The spin-only magnetic moment ($\mu_s$) of a complex ion is determined by the number of unpaired electrons (n) in the d-orbitals of the central metal ion. The formula is given by: $\mu_s = \sqrt{n(n+2)}$ B.M. (Bohr Magnetons). A larger value of 'n' results in a larger magnetic moment. We need to find 'n' for each complex ion P, Q, and R.

Step 1: Analyze Complex P, $[FeF_6]^{3-}$

1. Oxidation State of Fe: Let the oxidation state of Fe be 'x'. The charge on each fluoride ligand ($F^-$) is -1. The overall charge of the complex is -3. $x + 6(-1) = -3 \implies x = +3$ So, we have Iron in the +3 oxidation state ($Fe^{3+}$).

2. Electronic Configuration: The atomic number of Fe is 26. Its ground state configuration is $[Ar] 3d^6 4s^2$. For $Fe^{3+}$, three electrons are removed (two from 4s and one from 3d), so the configuration is $[Ar] 3d^5$.

3. Ligand Field and Electron Filling: The fluoride ion ($F^-$) is a weak-field ligand. In an octahedral complex with a weak-field ligand, the crystal field splitting energy ($\Delta_o$) is small, leading to a high-spin complex. The five d-electrons will occupy the orbitals to maximize spin. The configuration will be $t_{2g}^3 e_g^2$. (↑↑↑ in $t_{2g}$, ↑↑ in $e_g$). The number of unpaired electrons, $n_P = 5$.

4. Magnetic Moment: $\mu_P = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ B.M.

Step 2: Analyze Complex Q, $[V(H_2O)_6]^{2+}$

1. Oxidation State of V: Let the oxidation state of V be 'x'. Water ($H_2O$) is a neutral ligand (charge = 0). The overall charge is +2. $x + 6(0) = +2 \implies x = +2$ So, we have Vanadium in the +2 oxidation state ($V^{2+}$).

2. Electronic Configuration: The atomic number of V is 23. Its ground state configuration is $[Ar] 3d^3 4s^2$. For $V^{2+}$, two electrons are removed from the 4s orbital, so the configuration is $[Ar] 3d^3$.

3. Electron Filling: The three d-electrons will occupy the lower energy $t_{2g}$ orbitals with parallel spins according to Hund's rule, regardless of whether the ligand is strong or weak field. The configuration will be $t_{2g}^3 e_g^0$. (↑↑↑ in $t_{2g}$). The number of unpaired electrons, $n_Q = 3$.

4. Magnetic Moment: $\mu_Q = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ B.M.

Step 3: Analyze Complex R, $[Fe(H_2O)_6]^{2+}$

1. Oxidation State of Fe: Let the oxidation state of Fe be 'x'. Water ($H_2O$) is a neutral ligand. The overall charge is +2. $x + 6(0) = +2 \implies x = +2$ So, we have Iron in the +2 oxidation state ($Fe^{2+}$).

2. Electronic Configuration: The ground state configuration of Fe (Z=26) is $[Ar] 3d^6 4s^2$. For $Fe^{2+}$, two electrons are removed from the 4s orbital, so the configuration is $[Ar] 3d^6$.

3. Ligand Field and Electron Filling: Water ($H_2O$) is a weak-field ligand. This results in a high-spin octahedral complex. The six d-electrons are filled as $t_{2g}^4 e_g^2$. (↑↓↑↑ in $t_{2g}$, ↑↑ in $e_g$). The number of unpaired electrons, $n_R = 4$.

4. Magnetic Moment: $\mu_R = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$ B.M.

Step 4: Compare and Order

We have the number of unpaired electrons for each complex:

• P: $n_P = 5$
• Q: $n_Q = 3$
• R: $n_R = 4$

The order of the number of unpaired electrons is $n_Q < n_R < n_P$. Since the spin-only magnetic moment increases with the number of unpaired electrons, the order of the magnetic moments is: $\mu_Q < \mu_R < \mu_P$

Substituting the complex ion labels, the correct order is: $Q < R < P$ This corresponds to option B.