Given complexes are low spin complexes `[V(CN)_(6)]^(4-)` `[Cr(NH_(3))_(6)]^(2+)` `[Ru(NH_(3))_(6)]^(3+)` `[Fe(CN)_(6)]^(4-)` Then of magnetic `(mu)` for `Fe^(2+),Cr^(+3),Ru^(3+)` is

Options

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

Explanation

To solve the problem, we need to determine the magnetic moments of the given transition metal complexes based on their oxidation states and the number of unpaired electrons. Since we are told that these complexes are low-spin complexes, we will consider the pairing of electrons in the d-orbitals.

### Step-by-Step Solution:

1. **Identify the oxidation states and electronic configurations**:
- For each complex, we need to find the oxidation state of the metal and its corresponding electronic configuration.

a. **For `[V(CN)_(6)]^(4-)`**:
- Vanadium in this complex is in the +2 oxidation state (since CN is a -1 ligand, and there are 6 of them).
- Electronic configuration of V: \( [Ar] 3d^3 \).

b. **For `[Cr(NH_(3))_(6)]^(2+)`**:
- Chromium is in the +2 oxidation state.
- Electronic configuration of Cr: \( [Ar] 3d^4 \).

c. **For `[Ru(NH_(3))_(6)]^(3+)`**:
- Ruthenium is in the +3 oxidation state.
- Electronic configuration of Ru: \( [Kr] 4d^5 \).

d. **For `[Fe(CN)_(6)]^(4-)`**:
- Iron is in the +2 oxidation state.
- Electronic configuration of Fe: \( [Ar] 3d^6 \).

2. **Determine the number of unpaired electrons**:
- Since these are low-spin complexes, we will fill the lower energy \( t_{2g} \) orbitals first and then pair the electrons.

a. **For V\(^{2+}\) (3d\(^3\))**:
- Configuration: \( t_{2g}^3 \) (3 unpaired electrons).

b. **For Cr\(^{2+}\) (3d\(^4\))**:
- Configuration: \( t_{2g}^4 \) (2 unpaired electrons, as the fourth electron pairs up).

c. **For Ru\(^{3+}\) (4d\(^5\))**:
- Configuration: \( t_{2g}^5 \) (1 unpaired electron, as the first five electrons fill \( t_{2g} \) and do not pair).

d. **For Fe\(^{2+}\) (3d\(^6\))**:
- Configuration: \( t_{2g}^6 \) (0 unpaired electrons, as all electrons pair up in the \( t_{2g} \) orbitals).

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

a. **For V\(^{2+}\)**:
- \( n = 3 \)
- \( \mu = \sqrt{3(3 + 2)} = \sqrt{15} \).

b. **For Cr\(^{2+}\)**:
- \( n = 2 \)
- \( \mu = \sqrt{2(2 + 2)} = \sqrt{8} = 2.83 \).

c. **For Ru\(^{3+}\)**:
- \( n = 1 \)
- \( \mu = \sqrt{1(1 + 2)} = \sqrt{3} \).

d. **For Fe\(^{2+}\)**:
- \( n = 0 \)
- \( \mu = 0 \).

4. **Summarize the results**:
- V\(^{2+}\): 3 unpaired electrons, \( \mu = \sqrt{15} \)
- Cr\(^{2+}\): 2 unpaired electrons, \( \mu = 2.83 \)
- Ru\(^{3+}\): 1 unpaired electron, \( \mu = \sqrt{3} \)
- Fe\(^{2+}\): 0 unpaired electrons, \( \mu = 0 \)

### Final Answer:
The magnetic moments for \( Fe^{2+}, Cr^{3+}, Ru^{3+} \) are \( 0, 2.83, \sqrt{3} \) respectively.