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

To find the pair having the same magnetic moment, we need to determine the number of unpaired electrons ($$n$$) for each complex ion. The magnetic moment is determined by the formula $$\mu = \sqrt{n(n+2)}\text{ BM}$$, which means complexes with the same number of unpaired electrons will have the same magnetic moment.

The atomic numbers provided are $$\text{Cr} = 24$$, $$\text{Mn} = 25$$, $$\text{Fe} = 26$$, and $$\text{Co} = 27$$.

Let us analyze the complex ions in option D:

For $$[Cr(H_2O)_6]^{2+}$$, chromium is in the $$+2$$ oxidation state with a $$3d^4$$ valence configuration. Since $$H_2O$$ is a weak field ligand in an octahedral crystal field, the electrons distribute as $$t_{2g}^3 e_g^1$$, giving $$n = 4$$ unpaired electrons.

For $$[Fe(H_2O)_6]^{2+}$$, iron is in the $$+2$$ oxidation state with a $$3d^6$$ valence configuration. In an octahedral weak field, the configuration is $$t_{2g}^4 e_g^2$$, which also results in $$n = 4$$ unpaired electrons.

Since both $$[Cr(H_2O)_6]^{2+}$$ and $$[Fe(H_2O)_6]^{2+}$$ have $$4$$ unpaired electrons, their magnetic moments are identical.

Hence the correct option is D.