Textbook Question
In octahedral complexes, the choice between high-spin and low-spin electron configurations arises only for d4 - d7 complexes. Explain.
Ch.21 - Transition Elements and Coordination Chemistry
McMurry8th EditionChemistryISBN: 9781292336145
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Table of contents
- Ch.1 - Chemical Tools: Experimentation & Measurement170
- Ch.2 - Atoms, Molecules & Ions160
- Ch.3 - Mass Relationships in Chemical Reactions169
- Ch.4 - Reactions in Aqueous Solution199
- Ch.5 - Periodicity & Electronic Structure of Atoms102
- Ch.6 - Ionic Compounds: Periodic Trends and Bonding Theory92
- Ch.7 - Covalent Bonding and Electron-Dot Structures61
- Ch.8 - Covalent Compounds: Bonding Theories and Molecular Structure59
- Ch.9 - Thermochemistry: Chemical Energy122
- Ch.10 - Gases: Their Properties & Behavior155
- Ch.11 - Liquids & Phase Changes54
- Ch.12 - Solids and Solid-State Materials73
- Ch.13 - Solutions & Their Properties120
- Ch.14 - Chemical Kinetics98
- Ch.15 - Chemical Equilibrium125
- Ch.16 - Aqueous Equilibria: Acids & Bases110
- Ch.17 - Applications of Aqueous Equilibria147
- Ch.18 - Thermodynamics: Entropy, Free Energy & Equilibrium86
- Ch.19 - Electrochemistry154
- Ch.20 - Nuclear Chemistry96
- Ch.21 - Transition Elements and Coordination Chemistry156
- Ch.22 - The Main Group Elements187
- Ch.23 - Organic and Biological Chemistry51
Chapter 21, Problem 109
The Cr3+(aq) cation is violet, but Y3+(aq) is colorless. Explain why this difference in color occurs.
Verified step by step guidance1
Step 1: Understand the concept of color in transition metal complexes. The color of a transition metal complex is due to the d-d electron transitions within the d-orbitals, which are split in energy in the presence of a ligand field.
Step 2: Recognize that Cr3+ is a transition metal ion. Chromium in the +3 oxidation state has a partially filled d-orbital (specifically, it has 3 electrons in the 3d orbitals), which allows for d-d transitions when light is absorbed, resulting in the violet color.
Step 3: Note that Y3+ is not a transition metal ion. Yttrium in the +3 oxidation state has a completely empty d-orbital (it has no electrons in the 4d orbitals), so there are no d-d transitions possible, leading to it being colorless.
Step 4: Consider the role of ligand field theory. In Cr3+, the presence of ligands causes the d-orbitals to split into different energy levels, allowing for the absorption of specific wavelengths of light, which is perceived as color.
Step 5: Conclude that the difference in color between Cr3+ and Y3+ is due to the presence of d-electrons in Cr3+ that can undergo electronic transitions, whereas Y3+ lacks d-electrons, preventing such transitions and resulting in a colorless appearance.
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
d-Block Elements and Color
Transition metals, such as chromium (Cr) and yttrium (Y), have partially filled d-orbitals. The color of a transition metal ion in solution is primarily due to the presence of d-electrons, which can absorb specific wavelengths of light. When these d-electrons transition between energy levels, they absorb certain colors, resulting in the observed color of the solution.
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Guided course
02:06
d Orbital Orientations
Crystal Field Theory
Crystal Field Theory explains how the arrangement of ligands around a metal ion affects the energy levels of its d-orbitals. In the case of Cr3+, the presence of ligands causes a splitting of the d-orbitals, allowing for electronic transitions that absorb visible light, leading to the violet color. In contrast, Y3+ has a full d-subshell, resulting in no such transitions and thus no color.
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Guided course
01:18The study of ligand-metal interactions helped to form Ligand Field Theory which combines CFT with MO Theory.
Ligand Field Strength
The strength of the ligands surrounding a metal ion influences the extent of d-orbital splitting. Strong field ligands cause greater splitting, which can lead to more significant color differences. In the case of Cr3+, the ligands create a suitable environment for electronic transitions, while Y3+ does not exhibit such transitions due to its electronic configuration, resulting in a colorless solution.
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02:40Strong-Field Ligands result in a large Δ and Weak-Field Ligands result in a small Δ.
Related Practice
Textbook Question
The reaction of the octahedral complex Co(NH3)3(NO2)3 with HCl yields a complex [Co(NH3)3(H2O)Cl2]+ in which the two chloride ligands are trans to one another.
(a) Draw the two possible stereoisomers of the starting material [Co(NH3)3(NO2)3]. (All three NO2- ligands are bonded to Co through the N atom.)
(b) Assuming that the NH3 groups remain in place, which of the two starting isomers could give rise to the observed product?
Textbook Question
Consider the octahedral complex [Co(en)(dien)Cl]2+, where dien = H2NCH2CH2NHCH2CH2NH2, which can be abbreivated
(a) The dien (diethylenetriamine) ligand is a tridentate ligand. Explain what is meant by 'tridentate' and why dien can act as a tridentate ligand.
(b) Draw all possible stereoisomers of [Co(en)(dien)Cl]2+ (dien is a flexible ligand). Which stereoisomers are chiral, and which are achiral?
Textbook Question
For each of the following, (i) give the systematic name of the compound and specify the oxidation state of the transition metal, (ii) draw a crystal field energy-level diagram and assign the d electrons to orbitals, (iii) indicate whether the complex is high-spin or low-spin (for d4 - d7 complexes), and (iv) specify the number of unpaired electrons.
(a) (NH4)[Cr(H2O)6](SO4)2