LIST-I contains metal species and LIST-II contains their properties.

LIST-I	LIST-II
(I) $$[Cr(CN)_6]^{4-}$$	(P) $$t_{2g}$$ orbitals contain 4 electrons
(II) $$[RuCl_6]^{2-}$$	(Q) $$\mu$$(spin-only) = 4.9 BM
(III) $$[Cr(H_2O)_6]^{2+}$$	(R) low spin complex ion
(IV) $$[Fe(H_2O)_6]^{2+}$$	(S) metal ion in 4+ oxidation state
	(T) $$d^4$$ species

[Given : Atomic number of Cr = 24, Ru = 44, Fe = 26]

Match each metal species in LIST-I with their properties in LIST-II, and choose the correct option

First find for every complex
  1. the oxidation state of the metal,
  2. the corresponding $$d^{n}$$ configuration, and
  3. whether the complex will be high-spin or low-spin.
From this we can decide the number of electrons in the $$t_{2g}$$ set, the spin-only magnetic moment $$\mu$$ (in Bohr magneton, BM) and any special oxidation state information.

Case I : $$[Cr(CN)_6]^{4-}$$
Let the oxidation state of chromium be $$x$$.
$$x + 6(-1) = -4 \;\Rightarrow\; x = +2$$
Cr has atomic number 24, so neutral chromium is $$[Ar]\,3d^5 4s^1$$.
Cr$$^{2+}$$ loses the two $$4s$$ electrons ⇒ $$d^{4}$$ species (property T).
$$CN^-$$ is a strong-field ligand; for a $$d^{4}$$ metal this gives the low-spin arrangement $$t_{2g}^4 e_g^0$$ (property R).
Hence I → R, T.

Case II : $$[RuCl_6]^{2-}$$
Let the oxidation state of ruthenium be $$x$$.
$$x + 6(-1) = -2 \;\Rightarrow\; x = +4$$ ⇒ property S.
Ru (Z = 44) : neutral configuration $$[Kr]\,4d^7 5s^1$$.
Removing four electrons (one 5s + three 4d) gives Ru$$^{4+}$$ = $$4d^{4}$$ ⇒ again a $$d^{4}$$ ion.
For 4d and 5d metals, even halide ligands usually generate a sufficiently large $$\Delta_o$$ to give low-spin complexes, so the electronic distribution is $$t_{2g}^4 e_g^0$$ (four electrons in $$t_{2g}$$ → property P).
Hence II → P, S.

Case III : $$[Cr(H_2O)_6]^{2+}$$
Water is neutral, so Cr remains in +2 oxidation state: $$d^{4}$$ ⇒ property T.
$$H_2O$$ is a weak-field ligand; 3d ions therefore adopt the high-spin pattern $$t_{2g}^3 e_g^1$$.
Number of unpaired electrons $$n = 4$$, so
$$\mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9\ \text{BM}$$ ⇒ property Q.
Hence III → Q, T.

Case IV : $$[Fe(H_2O)_6]^{2+}$$
Water is neutral, so Fe is in +2 state. Fe (Z = 26) : neutral $$[Ar]\,3d^6 4s^2$$; Fe$$^{2+}$$ = $$3d^{6}$$.
With the weak-field ligand $$H_2O$$, the complex is high-spin: $$t_{2g}^4 e_g^2$$, so the $$t_{2g}$$ orbitals carry 4 electrons (property P).
Unpaired electrons $$n = 4$$, giving the same magnetic moment $$\mu \approx 4.9\ \text{BM}$$ (property Q).
Hence IV → P, Q.

Collecting all matches:
I → R, T      II → P, S      III → Q, T      IV → P, Q

The option that lists exactly these pairings is Option A.