Question

The coordination complex \(\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]\) forms colorless, diamagnetic crystals that melt at \(90{ }^{\circ} \mathrm{C}\). (a) What is the oxidation number of chromium in this compound? (b) Given that \(\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]\) is diamagnetic, what is the electron configuration of chromium in this compound? (c) Given that \(\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]\) is colorless, would you expect CO to be a weak-field or strongfield ligand? (d) Write the name for \(\left[\mathrm{Cr}(\mathrm{CO})_{4}\right]\) using the nomenclature rules for coordination compounds.

Short answer

(a) The oxidation number of chromium in \(\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]\) is 0. (b) The electron configuration of chromium in this compound is \([Ar] 3d^4 4s^0\). (c) CO is a strong-field ligand. (d) The name of the complex \(\left[\mathrm{Cr}(\mathrm{CO})_{4}\right]\) is Tetracarbonylchromium(0).

Step-by-step solution

Part (a)

To find the oxidation number of chromium, we need to consider the charge on each ligand (CO in this case) and the overall charge of the complex. Carbonyl ligands (CO) are neutral ligands (no charge), and since the coordination complex is also neutral, the oxidation number of the chromium will be 0: Oxidation number of Cr = Overall charge - (number of ligands * charge on each ligand) Oxidation number of Cr = 0 - (6 * 0) = 0

Part (b)

The coordination complex is diamagnetic. Diamagnetic compounds have all electrons paired. The electron configuration of Cr in the ground state is \([Ar] 3d^5 4s^1\). To achieve a diamagnetic state, chromium must lose one electron from the 3d orbitals and one electron from the 4s orbitals. So, the electron configuration of chromium in the compound will be: \([Ar] 3d^4 4s^0\)

Part (c)

As the coordination complex is colorless, it means that there is no absorption of visible light. So, there must be a large energy gap between the d orbitals involved in bonding. This large energy gap is usually associated with strong-field ligands. Hence, CO is a strong-field ligand.

Part (d)

To name the complex \(\left[\mathrm{Cr}(\mathrm{CO})_{4}\right]\), we first need to consider the oxidation state of chromium, which is 0, as determined in part (a). The name follows this order: ligand name, cation name (metal), followed by the oxidation state in Roman numerals. The ligand has the name carbonyl, and the metal is chromium. Thus, the name of the complex is: Tetracarbonylchromium(0)

Key concepts

Oxidation number

The oxidation number is a crucial concept in understanding coordination complexes. It represents the hypothetical charge an atom would have if all bonds were ionic. To calculate the oxidation number of a central metal in a complex, you must consider both the overall charge of the complex and the charges on any ligands.
For example, in the complex oindent\(\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]\), carbon monoxide (CO) is a neutral ligand, meaning it carries no charge. Therefore, the sum of the charges of the ligands and the complex must equal the oxidation number of chromium. Given that the complex is neutral: \( \text{Oxidation number of Cr} = 0 - (6 \times 0) = 0\).

• The oxidation number is zero.
• The metal does not gain or lose electrons.

Identifying the oxidation number helps predict the compound's electronic structure and reactivity.

Diamagnetic compounds

Diamagnetic compounds have a fascinating characteristic: all of their electrons are paired. In the case of the compound \(\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]\), the diamagnetism indicates a fully paired electron arrangement. The electron configuration of chromium in its ground state is \([Ar] 3d^5 4s^1\).

In a diamagnetic state, electrons rearrange to ensure all are paired. For chromium in \(\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]\), it involves losing one electron from 3d and 4s each, resulting in an electron configuration of \([Ar] 3d^4 4s^0\).

• Electrons are rearranged within the d-orbitals.
• Results in no net magnetic moment.

Fully paired electrons minimize the magnetic interaction with external magnetic fields, making the compound diamagnetic.

Strong-field ligands

Strong-field ligands are known for their significant ability to split the d-orbitals in a coordination complex. This splitting influences both the color and magnetic properties of the complex. In \(\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]\), the ligand CO is considered a strong-field ligand. This is verified by the compound's colorless nature, indicating no absorption in the visible spectrum due to a large energy gap in the d-orbitals.

• Larger energy gap means no visible light absorption.
• Results in stronger bond with the central metal atom.

Strong-field ligands like CO lead to a low-spin state, which often results in paired electrons. This is why such complexes can be both diamagnetic and colorless.

Naming coordination compounds

Naming coordination compounds follows a systematic approach ensuring easy identification and understanding. The naming starts with the ligands, followed by the central metal atom, and finally, the metal's oxidation state in Roman numerals, if necessary. For example, in\(\left[\mathrm{Cr}(\mathrm{CO})_{4}\right]\):

• Ligands are named first: 'carbonyl'.
• Cation or central metal is named next: 'chromium'.
• The oxidation number follows: '(0)' since chromium has an oxidation state of zero.

Thus, the name for this coordination compound becomes 'Tetracarbonylchromium(0)'. Using consistent rules ensures clarity across diverse and complex chemistry discussions.