Which of the following complex ions has electrons that are symmetrically filled in both $$t_{2g}$$ and $$e_g$$ orbitals?

To determine which complex ion has electrons symmetrically filled in both $$t_{2g}$$ and $$e_g$$ orbitals, we need to analyze the electron configuration of each complex. Symmetric filling means that within each set of orbitals ($$t_{2g}$$ or $$e_g$$), the electrons are distributed evenly—either all orbitals are half-filled (one electron per orbital) or fully filled (two electrons per orbital). This ensures a symmetric electron distribution.

We will evaluate each option by finding the oxidation state of the central metal ion, the number of d-electrons, and the spin state (high or low) based on the ligands. Then, we will distribute the electrons in the octahedral crystal field splitting diagram ($$t_{2g}$$ and $$e_g$$ orbitals) and check for symmetric filling.

Option A: $$[Co(NH_3)_6]^{2+}$$

Ammonia ($$NH_3$$) is a neutral ligand, so the charge comes from the metal ion. The complex has a +2 charge, so cobalt (Co) must be in the +2 oxidation state. Cobalt has an atomic number of 27, so $$Co^{2+}$$ has the electron configuration $$[Ar] 3d^7$$. Ammonia is a strong field ligand, so the complex is low spin. In a low spin octahedral complex, electrons fill the lower energy $$t_{2g}$$ orbitals first. For $$d^7$$:

• $$t_{2g}$$ orbitals (three orbitals) hold six electrons (fully filled, paired).
• The remaining one electron goes into the $$e_g$$ orbitals (two orbitals).

So, $$t_{2g}^6 e_g^1$$. The $$t_{2g}$$ set is fully filled (symmetric), but the $$e_g$$ set has one electron in one orbital and an empty orbital, which is asymmetric. Therefore, this complex does not have symmetric filling in both sets.

Option B: $$[Mn(CN)_6]^{4-}$$

Cyanide ($$CN^-$$) has a -1 charge. The complex has a -4 charge, so manganese (Mn) must be in the +2 oxidation state (since $$x + 6 \times (-1) = -4$$ gives $$x = +2$$). Manganese has an atomic number of 25, so $$Mn^{2+}$$ has the configuration $$[Ar] 3d^5$$. Cyanide is a strong field ligand, so the complex is low spin. For low spin $$d^5$$:

• All five electrons go into the $$t_{2g}$$ orbitals, giving $$t_{2g}^5 e_g^0$$.

The $$t_{2g}$$ set has five electrons: two orbitals are fully filled (two electrons each), and one orbital has one electron. This is asymmetric because the orbitals have different occupancies. The $$e_g$$ set is empty (symmetric), but since $$t_{2g}$$ is asymmetric, the complex does not have symmetric filling in both sets.

Option C: $$[CoF_6]^{3-}$$

Fluoride ($$F^-$$) has a -1 charge. The complex has a -3 charge, so cobalt (Co) must be in the +3 oxidation state (since $$x + 6 \times (-1) = -3$$ gives $$x = +3$$). Cobalt has an atomic number of 27, so $$Co^{3+}$$ has the configuration $$[Ar] 3d^6$$. Fluoride is a weak field ligand, so the complex is high spin. For high spin $$d^6$$:

• The $$t_{2g}$$ orbitals hold four electrons: two orbitals have one electron each, and one orbital has two electrons.
• The $$e_g$$ orbitals hold two electrons: one electron in each orbital.

So, $$t_{2g}^4 e_g^2$$. The $$t_{2g}$$ set is asymmetric (different occupancies), while the $$e_g$$ set is half-filled (symmetric). Since $$t_{2g}$$ is asymmetric, the complex does not have symmetric filling in both sets.

Option D: $$[FeF_6]^{3-}$$

Fluoride ($$F^-$$) has a -1 charge. The complex has a -3 charge, so iron (Fe) must be in the +3 oxidation state (since $$x + 6 \times (-1) = -3$$ gives $$x = +3$$). Iron has an atomic number of 26, so $$Fe^{3+}$$ has the configuration $$[Ar] 3d^5$$. Fluoride is a weak field ligand, so the complex is high spin. For high spin $$d^5$$:

• The $$t_{2g}$$ orbitals hold three electrons: one electron in each orbital (half-filled).
• The $$e_g$$ orbitals hold two electrons: one electron in each orbital (half-filled).

So, $$t_{2g}^3 e_g^2$$. Both sets are half-filled: $$t_{2g}$$ has one electron per orbital (symmetric), and $$e_g$$ has one electron per orbital (symmetric). Therefore, this complex has symmetric filling in both sets.

Hence, the correct answer is Option D.