1. Find oxidation state and d-electron count of the metal in each complex

Given:

• $P = [\mathrm{FeF}_6]^{3-}$
• $Q = [\mathrm{V(H_2O)}_6]^{2+}$
• $R = [\mathrm{Fe(H_2O)}_6]^{2+}$

For $P = [\mathrm{FeF}_6]^{3-}$

Let oxidation state of Fe be $x$.

$x + 6(-1) = -3$ $x = +3$

So metal is $\mathrm{Fe^{3+}}$.

Fe: $Z=26$, ground state $[Ar]3d^64s^2$

$\mathrm{Fe^{3+}} : 3d^5$

Since $\mathrm{F^-}$ is a weak field ligand, octahedral complex is high spin. Thus number of unpaired electrons:

$n = 5$

For $Q = [\mathrm{V(H_2O)}_6]^{2+}$

Let oxidation state of V be $x$.

Water is neutral, so:

$x = +2$

So metal is $\mathrm{V^{2+}}$.

V: $Z=23$, ground state $[Ar]3d^34s^2$

$\mathrm{V^{2+}} : 3d^3$

In octahedral field, $d^3$ gives:

$t_{2g}^3 e_g^0$

Number of unpaired electrons:

$n = 3$

For $R = [\mathrm{Fe(H_2O)}_6]^{2+}$

Let oxidation state of Fe be $x$.

Water is neutral, so:

$x = +2$

So metal is $\mathrm{Fe^{2+}}$.

$\mathrm{Fe^{2+}} : 3d^6$

$\mathrm{H_2O}$ is a weak field ligand, so octahedral complex is high spin. Hence configuration:

$t_{2g}^4 e_g^2$

Number of unpaired electrons:

$n = 4$

2. Use spin-only magnetic moment formula

$\mu = \sqrt{n(n+2)}\ \text{B.M.}$

For $Q$:

$\mu_Q = \sqrt{3(3+2)} = \sqrt{15}$

For $R$:

$\mu_R = \sqrt{4(4+2)} = \sqrt{24}$

For $P$:

$\mu_P = \sqrt{5(5+2)} = \sqrt{35}$

Thus,

$\sqrt{15} < \sqrt{24} < \sqrt{35}$

So the order is:

$Q < R < P$

3. Match with options

This corresponds to Option C.

4. Comparison with stored correct answer

Stored correct answer: C

Derived answer: C

Hence, the derived answer agrees with the stored answer.