The pair in which both the species have the same magnetic moment (spin only) is:

For comparing the magnetic moments we shall use the spin-only formula

$$\mu_\text{spin}=\sqrt{n(n+2)}\;\text{BM},$$

where $$n$$ is the number of unpaired electrons present in the metal ion inside the complex. Hence, for every complex we must first decide

(i) the oxidation state of the metal,

(ii) the 3d electron count of that oxidation state, and

(iii) whether the complex is high-spin or low-spin (decided by the strength of the ligand and the geometry).

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

Water is a neutral and weak field ligand, so both complexes are high-spin octahedral.

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

Charge balance gives $$\text{Oxidation state of Cr}=+2.$$

Atomic number of Cr = 24, so $$Cr^{2+}$$ has the electronic configuration $$[Ar]\,3d^4.$$

In a weak field octahedron the d4 distribution is $$t_{2g}^{3}e_g^{1},$$ containing

$$n=4$$ unpaired electrons.

Hence $$\mu=\sqrt{4(4+2)}=\sqrt{24}=4.90\;\text{BM}.$$

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

Again, $$\text{Oxidation state of Fe}=+2.$$

Atomic number of Fe = 26, so $$Fe^{2+}$$ is $$[Ar]\,3d^6.$$

With water (weak field) the octahedral d6 arrangement is high-spin $$t_{2g}^{4}e_g^{2},$$ giving

$$n=4$$ unpaired electrons.

Therefore $$\mu=\sqrt{4(4+2)}=\sqrt{24}=4.90\;\text{BM}.$$

Both complexes possess the same magnetic moment (4.90 BM).

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

$$[Co(OH)_4]^{2-}$$ is tetrahedral. Total ligand charge $$=-4$$, overall charge $$=-2$$, hence $$\text{Co}=+2\;(d^7).$$ In a tetrahedral, weak-field environment the complex is high-spin and the d7 distribution gives

$$n=3$$ unpaired electrons.

$$\mu=\sqrt{3(3+2)}=\sqrt{15}=3.87\;\text{BM}.$$

$$[Fe(NH_3)_6]^{2+}$$ is octahedral with NH3 a borderline/weak ligand, so high-spin $$d^6$$ with

$$n=4$$ and $$\mu=4.90\;\text{BM}.$$

Since 3.87 BM ≠ 4.90 BM, the pair does not match.

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

For $$Mn^{2+}$$ (water, high-spin) we have $$d^5$$ with

$$n=5,\quad \mu=\sqrt{5(5+2)}=\sqrt{35}=5.92\;\text{BM}.$$

We already found $$[Cr(H_2O)_6]^{2+}$$ to be 4.90 BM, so the moments differ.

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

$$[CoCl_4]^{2-}$$ is tetrahedral, Cl− weak field, giving high-spin $$Co^{2+}$$ ($$d^7$$) with

$$n=3,\;\mu=3.87\;\text{BM}.$$

This does not equal 4.90 BM of $$[Cr(H_2O)_6]^{2+}.$$

Only Option A contains two complexes each having $$n=4$$ unpaired electrons and therefore the same spin-only magnetic moment.

Hence, the correct answer is Option A.