1. Use spin-only magnetic moment formula

For a coordination compound, the spin-only magnetic moment is

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

where $n$ is the number of unpaired electrons.

So, we only need to find the number of unpaired electrons in each complex.

2. Option A: $[Cr(H_2O)_6]^{2+}$ and $[CoCl_4]^{2-}$

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

• Oxidation state of Cr = $+2$
• $Cr: [Ar]3d^54s^1$
• $Cr^{2+}: 3d^4$
• $H_2O$ is a weak field ligand, octahedral high-spin complex
• Hence configuration: $t_{2g}^3e_g^1$
• Number of unpaired electrons = $4$

So, $\mu = \sqrt{4(4+2)} = \sqrt{24}$

(ii) $[CoCl_4]^{2-}$

• Oxidation state of Co = $+2$
• $Co^{2+}: 3d^7$
• $Cl^-$ is weak field, tetrahedral complex is always high spin
• For tetrahedral $d^7$, number of unpaired electrons = $3$

So, $\mu = \sqrt{3(3+2)} = \sqrt{15}$

These are not same.

3. Option B: $[Co(OH)_4]^{2-}$ and $[Fe(NH_3)_6]^{2+}$

(i) $[Co(OH)_4]^{2-}$

• Oxidation state of Co = $+2$
• $Co^{2+}: 3d^7$
• Tetrahedral with weak field ligand $OH^-$, hence high spin
• Tetrahedral $d^7$ has $3$ unpaired electrons

So, $\mu = \sqrt{15}$

(ii) $[Fe(NH_3)_6]^{2+}$

• Oxidation state of Fe = $+2$
• $Fe^{2+}: 3d^6$
• $NH_3$ is not strong enough here to pair $Fe^{2+}$ in octahedral complex in typical JEE treatment, so high spin
• Octahedral high-spin $d^6$: $t_{2g}^4e_g^2$
• Number of unpaired electrons = $4$

So, $\mu = \sqrt{24}$

These are not same.

4. Option C: $[Mn(H_2O)_6]^{2+}$ and $[Cr(H_2O)]^{2+}$

The second complex appears to be written as $[Cr(H_2O)]^{2+}$, which is likely a typo for a hydrated chromium(II) species; but taking the given species as chromium(II), its $d$-count remains $d^4$.

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

• Oxidation state of Mn = $+2$
• $Mn^{2+}: 3d^5$
• $H_2O$ is weak field, octahedral high spin
• Number of unpaired electrons = $5$

So, $\mu = \sqrt{5(5+2)} = \sqrt{35}$

(ii) Chromium(II) aqua complex

• $Cr^{2+}: 3d^4$
• Weak field aqua ligand gives high-spin configuration
• Number of unpaired electrons = $4$

So, $\mu = \sqrt{24}$

These are not same.

5. Option D: $[Cr(H_2O)_6]^{2+}$ and $[Fe(H_2O)_6]^{2+}$

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

• $Cr^{2+}: 3d^4$
• Octahedral, weak field, high spin
• Unpaired electrons = $4$

Thus, $\mu = \sqrt{24}$

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

• $Fe^{2+}: 3d^6$
• $H_2O$ is weak field, octahedral high spin
• Configuration: $t_{2g}^4e_g^2$
• Unpaired electrons = $4$

Thus, $\mu = \sqrt{24}$

Both have the same magnetic moment.

6. Conclusion

The correct pair is:

$\boxed{D}$