Match List I with List II

Choose the correct answer from the options given below:

For every complex in List I we will:

(i) find the oxidation state of the central metal ion,
(ii) write its ground-state $$d$$-electron configuration,
(iii) decide whether the ion is high-spin or low-spin (strong-field vs weak-field ligand),
(iv) count the number of unpaired electrons and then match with List II.

Complex A : $$[Cr(H_2O)_6]^{3+}$$

Oxidation state of Cr = +3  ($$0-6(0)+x=+3\Rightarrow x=+3$$).
$$Cr^{3+}$$ ⇒ $$d^3$$ electronic configuration.
$$H_2O$$ is a weak-field ligand, so a high-spin octahedral ion is obtained:
$$t_{2g}^{3}\,e_{g}^{0}$$ - three unpaired electrons.
Hence A corresponds to “3 unpaired electrons” → List II item III.

Complex B : $$[MnF_6]^{3-}$$

Total charge: $$x+6(-1)=-3 \; \Rightarrow \; x=+3$$, so the metal is $$Mn^{3+}\;(d^4)$$.
$$F^-$$ is a weak-field ligand, therefore the complex is high-spin:
$$t_{2g}^{3}\,e_{g}^{1}$$ - four unpaired electrons.
Thus B matches List II item IV (“4 unpaired electrons”).

Complex C : $$[Fe(CN)_6]^{4-}$$

Oxidation state: $$x+6(-1)=-4 \; \Rightarrow \; x=+2$$, so ion is $$Fe^{2+}\;(d^6)$$.
$$CN^-$$ is a strong-field ligand, giving a low-spin octahedral ion:
$$t_{2g}^{6}\,e_{g}^{0}$$ - all electrons paired, zero unpaired electrons.
Therefore C corresponds to List II item I (“0 unpaired electrons”).

Complex D : $$[FeF_6]^{3-}$$

Oxidation state: $$x+6(-1)=-3 \; \Rightarrow \; x=+3$$, so ion is $$Fe^{3+}\;(d^5)$$.
With weak-field $$F^-$$ ligands, the complex is high-spin:
$$t_{2g}^{3}\,e_{g}^{2}$$ - five unpaired electrons.
Hence D matches List II item II (“5 unpaired electrons”).

Putting the four results together:

A → III, B → IV, C → I, D → II.

The combination “A-III, B-IV, C-I, D-II” is Option B.

Final Answer: Option B which is A-III, B-IV, C-I, D-II.