The correct order of the spin-only magnetic moment of metal ions in the following low-spin complexes, `[V(CN)_(6)]^(4-),[Fe(CN)_(6)]^(4-),[Ru(NH_(3))_(6)]^(3+)`, and `[Cr(NH_(3))_(6)]^(2+)` , is :

Options

A. `Cr^(2+)gtRu^(3+)gtFe^(2+)gtV^(2+)`
B. `V^(2+)gtCr^(2)gtRu^(3+)gtFe^(2+)`
C. `V^(2+)gtRu^(3+)gtCr^(2+)gtFe^(2+)`
D. `Cr^(2+)gtV^(2+)gtRu^(3+)gtFe^(2+)`

Answer

To determine the correct order of the spin-only magnetic moment of the given low-spin complexes, we will follow these steps:

### Step 1: Identify the oxidation states and electron configurations of the metal ions.

1. **Vanadium in [V(CN)₆]⁴⁻**:
- Oxidation state: +2 (since CN is -1 and there are 6 CN ligands: 6*(-1) + x = -4 → x = +2)
- Atomic number of Vanadium (V) = 23
- Electron configuration for V: [Ar] 3d³ (for V²⁺, it loses 2 electrons from the 4s and 1 from 3d)

2. **Iron in [Fe(CN)₆]⁴⁻**:
- Oxidation state: +2 (similar calculation)
- Atomic number of Iron (Fe) = 26
- Electron configuration for Fe: [Ar] 3d⁶ (for Fe²⁺, it loses 2 electrons from 4s)

3. **Ruthenium in [Ru(NH₃)₆]³⁺**:
- Oxidation state: +3 (since NH₃ is neutral)
- Atomic number of Ruthenium (Ru) = 44
- Electron configuration for Ru: [Kr] 4d⁵ (for Ru³⁺, it loses 3 electrons from 5s and 4d)

4. **Chromium in [Cr(NH₃)₆]²⁺**:
- Oxidation state: +2 (similar calculation)
- Atomic number of Chromium (Cr) = 24
- Electron configuration for Cr: [Ar] 3d⁵ (for Cr²⁺, it loses 2 electrons from 4s)

### Step 2: Determine the number of unpaired electrons in each complex.

1. **[V(CN)₆]⁴⁻**:
- Configuration: 3d³
- In a low-spin complex, all three electrons remain unpaired.
- Unpaired electrons (N) = 3

2. **[Fe(CN)₆]⁴⁻**:
- Configuration: 3d⁶
- In a low-spin complex, the six electrons will pair up, resulting in no unpaired electrons.
- Unpaired electrons (N) = 0

3. **[Ru(NH₃)₆]³⁺**:
- Configuration: 4d⁵
- In a low-spin complex, there will be one unpaired electron.
- Unpaired electrons (N) = 1

4. **[Cr(NH₃)₆]²⁺**:
- Configuration: 3d⁵
- In a low-spin complex, there will be three unpaired electrons.
- Unpaired electrons (N) = 4 (two will pair up)

### Step 3: Calculate the spin-only magnetic moment (μ) for each complex using the formula:
\[ \mu = \sqrt{N(N + 2)} \]

1. **For [V(CN)₆]⁴⁻**:
- μ = √(3(3 + 2)) = √(15) ≈ 3.87 BM

2. **For [Fe(CN)₆]⁴⁻**:
- μ = √(0(0 + 2)) = 0 BM

3. **For [Ru(NH₃)₆]³⁺**:
- μ = √(1(1 + 2)) = √(3) ≈ 1.73 BM

4. **For [Cr(NH₃)₆]²⁺**:
- μ = √(4(4 + 2)) = √(24) ≈ 4.89 BM

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

- [Fe(CN)₆]⁴⁻: 0 BM
- [Ru(NH₃)₆]³⁺: 1.73 BM
- [V(CN)₆]⁴⁻: 3.87 BM
- [Cr(NH₃)₆]²⁺: 4.89 BM

### Final Order of Spin-Only Magnetic Moment:
**[Fe(CN)₆]⁴⁻ < [Ru(NH₃)₆]³⁺ < [V(CN)₆]⁴⁻ < [Cr(NH₃)₆]²⁺**