complex I: \([Cr(H_2O)_5]Br_2\)**
- \(Cr^{2+}\) has 4 d-electrons: \(3d^4\) (4 unpaired electrons).
- \(\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \, \text{BM}\).

2. **Complex II: \(Na_4[Fe(CN)_6]\)**
- \(Fe^{2+}\) has 6 d-electrons: \(3d^6\) (0 unpaired electrons due to strong field ligand).
- \(\mu = \sqrt{0(0+2)} = 0 \, \text{BM}\).

3. **Complex III: \(Na_3[Fe(C_2O_4)_3](\Delta_0 > P)\)**
- \(Fe^{3+}\) has 5 d-electrons: \(3d^5\) (5 unpaired electrons).
- \(\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9 \, \text{BM}\).

4. **Complex IV: \((Et_4N)_4[CoCl_4]\)**
- \(Co^{4+}\) has 6 d-electrons: \(3d^6\) (0 unpaired electrons).
- \(\mu = \sqrt{0(0+2)} = 0 \, \text{BM}\).

### Step 3: Order the magnetic moments

Now we can order the complexes based on their calculated magnetic moments:
1. **Complex III: \(Na_3[Fe(C_2O_4)_3](\Delta_0 > P)\)** - \(5.9 \, \text{BM}\)
2. **Complex I: \([Cr(H_2O)_5]Br_2\)** - \(4.9 \, \text{BM}\)
3. **Complex II: \(Na_4[Fe(CN)_6]\)** - \(0 \, \text{BM}\)
4. **Complex IV: \((Et_4N)_4[CoCl_4]\)** - \(0 \, \text{BM}\)

### Final Order of Spin-Only Magnetic Moments
The correct order of the spin-only magnetic moments is:
\[ III > I > II = IV \]