Match the electronic configurations in List-I with appropriate metal complex ions in List-II and choose the correct option.

[Atomic Number: Fe = 26, Mn = 25, Co = 27]

List-I		List-II
(P)	$$t_{2g}^6 e_g^0$$	(1)	$$[Fe(H_2O)_6]^{2+}$$
(Q)	$$t_{2g}^3 e_g^2$$	(2)	$$[Mn(H_2O)_6]^{2+}$$
(R)	$$e^2 t_2^3$$	(3)	$$[Co(NH_3)_6]^{3+}$$
(S)	$$t_{2g}^4 e_g^2$$	(4)	$$[FeCl_4]^{-}$$
		(5)	$$[CoCl_4]^{2-}$$

First find the oxidation state and hence the number of $$d$$-electrons present in every metal centre.

$$[Fe(H_2O)_6]^{2+}:$$
$$Fe = +2 \;\Rightarrow\; d^{6}$$

$$[Mn(H_2O)_6]^{2+}:$$
$$Mn = +2 \;\Rightarrow\; d^{5}$$

$$[Co(NH_3)_6]^{3+}:$$
$$Co = +3 \;\Rightarrow\; d^{6}$$

$$[FeCl_4]^{-}:$$
Total charge $$= -1$$, each $$Cl^{-}$$ contributes $$-1$$, therefore
$$x - 4 = -1 \;\Rightarrow\; x = +3$$
$$Fe = +3 \;\Rightarrow\; d^{5}$$

Next decide the geometry and the ligand field strength.

• $$[Fe(H_2O)_6]^{2+}$$ and $$[Mn(H_2O)_6]^{2+}$$ are octahedral with the weak-field ligand $$H_2O$$ ⇒ high-spin.
• $$[Co(NH_3)_6]^{3+}$$ is octahedral, but $$NH_3$$ is a stronger field ligand and, with the highly charged $$Co^{3+}$$, gives a low-spin complex.
• $$[FeCl_4]^{-}$$ is tetrahedral because the large ligand $$Cl^{-}$$ prefers tetrahedral geometry; $$Cl^{-}$$ is weak field, so the complex is always high-spin.

Write the electronic arrangements:

Case 1 - Octahedral high-spin $$d^{6}$$
$$t_{2g}^{4}\,e_g^{2}$$

Case 2 - Octahedral high-spin $$d^{5}$$
$$t_{2g}^{3}\,e_g^{2}$$

Case 3 - Octahedral low-spin $$d^{6}$$
$$t_{2g}^{6}\,e_g^{0}$$

Case 4 - Tetrahedral high-spin $$d^{5}$$
Level order reverses, so the lower set is $$e$$ and the higher set is $$t_2$$:
$$e^{2}\,t_{2}^{3}$$

Now match each complex with the patterns listed in List-I:

P : $$t_{2g}^{6} e_g^{0}$$ ⇒ octahedral low-spin $$d^{6}$$ ⇒ $$[Co(NH_3)_6]^{3+}\;(3)$$

Q : $$t_{2g}^{3} e_g^{2}$$ ⇒ octahedral high-spin $$d^{5}$$ ⇒ $$[Mn(H_2O)_6]^{2+}\;(2)$$

R : $$e^{2} t_2^{3}$$ ⇒ tetrahedral high-spin $$d^{5}$$ ⇒ $$[FeCl_4]^{-}\;(4)$$

S : $$t_{2g}^{4} e_g^{2}$$ ⇒ octahedral high-spin $$d^{6}$$ ⇒ $$[Fe(H_2O)_6]^{2+}\;(1)$$

Therefore
$$P \to 3,\; Q \to 2,\; R \to 4,\; S \to 1$$

The option showing this combination is
Option D which is: P → 3; Q → 2; R → 4; S → 1.