The correct statement about the magnetic properties of [Fe(CN)$$_6$$]$$^{3-}$$ and [FeF$$_6$$]$$^{3-}$$ is: (Z = 26).

To determine the magnetic properties of the complexes [Fe(CN)$$_6$$]$$^{3-}$$ and [FeF$$_6$$]$$^{3-}$$, we need to analyze the number of unpaired electrons in each complex. Paramagnetic substances have unpaired electrons, while diamagnetic substances have all electrons paired. The central metal ion is iron (Fe) with atomic number Z = 26, so its electron configuration is [Ar] 4s$$^2$$ 3d$$^6$$.

First, we find the oxidation state of iron in both complexes. For [Fe(CN)$$_6$$]$$^{3-}$$, let the oxidation state of Fe be $$x$$. Cyanide (CN$$^-$$) is a ligand with charge -1, and the complex has an overall charge of -3. So, $$x + 6 \times (-1) = -3$$, which gives $$x - 6 = -3$$, and solving for $$x$$, we get $$x = +3$$. Similarly, for [FeF$$_6$$]$$^{3-}$$, fluoride (F$$^-$$) also has charge -1, so $$y + 6 \times (-1) = -3$$, giving $$y - 6 = -3$$, and $$y = +3$$. Thus, in both complexes, iron is in the +3 oxidation state.

The electron configuration of Fe$$^{3+}$$ is derived by removing three electrons from Fe (which has configuration [Ar] 4s$$^2$$ 3d$$^6$$). The 4s electrons are removed first, followed by one from 3d, so Fe$$^{3+}$$ has configuration [Ar] 3d$$^5$$.

Now, we consider the ligands and their field strengths. Cyanide (CN$$^-$$) is a strong field ligand, causing large splitting of the d-orbitals, leading to a low-spin complex. Fluoride (F$$^-$$) is a weak field ligand, causing small splitting, leading to a high-spin complex.

For [Fe(CN)$$_6$$]$$^{3-}$$ (low-spin d$$^5$$ complex in octahedral field):

• The d-orbitals split into t$$_{2g}$$ (lower energy, three orbitals) and e$$_g$$ (higher energy, two orbitals).
• In low-spin complexes, electrons pair up in the t$$_{2g}$$ orbitals before occupying e$$_g$$.
• With five electrons, the configuration is $$t_{2g}^5 \, e_g^0$$.
• Three t$$_{2g}$$ orbitals hold five electrons: two orbitals have paired electrons (two pairs, four electrons), and one orbital has one unpaired electron.
• Thus, there is one unpaired electron, making the complex paramagnetic.

For [FeF$$_6$$]$$^{3-}$$ (high-spin d$$^5$$ complex in octahedral field):

• In high-spin complexes, electrons occupy all orbitals singly before pairing.
• With five electrons, each d-orbital gets one electron: three in t$$_{2g}$$ and two in e$$_g$$.
• The configuration is $$t_{2g}^3 \, e_g^2$$.
• All five electrons are unpaired.
• Thus, there are five unpaired electrons, making the complex paramagnetic.

Therefore, both complexes are paramagnetic.

Now, evaluating the options:

• A: Both are paramagnetic. (Matches our finding)
• B: Both are diamagnetic. (Incorrect, as both have unpaired electrons)
• C: [Fe(CN)$$_6$$]$$^{3-}$$ is paramagnetic, [FeF$$_6$$]$$^{3-}$$ is diamagnetic. (Incorrect, as both are paramagnetic)
• D: [Fe(CN)$$_6$$]$$^{3-}$$ is diamagnetic, [FeF$$_6$$]$$^{3-}$$ is paramagnetic. (Incorrect, as both are paramagnetic)

Hence, the correct answer is Option A.