The coordination complex [Cr(CO) 6 ] forms colorless, diamagnetic crystals that melt at 90 °C c. Given that [Cr(CO) 6 ] is colorless, would you expect CO to be a weak-field or strong-field ligand?

Hello, everyone. Today, we have the following problem. The following complex is observed to form a light blue solution would water be considered a weak field or strong field ligand, a weak field ligand or B strong field ligand. So this question draws on the crystal field theory. And this theory basically states that the the electronic or the electrons transition from a high energy or they transition, excuse me, from a lower energy to a higher energy. And these consist or these concern D orbitals. And these also help determine the color of coordinator complexes. So this complex here has six L ins, six water L ins. So it has a coordinating number that is equal to six giving it an octahedral geometry for octahedral complexes. The crystal field splitting energy of the metal ion is the following. So we have it these lower energy de orbitals as the following. And then it's followed by two higher energy de orbitals with the following configurations. No, a strong field like in will have a larger energy value. So the higher the energy value will directly correlate or equal a strong filled it like again. And the same can be said for the reverse, a decreased energy value can correlate with a weak field Lian. So the complex is colored that we see it's a light blue solution. And therefore D to D electron transitions are possible. And there must actually be a small gap between the de orbitals and the splitting energy must have a relatively low value, meaning that water is a weak field ligand. And so with that, we have solved the problem overall, I hope this helped and not until next time.

Ch.23 - Transition Metals and Coordination Chemistry

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Brown 14th Edition

Ch.23 - Transition Metals and Coordination Chemistry

Problem 91c

Chapter 23, Problem 91c

The coordination complex [Cr(CO)₆] forms colorless, diamagnetic crystals that melt at 90 °C
c. Given that [Cr(CO)₆] is colorless, would you expect CO to be a weak-field or strong-field ligand?

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1. The color of a coordination complex is determined by the energy difference between the d-orbitals. When light is absorbed, an electron is excited from a lower energy d-orbital to a higher energy d-orbital. The color we see is the complementary color of the light absorbed.

2. If the complex is colorless, it means that the energy difference between the d-orbitals is too large to be overcome by the energy of visible light. Therefore, no light is absorbed and no color is observed.

3. Ligands can be classified as weak-field or strong-field based on their ability to split the d-orbitals. Weak-field ligands cause a small split in the d-orbitals, while strong-field ligands cause a large split.

4. Since [Cr(CO)<sub>6</sub>] is colorless, it suggests that the energy difference between the d-orbitals is large. Therefore, CO is likely a strong-field ligand because it causes a large split in the d-orbitals.

5. Additionally, the complex is diamagnetic, which means all of its electrons are paired. This is another characteristic of complexes with strong-field ligands, as they tend to cause pairing of electrons in the d-orbitals.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Ligand Field Theory

Ligand Field Theory explains how ligands affect the energy levels of d-orbitals in transition metal complexes. Strong-field ligands cause a larger splitting of the d-orbitals, leading to lower energy configurations and often resulting in low-spin complexes. Conversely, weak-field ligands cause smaller splitting, leading to high-spin configurations. Understanding this theory is crucial for predicting the magnetic and color properties of coordination complexes.

Color and Electronic Transitions

The color of a coordination complex is determined by the wavelengths of light absorbed during electronic transitions between d-orbitals. A colorless complex indicates that it does not absorb visible light, which can suggest that the d-orbital splitting is either very small or that the electronic transitions do not fall within the visible spectrum. This concept is essential for understanding the relationship between ligand strength and the resulting color of the complex.

Diamagnetism

Diamagnetism is a property of materials that are not attracted to a magnetic field and occurs when all electrons are paired. In coordination complexes, the presence of paired electrons in the d-orbitals, often due to strong-field ligands, results in diamagnetism. Recognizing the relationship between ligand strength, electron pairing, and magnetic properties helps in predicting the behavior of complexes like [Cr(CO)6].

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Related Practice

Open Question

Complete the exercises below. When Alfred Werner was developing the field of coordination chemistry, it was argued by some that the optical activity he observed in the chiral complexes he had prepared was due to the presence of carbon atoms in the molecule. To disprove this argument, Werner synthesized a chiral complex of cobalt that had no carbon atoms in it, and he was able to resolve it into its enantiomers. Design a cobalt(III) complex that would be chiral if it could be synthesized and that contains no carbon atoms. (It may not be possible to synthesize the complex you design, but we will not worry about that for now.)

Open Question

Complete the exercises below. Generally speaking, for a given metal and ligand, the stability of a coordination compound is greater for the metal in the +3 oxidation state rather than in the +2 oxidation state (for metals that form stable +3 ions in the first place). Suggest an explanation, keeping in mind the Lewis acid–base nature of the metal–ligand bond.

Textbook Question

The coordination complex [Cr(CO)₆] forms colorless, diamagnetic crystals that melt at 90 °C

a. What is the oxidation number of chromium in this compound?

d. Write the name for [Cr(CO)₆] using the nomenclature rules for coordination compounds.