Problem 57 - General Chemistry: Principles and Modern Applications

Question

Provide a valence bond description of the bonding in the \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) ion. According to the valence bond description, how many unpaired electrons are there in the \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) complex? How does this prediction compare with that of crystal field theory?

Step-by-Step Solution

Answer

In Cr(NH3)6^3+ ion, as per valence bond theory, there will be no unpaired electrons. The prediction from valence bond is different from that of crystal field theory, which predicts that there are no unpaired electrons.

Step 1: Determine Chromium Electron Configuration

The atomic number of Chromium (Cr) is 24. So, its electron configuration is \([Ar] 3d^5 4s^1\). In the case of the \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) ion, Chromium loses 3 electrons. This leads to an electron configuration of \([Ar] 3d^3\), which means it has 3 electrons in the 3d orbital.

Step 2: Apply Valence Bond Theory

According to the valence bond theory, the presence of six NH3 ligands results in the hybridization of Cr. The hybridization is \(d^2sp^3\) since we have 6 ligands forming bonds with the Cr^3+ center. Three electrons in the d orbital get distributed among the hybrid orbitals, so one unpaired electron will be paired up with one ligand electron each, in three hybrid orbitals.

Step 3: Determine the Number of Unpaired Electrons

As per the valence bond theory, unpaired electrons are the result of empty orbitals. Since three ligands each bonded with an electron in the hybrid orbitals, this leaves three hybrid orbitals with no unpaired electron.

Step 4: Compare with Crystal Field Theory

If we apply crystal field theory instead, we would conclude there are no unpaired electrons. This theory suggests that strong field ligands like NH3 result in the pairing up of all the 3d electrons in lower t2g levels, leaving no unpaired electron.

Key Concepts

Chromium Electron Configuration

When learning about the electron configuration of chromium, it's important to note that it has an unusual setup. Chromium (Cr) is element number 24 on the periodic table. Typically, elements follow a strict order in filling orbitals; however, chromium is an exception. Its ground state electron configuration is

• \([Ar] 3d^5 4s^1\).

This means it has five electrons in the 3d subshell and one in the 4s subshell, rather than the expected

• \([Ar] 3d^4 4s^2\).

This behavior is due to stability gained by having a half-filled 3d subshell, which reduces electron repulsion more efficiently.
For the \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) complex, chromium loses three electrons, typically from

• the 4s and two 3d orbitals,

resulting in the configuration

• \([Ar] 3d^3\).

This sets the stage for understanding bonding interactions in coordination chemistry.

Hybridization

In coordination complexes like \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\), hybridization is a key concept for understanding how bonds form. Hybridization is a model used in valence bond theory to describe the merging of atomic orbitals into new hybrid orbitals. These are used by the central metal ion to bond with ligands.For \(\mathrm{Cr}^{3+}\) in this complex, six \(\mathrm{NH}_{3}\) ligands approach, leading to

• \(d^2sp^3\)

hybridization. Here, two 3d orbitals, one 4s, and three 4p orbitals combine, forming six equivalent hybrid orbitals.These hybrid orbitals can then pair with electrons from the ligands' electron pairs. In valence bond theory, the pairing with ligand electrons results in the central chromium having all electrons paired. So, according to this theory, you see no unpaired electrons remaining, indicative of the electron pairing that occurs when strong field ligands like \(\mathrm{NH}_{3}\) are present.

Crystal Field Theory

Crystal Field Theory (CFT) offers a different perspective compared to valence bond theory. It's a model that explains the electronic structure of transition metal complexes by considering ligand interactions. In CFT, ligands are treated as electric point charges that create an electrostatic field when approaching the central metal ion. This affects the d orbitals, splitting them into different energy levels. For octahedral complexes like \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\), the d orbitals split into two groups:

• Lower energy \(t_{2g}\) orbitals,
• Higher energy \(e_g\) orbitals.

With strong field ligands such as \(\mathrm{NH}_{3}\), electrons fill the lower energy \(t_{2g}\) levels first, pairing up before occupying higher levels. This results in no unpaired electrons in this configuration.Crystal Field Theory aligns with valence bond theory in predicting no unpaired electrons for \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\), showing consistency between these analytical approaches.