1. Find oxidation state and d-electron count for each complex

Recall:

• $\mathrm{Cr}: [Ar]3d^5 4s^1$
• $\mathrm{Fe}: [Ar]3d^6 4s^2$
• $\mathrm{Ru}: [Kr]4d^7 5s^1$

For cations, electrons are removed first from $s$ orbital.

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2. Complex (I): $[\mathrm{Cr}(\mathrm{CN})_6]^{4-}$

Let oxidation state of Cr be $x$.

$x+6(-1)=-4$ $x=+2$

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

Electronic configuration of $\mathrm{Cr}^{2+}$:

$\mathrm{Cr}: 3d^5 4s^1 \Rightarrow \mathrm{Cr}^{2+}: 3d^4$

Thus it is a $d^4$ species $\Rightarrow (T)$.

Since $\mathrm{CN}^-$ is a strong field ligand, the octahedral complex is low spin:

$t_{2g}^4 e_g^0$

Hence it is a low spin complex ion $\Rightarrow (R)$.

So,

$\text{(I)} \rightarrow R, T$

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3. Complex (II): $[\mathrm{RuCl}_6]^{2-}$

Let oxidation state of Ru be $x$.

$x+6(-1)=-2$ $x=+4$

So metal ion is in $4+$ oxidation state $\Rightarrow (S)$.

Now find d-count:

$\mathrm{Ru}: 4d^7 5s^1 \Rightarrow \mathrm{Ru}^{4+}: 4d^4$

So it is a $d^4$ species $\Rightarrow (T)$.

For octahedral $d^4$, with 4d metal, pairing tendency is higher and generally low spin behavior is favored more than 3d series, but among the listed options we need exact matching. Let us test options carefully.

For octahedral $d^4$ low spin: $t_{2g}^4 e_g^0$, so $t_{2g}$ contains 4 electrons.

Thus property $(P)$ also looks possible in principle. But let us continue with all species and then compare with options.

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4. Complex (III): $[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]^{2+}$

Let oxidation state of Cr be $x$.

$x+6(0)=+2$ $x=+2$

So metal ion is $\mathrm{Cr}^{2+}$, i.e. again

$\mathrm{Cr}^{2+}: 3d^4$

Hence it is a $d^4$ species $\Rightarrow (T)$.

$\mathrm{H_2O}$ is a weak field ligand, so this is high spin octahedral:

$t_{2g}^3 e_g^1$

Number of unpaired electrons $n=4$.

Spin-only magnetic moment:

$\mu = \sqrt{n(n+2)} = \sqrt{4(6)}=\sqrt{24}\approx 4.9\,\mathrm{BM}$

Hence $\Rightarrow (Q)$.

So,

$\text{(III)} \rightarrow Q, T$

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5. Complex (IV): $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{2+}$

Oxidation state of Fe is $+2$.

Electronic configuration:

$\mathrm{Fe}: 3d^6 4s^2 \Rightarrow \mathrm{Fe}^{2+}: 3d^6$

$\mathrm{H_2O}$ is weak field, so octahedral high spin:

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

Thus $t_{2g}$ contains 4 electrons $\Rightarrow (P)$.

Number of unpaired electrons $n=4$, so

$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9\,\mathrm{BM}$

Hence $\Rightarrow (Q)$.

So,

$\text{(IV)} \rightarrow P, Q$

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6. Match with options

We have obtained:

• (I) $\rightarrow R, T$
• (III) $\rightarrow Q, T$
• (IV) $\rightarrow P, Q$

Only Option A matches these three exactly, and for (II) it gives $P, S$.

Let us verify (II):

• oxidation state is definitely $+4$ $\Rightarrow S$
• $\mathrm{Ru}^{4+}$ is $4d^4$
• for octahedral $4d$ complexes, low spin arrangement is favored, so configuration is $t_{2g}^4 e_g^0$ hence $t_{2g}$ contains 4 electrons $\Rightarrow P$

Thus for (II):

$\text{(II)} \rightarrow P, S$

Therefore the correct matching is:

$\boxed{\text{A}}$