The pair having the same magnetic moment is: [At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27]

The magnetic moment of a coordination complex is decided only by the number of unpaired electrons present on the central metal ion. The spin-only formula is first written:

$$\mu = \sqrt{n(n+2)} \; \text{BM}$$

where $$n$$ is the number of unpaired (उक्त) electrons.

So, for each given complex we will

(i) find the oxidation state of the metal,

(ii) write the d-electron configuration of that oxidation state,

(iii) decide whether the complex is high-spin or low-spin (जल $$H_2O$$ and $$Cl^-$$ are weak-field ligands, therefore high-spin), and

(iv) count $$n$$ and calculate $$\mu$$ using the above formula.

1. $$[Mn(H_2O)_6]^{2+}$$

The charge on the complex is $$+2$$, water is neutral, so the oxidation state of Mn is $$+2$$.

Atomic number of Mn is 25, therefore

$$Mn: [Ar]\,3d^5\,4s^2 \quad \Longrightarrow \quad Mn^{2+}: [Ar]\,3d^5$$

Water gives a weak field, hence the complex is high-spin and all five d electrons remain unpaired.

Thus $$n = 5$$.

Now substitute into the formula:

$$\mu_{Mn^{2+}} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM}$$

2. $$[Cr(H_2O)_6]^{2+}$$

Again, water is neutral, overall charge is $$+2$$, so Cr is in $$+2$$ oxidation state.

Atomic number of Cr is 24, hence

$$Cr: [Ar]\,3d^5\,4s^1 \quad \Longrightarrow \quad Cr^{2+}: [Ar]\,3d^4$$

With a weak-field ligand the complex is high-spin; four d electrons stay unpaired.

Therefore $$n = 4$$.

Magnetic moment:

$$\mu_{Cr^{2+}} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM}$$

3. $$[CoCl_4]^{2-}$$

The overall charge is $$-2$$. There are four $$Cl^-$$ ligands (each $$-1$$), so let the oxidation state of Co be $$x$$:

$$x + 4(-1) = -2 \;\; \Longrightarrow \;\; x = +2$$

Thus we have $$Co^{2+}$$. Atomic number of Co is 27, so

$$Co: [Ar]\,3d^7\,4s^2 \quad \Longrightarrow \quad Co^{2+}: [Ar]\,3d^7$$

$$Cl^-$$ is a very weak-field ligand and the geometry is tetrahedral (which is always high-spin). The d7 configuration therefore keeps three unpaired electrons.

Hence $$n = 3$$.

Magnetic moment:

$$\mu_{Co^{2+}} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM}$$

4. $$[Fe(H_2O)_6]^{2+}$$

Water is neutral, overall charge is $$+2$$, so Fe is in $$+2$$ oxidation state.

Atomic number of Fe is 26, therefore

$$Fe: [Ar]\,3d^6\,4s^2 \quad \Longrightarrow \quad Fe^{2+}: [Ar]\,3d^6$$

With the weak-field $$H_2O$$ ligand we get a high-spin d6 configuration, which contains four unpaired electrons.

Thus $$n = 4$$.

Magnetic moment:

$$\mu_{Fe^{2+}} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM}$$

Comparison of the magnetic moments

$$[Mn(H_2O)_6]^{2+} \; : \; 5.92 \text{ BM}$$

$$[Cr(H_2O)_6]^{2+} \; : \; 4.90 \text{ BM}$$

$$[CoCl_4]^{2-} \; : \; 3.87 \text{ BM}$$

$$[Fe(H_2O)_6]^{2+} \; : \; 4.90 \text{ BM}$$

We observe that the values for $$[Cr(H_2O)_6]^{2+}$$ and $$[Fe(H_2O)_6]^{2+}$$ are identical (each $$\approx 4.90 \text{ BM}$$). No other pair matches.

Hence, the correct answer is Option D.