The calculated spin only magnetic moments of $$[Cr(NH_3)_6]^{3+}$$ and $$[CuF_6]^{3-}$$ in BM, respectively, are (Atomic numbers of Cr and Cu are 24 and 29, respectively)

For any transition-metal complex, the spin-only magnetic moment is calculated from
$$\mu_{\text{spin}} = \sqrt{n\,(n+2)}\;\text{BM}$$
where $$n$$ is the number of unpaired electrons.

Case 1: $$[Cr(NH_3)_6]^{3+}$$

Step-1: Oxidation state of the metal.
Let the oxidation state of Cr be $$x$$.
$$x + 6(0) = +3 \;\Rightarrow\; x = +3$$
So the metal ion is $$Cr^{3+}$$.

Step-2: d-electron count of $$Cr^{3+}$$.
Atomic number of Cr = 24, ground state = $$[Ar]\,3d^{5}4s^{1}$$.
Removing three electrons (to form $$Cr^{3+}$$) gives $$3d^{3}$$ (i.e. $$d^3$$ configuration).

Step-3: Nature of ligand and electron arrangement.
$$NH_3$$ is a weak-field ligand in 3d series; hence a high-spin octahedral complex is formed.
For $$d^3$$ in an octahedral field: $$t_{2g}^{3}\,e_g^{0}$$ → all three electrons remain unpaired.
Therefore, $$n = 3$$.

Step-4: Magnetic moment.
$$\mu = \sqrt{3(3+2)} = \sqrt{15} = 3.87\;\text{BM}$$

Case 2: $$[CuF_6]^{3-}$$

Step-1: Oxidation state of the metal.
Let the oxidation state of Cu be $$x$$.
$$x + 6(-1) = -3 \;\Rightarrow\; x = +3$$
So the metal ion is $$Cu^{3+}$$.

Step-2: d-electron count of $$Cu^{3+}$$.
Atomic number of Cu = 29, ground state = $$[Ar]\,3d^{10}4s^{1}$$.
Removing three electrons gives $$3d^{8}$$ (i.e. $$d^8$$ configuration).

Step-3: Nature of ligand and electron arrangement.
$$F^-$$ is a weak-field ligand, so the complex is high-spin octahedral.
For high-spin $$d^8$$ in an octahedral field: $$t_{2g}^{6}\,e_g^{2}$$ with one electron in each $$e_g$$ orbital.
Hence, $$n = 2$$ unpaired electrons.

Step-4: Magnetic moment.
$$\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.83 \approx 2.84\;\text{BM}$$

Thus, the calculated spin-only magnetic moments are 3.87 BM for $$[Cr(NH_3)_6]^{3+}$$ and 2.84 BM for $$[CuF_6]^{3-}$$.

Therefore, the correct choice is:
Option A which is: 3.87 and 2.84.