1. Find oxidation states and d-electron counts

• For $[Cr(NH_3)_6]^{3+}$:

  • $NH_3$ is a neutral ligand.
  • So oxidation state of Cr is $+3$.
  • Cr: $Z=24 \Rightarrow [Ar]3d^54s^1$
  • $Cr^{3+}$ loses 3 electrons $\Rightarrow 3d^3$

• For $[CuF_6]^{3-}$:

  • Each $F^-$ has charge $-1$.
  • Let oxidation state of Cu be $x$: $x+6(-1)=-3 \Rightarrow x=+3$
  • Cu: $Z=29 \Rightarrow [Ar]3d^{10}4s^1$
  • $Cu^{3+}$ loses 3 electrons $\Rightarrow 3d^8$

2. Determine number of unpaired electrons

• $[Cr(NH_3)_6]^{3+}$:

  • Octahedral complex with $d^3$ configuration.
  • Electron arrangement: $t_{2g}^3e_g^0$
  • Number of unpaired electrons, $n=3$

• $[CuF_6]^{3-}$:

  • Octahedral complex with $d^8$ configuration.
  • In octahedral field: $t_{2g}^6e_g^2$
  • Number of unpaired electrons, $n=2$

3. Use spin-only magnetic moment formula

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

• For $[Cr(NH_3)_6]^{3+}$: $\mu = \sqrt{3(3+2)}=\sqrt{15}\approx 3.87\;\text{BM}$

• For $[CuF_6]^{3-}$: $\mu = \sqrt{2(2+2)}=\sqrt{8}\approx 2.84\;\text{BM}$

4. Match with options

Thus the calculated spin-only magnetic moments are: $3.87\;\text{BM and }2.84\;\text{BM}$

So the correct option is A.