Match the LIST-I with LIST-II.

choose the correct answer from the options given below

We first copy the two lists exactly as given in the paper.

LIST-I (Complex)         LIST-II (Number of unpaired electrons)
A. $$\left[Fe(H_2O)_6\right]^{2+}$$     I. 3
B. $$\left[Cr(H_2O)_6\right]^{3+}$$    II. 1
C. $$\left[Fe(CN)_6\right]^{4-}$$      III. 4
D. $$\left[Cu(NH_3)_4\right]^{2+}$$   IV. 0

To match each complex with its correct entry in LIST-II we must:

(i) Find the oxidation state of the central metal ion.
(ii) Write the electronic configuration of that oxidation state in the ground state (gas-phase).
(iii) Decide whether the ligand present is weak-field or strong-field. That tells us whether the complex is high-spin or low-spin.
(iv) Fill electrons into the crystal-field or molecular-orbital diagram and count the unpaired electrons.

Complex  $$\left[Fe(H_2O)_6\right]^{2+}$$

Oxidation state: $$x + 6(0) = +2 \;\Rightarrow\; x = +2$$, so the metal is $$Fe^{2+}$$.

Gas-phase configuration of $$Fe$$ is $$[Ar]\,3d^{6}4s^{2}$$. Removing two electrons gives $$Fe^{2+}: [Ar]\,3d^{6}$$  (i.e. $$d^{6}$$).

$$H_2O$$ is a weak-field ligand, so the complex is high-spin octahedral. In the high-spin octahedral crystal field the splitting is $$t_{2g}^{4}e_g^{2}$$ with four unpaired electrons.

Hence A matches III (4 unpaired electrons).

Complex  $$\left[Cr(H_2O)_6\right]^{3+}$$

Oxidation state: $$x + 6(0) = +3 \;\Rightarrow\; x = +3$$, so the metal is $$Cr^{3+}$$.

Ground configuration of $$Cr$$ is $$[Ar]\,3d^{5}4s^{1}$$. Removing three electrons gives $$Cr^{3+}: [Ar]\,3d^{3}$$  (i.e. $$d^{3}$$).

Again $$H_2O$$ is weak-field, but for a $$d^{3}$$ ion the splitting pattern is the same in high-spin and low-spin: $$t_{2g}^{3}e_g^{0}$$. That contains three unpaired electrons.

Hence B matches I (3 unpaired electrons).

Complex  $$\left[Fe(CN)_6\right]^{4-}$$

Oxidation state: $$x + 6(-1) = -4 \;\Rightarrow\; x = +2$$, so again the metal is $$Fe^{2+}$$ ($$d^{6}$$).

$$CN^-$$ is a very strong-field (π-acceptor) ligand, so the complex is low-spin octahedral. Electron filling is $$t_{2g}^{6}e_g^{0}$$ with zero unpaired electrons.

Hence C matches IV (0 unpaired electrons).

Complex  $$\left[Cu(NH_3)_4\right]^{2+}$$ (tetra-ammine-copper(II))

Oxidation state: $$x + 4(0) = +2 \;\Rightarrow\; x = +2$$, so the metal is $$Cu^{2+}$$.

Ground configuration of $$Cu$$ is $$[Ar]\,3d^{10}4s^{1}$$. Removing two electrons gives $$Cu^{2+}: [Ar]\,3d^{9}$$  (i.e. $$d^{9}$$).

The complex is square-planar (commonly written with coordination number 4). In both square-planar and octahedral fields a $$d^{9}$$ ion ends up with exactly one unpaired electron (Jahn-Teller distortion in octahedral, or direct splitting in $$D_{4h}$$ symmetry).

Hence D matches II (1 unpaired electron).

Collecting all four matches:
A → III,  B → I,  C → IV,  D → II

This correspondence is exactly Option D.

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