Consider the follwing complexes ion `P,Q` and `R`
`P =[FeF_(6)]^(3-), Q=[V(H_(2)O)_(6)]^(2+)` and `R=[Fe(H_(2)O)_(6)]^(2+)`
The correct order of the complex ions, according to their spin only magnetic moment values (inBM) is .

Options

A. `RltQgtR`
B. `QltRltP`
C. `RltPltQ`
D. `QltPltR`

Explanation

To determine the correct order of the complex ions \( P, Q, \) and \( R \) according to their spin-only magnetic moment values, we will follow these steps:

### Step 1: Determine the oxidation state of the central metal ion in each complex.

1. **For \( P = [FeF_6]^{3-} \)**:
- Let the oxidation state of Fe be \( x \).
- The charge of fluorine is \( -1 \) and there are 6 fluorine atoms.
- The equation is:
\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]
- Therefore, the oxidation state of Fe in \( P \) is \( +3 \).

2. **For \( Q = [V(H_2O)_6]^{2+} \)**:
- Let the oxidation state of V be \( x \).
- The charge of water is \( 0 \) and there are 6 water molecules.
- The equation is:
\[ x + 6(0) = +2 \implies x = +2 \]
- Therefore, the oxidation state of V in \( Q \) is \( +2 \).

3. **For \( R = [Fe(H_2O)_6]^{2+} \)**:
- Let the oxidation state of Fe be \( x \).
- The charge of water is \( 0 \) and there are 6 water molecules.
- The equation is:
\[ x + 6(0) = +2 \implies x = +2 \]
- Therefore, the oxidation state of Fe in \( R \) is \( +2 \).

### Step 2: Determine the electron configuration and number of unpaired electrons for each complex.

1. **For \( P = [FeF_6]^{3-} \)**:
- The electron configuration of neutral Fe is \( [Ar] 4s^2 3d^6 \).
- For \( Fe^{3+} \), it loses 3 electrons (2 from \( 4s \) and 1 from \( 3d \)):
\[ [Ar] 3d^5 \]
- The \( 3d^5 \) configuration has 5 unpaired electrons.

2. **For \( Q = [V(H_2O)_6]^{2+} \)**:
- The electron configuration of neutral V is \( [Ar] 4s^2 3d^3 \).
- For \( V^{2+} \), it loses 2 electrons (both from \( 4s \)):
\[ [Ar] 3d^3 \]
- The \( 3d^3 \) configuration has 3 unpaired electrons.

3. **For \( R = [Fe(H_2O)_6]^{2+} \)**:
- The electron configuration of neutral Fe is \( [Ar] 4s^2 3d^6 \).
- For \( Fe^{2+} \), it loses 2 electrons (both from \( 4s \)):
\[ [Ar] 3d^6 \]
- The \( 3d^6 \) configuration has 4 unpaired electrons.

### Step 3: Calculate the spin-only magnetic moment for each complex.

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

1. **For \( P \)**:
- \( n = 5 \):
\[ \mu_P = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM} \]

2. **For \( Q \)**:
- \( n = 3 \):
\[ \mu_Q = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM} \]

3. **For \( R \)**:
- \( n = 4 \):
\[ \mu_R = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM} \]

### Step 4: Order the complexes based on their magnetic moment values.

- \( P \) has the highest magnetic moment (\( \approx 5.92 \text{ BM} \)).
- \( R \) comes next (\( \approx 4.90 \text{ BM} \)).
- \( Q \) has the lowest magnetic moment (\( \approx 3.87 \text{ BM} \)).

### Final Order:
The correct order of the complex ions according to their spin-only magnetic moment values is:
\[ P > R > Q \]