The \(\mathrm{CrF}_{6}{ }^{4-}\) ion is known to have four unpaired electrons. Does the \(\mathrm{F}^{-}\) ligand produce a strong or weak field?

The complex ion, CrF₆⁴⁻, has chromium (Cr) as the central metal ion. To determine its oxidation state, we can use the fact that the overall charge of the complex ion is -4 and the charge ofeach fluoride ligand is -1. Let the oxidation state of Cr be x. Then, we can express the total charge as: x + 6(-1) = -4 Solving for x, we find that the oxidation state of Cr is +2.

The atomic number of chromium is 24, which means its ground state electron configuration is [Ar] 3d⁵ 4s¹. Since Cr²⁺ has a +2 oxidation state, it loses two electrons, one from the 4s and one from the 3d orbitals. The electron configuration of Cr²⁺ is [Ar] 3d⁴.

To determine if F⁻ produces a strong or weak field, we need to analyze the ligand field splitting in the octahedral complex. In a strong field, the d orbitals split into two groups: the lower-energy t₂g orbitals and the higher-energy e_g orbitals. This splitting results in a large energy gap (Δ) between the orbital levels. Electrons preferentially occupy the lower-energy t₂g orbitals before pairing up in the e_g orbitals. In contrast, a weak field has a smaller energy gap (Δ), and electrons can more easily occupy the higher-energy e_g orbitals. We are given that the CrF₆⁴⁻ ion has four unpaired electrons. This information suggests a weak ligand field, since a strong field would cause electron pairing, resulting in fewer unpaired electrons.

Considering that the CrF₆⁴⁻ ion has four unpaired electrons, it indicates that the energy gap (Δ) between the t₂g and e_g orbitals is not large enough to force the electrons to pair up. Therefore, we can conclude that the F⁻ ligand in this complex produces a weak field.