In which of the following order the given complex ions are arranged correctly with respect to their decreasing spin only magnetic moment?
`(i) [FeF_6]^(3-) (ii) [Co(NH_3)_6]^(3+) (iii) [NiCl_4]^(2-) (iv) [Cu(NH_3)_4]^(2+)`

Options

A. `(i) gt (iii) gt (iv) gt (ii)`
B. `(ii) gt (iii) gt (i) gt (iv)`
C. `(iii) gt (iv) gt (ii) gt (i)`
D. `(ii) gt (i) gt (iii) gt (iv)`

Answer

To determine the order of the given complex ions with respect to their decreasing spin-only magnetic moment, we will follow these steps:

### Step 1: Identify the oxidation states and d-electron counts for each complex ion.

1. **For [FeF6]^(3-)**:
- Let the oxidation state of Fe be \( x \).
- The charge of F is -1, and there are 6 F ligands: \( 6(-1) + x = -3 \)
- Solving gives \( x = +3 \).
- Fe in +3 state has \( 3d^5 \) configuration (5 d-electrons).

2. **For [Co(NH3)6]^(3+)**:
- Let the oxidation state of Co be \( x \).
- NH3 is a neutral ligand, so \( x + 0 = +3 \).
- Co in +3 state has \( 3d^6 \) configuration (6 d-electrons).

3. **For [NiCl4]^(2-)**:
- Let the oxidation state of Ni be \( x \).
- The charge of Cl is -1, and there are 4 Cl ligands: \( 4(-1) + x = -2 \)
- Solving gives \( x = +2 \).
- Ni in +2 state has \( 3d^8 \) configuration (8 d-electrons).

4. **For [Cu(NH3)4]^(2+)**:
- Let the oxidation state of Cu be \( x \).
- NH3 is a neutral ligand, so \( x + 0 = +2 \).
- Cu in +2 state has \( 3d^9 \) configuration (9 d-electrons).

### Step 2: Determine the nature of the ligands and the resulting electron pairing.

- **[FeF6]^(3-)**: F is a weak field ligand, so there will be no pairing. Thus, all 5 d-electrons remain unpaired. \( n = 5 \).

- **[Co(NH3)6]^(3+)**: NH3 is a strong field ligand, which causes pairing. All 6 d-electrons will pair up. Thus, \( n = 0 \).

- **[NiCl4]^(2-)**: Cl is a weak field ligand, so there will be no pairing. Thus, 8 d-electrons will have 2 unpaired electrons. \( n = 2 \).

- **[Cu(NH3)4]^(2+)**: NH3 is a strong field ligand, so there will be no pairing. Thus, with 9 d-electrons, there will be 1 unpaired electron. \( n = 1 \).

### Step 3: Calculate the spin-only magnetic moment using the formula.

The formula for spin-only magnetic moment (\( \mu_s \)) is given by:
\[ \mu_s = \sqrt{n(n + 2)} \]

1. **For [FeF6]^(3-)**:
- \( n = 5 \)
- \( \mu_s = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35} \)

2. **For [Co(NH3)6]^(3+)**:
- \( n = 0 \)
- \( \mu_s = \sqrt{0(0 + 2)} = 0 \)

3. **For [NiCl4]^(2-)**:
- \( n = 2 \)
- \( \mu_s = \sqrt{2(2 + 2)} = \sqrt{2 \times 4} = \sqrt{8} \)

4. **For [Cu(NH3)4]^(2+)**:
- \( n = 1 \)
- \( \mu_s = \sqrt{1(1 + 2)} = \sqrt{1 \times 3} = \sqrt{3} \)

### Step 4: Arrange the complexes in order of decreasing magnetic moment.

- **[FeF6]^(3-)**: \( \sqrt{35} \) (highest)
- **[NiCl4]^(2-)**: \( \sqrt{8} \)
- **[Cu(NH3)4]^(2+)**: \( \sqrt{3} \)
- **[Co(NH3)6]^(3+)**: \( 0 \) (lowest)

### Final Order:
The correct order of the complex ions with respect to their decreasing spin-only magnetic moment is:
\[ [FeF_6]^{3-} > [NiCl_4]^{2-} > [Cu(NH_3)_4]^{2+} > [Co(NH_3)_6]^{3+} \]