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

First, we need to find the oxidation state of Chromium in the complex ion. We know that the overall charge of the complex ion is -4, and since there are six fluorine ligands (each with a -1 charge), the oxidation state of Chromium must be +2.

The atomic number of Chromium is 24, which means that its electronic configuration is \[1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5, 4s^1\]. Since the Chromium ion in the complex has an oxidation state of +2, it loses two electrons. The new electronic configuration for the Chromium(II) ion is \[1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^4\].

In an octahedral complex, the \(d\) orbitals of the transition metal ion undergo crystal-field splitting into two sets: \(t_{2g}\) and \(e_{g}\). The \(t_{2g}\) orbitals are lower in energy with three orbitals, while the \(e_{g}\) orbitals are higher in energy, containing two orbitals. The energy gap between these orbits is referred to as Δ. The size of Δ will help us determine if the \(\mathrm{F}^{-}\) ligand produces a strong or weak field.

We are given that the \(\mathrm{CrF}_{6}^{4-}\) ion has four unpaired electrons, and from the electronic configuration in Step 2, we know that Chromium(II) has 4 electrons in the \(3d\) orbitals. This suggests the electrons are placed as such on the crystal-field splitting diagram: - \(t_{2g}\) orbitals: Each of the three orbitals has one unpaired electron. - \(e_{g}\) orbitals: One of the two orbitals has an unpaired electron, while the other is empty. Based on this electron placement, if a strong field ligand were present, it would require a higher Δ, resulting in the electrons occupying the lower energy \(t_{2g}\) orbitals and pairing up before entering the higher energy \(e_{g}\) orbitals. However, the complex has four unpaired electrons, indicating that the energy gap between the \(t_{2g}\) and \(e_{g}\) orbitals (Δ) is relatively small.

Since the energy gap Δ between the \(t_{2g}\) and \(e_{g}\) orbitals is relatively small and there are four unpaired electrons present, we can conclude that the \(\mathrm{F}^{-}\) ligand in the \(\mathrm{CrF}_{6}^{4-}\) ion produces a weak field.