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

List-I	List-II
(I) [Cr(CN)6]4−	(P) t2g orbitals contain 4 electrons
(II) [RuCl6]2−	(Q) (spin-only) = 4.9 BM
(III) [Cr(H2O)6]2+	(R) low spin complex ion
(IV) [Fe(H2O)6]2+	(S) metal ion in 4 + oxidation state
	(T) d4 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

The correct option is:
I → R, T; II → P, S; III → Q, T; IV → P, Q

Let us analyze each complex ion given in List-I step-by-step to find its matching properties in List-II.

(I) [Cr(CN)₆]^4-:

Chromium (Cr) has an atomic number of 24, with outer electronic configuration 3d⁵ 4s¹.
In [Cr(CN)₆]^4-, the oxidation state of Cr is +2.
So, Cr²⁺ has a d⁴ electronic configuration (T).

Since cyanide (CN^-) is a strong field ligand, it causes pairing of d-electrons, resulting in a low spin complex (R).
The 4 electrons fill the lower energy t_2g orbitals as t_2g⁴ e_g⁰. Thus, t_2g orbitals contain 4 electrons (P).
Matching options for (I): P, R, T (specifically R, T matches the given option set).

(II) [RuCl₆]^2-:

Ruthenium (Ru) has an atomic number of 44, belonging to the 4d series.
In [RuCl₆]^2-, let the oxidation state of Ru be x:
x + 6(-1) = -2 ⇒ x = +4.
Thus, Ru is in the +4 oxidation state (S).

Ru⁴⁺ has a 4d⁴ electronic configuration.
For 4d series metal ions, even weak field ligands like Cl^{- exert strong field effects causing electron pairing (low spin).
The 4 d-electrons enter the t_2g orbitals: t_2g4 e_g0.
Therefore, t_2g orbitals contain 4 electrons (P).
Matching options for (II): P, S.}

(III) [Cr(H₂O)₆]²⁺:

In [Cr(H₂O)₆]²⁺, Cr is in the +2 oxidation state, which is a d⁴ species (T).

Water (H₂O) is a weak field ligand, so no pairing occurs (high spin complex).
The electronic distribution is t_2g³ e_g¹, which gives 4 unpaired electrons (n = 4).

The spin-only magnetic moment ($\mu_s$) is calculated as:

$\mu_s = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \text{ BM}$ (Q).

Matching options for (III): Q, T.

(IV) [Fe(H₂O)₆]²⁺:

Iron (Fe) has an atomic number of 26, with configuration 3d⁶ 4s².
In [Fe(H₂O)₆]²⁺, Fe is in the +2 oxidation state (3d⁶).

Since H₂O is a weak field ligand, it forms a high spin complex.
The 6 d-electrons fill the orbitals as t_2g⁴ e_g².
Hence, t_2g orbitals contain 4 electrons (P).

The number of unpaired electrons is 4 (2 in e_g and 2 in t_2g), giving a spin-only magnetic moment:

$\mu_s = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \text{ BM}$ (Q).

Matching options for (IV): P, Q.

Combining all the correct matches:
I → R, T
II → P, S
III → Q, T
IV → P, Q