/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 As shown in Figure 23.26, the \(... [FREE SOLUTION] | 91Ó°ÊÓ

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As shown in Figure 23.26, the \(d-d\) transition of \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) produces an absorption maximum at a wavelength of about \(500 \mathrm{~nm}\). (a) What is the magnitude of \(\Delta\) for \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) in \(\mathrm{kJ} / \mathrm{mol}\) ? (b) How would the magnitude of \(\Delta\) change if the \(\mathrm{H}_{2} \mathrm{O}\) ligands in \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) were replaced with \(\mathrm{NH}_{2}\) ligands?

Short Answer

Expert verified
(a) The magnitude of \(\Delta\) for \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) in \(\mathrm{kJ/mol}\) is approximately \(239.34 \mathrm{~kJ/mol}\). (b) The magnitude of \(\Delta\) will increase if the \(\mathrm{H}_{2} \mathrm{O}\) ligands in \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) were replaced with \(\mathrm{NH}_{2}\) ligands.

Step by step solution

01

Find the energy associated with the absorption maximum at the given wavelength.

We can use Planck's equation to find energy, which states that the energy of a photon, E, is equal to the product of Planck's constant, h, and the speed of light, c, divided by the wavelength, λ: $$ E =\frac{h \times c}{λ} $$ Here, h = \(6.63 \times 10^{-34} \mathrm{Js}\) (Planck's constant) and c = \(3 \times 10^{8} \mathrm{m/s}\) (speed of light). Given λ = \(500 \mathrm{~nm}\), which needs to be converted into meters: \(500 \mathrm{~nm} = 500 \times 10^{-9} \mathrm{~m}\) Insert the values and calculate the energy E: $$ E = \frac{(6.63 \times 10^{-34}\mathrm{Js}) \times (3 \times 10^{8}\mathrm{m/s})}{(500 \times 10^{-9}\mathrm{~m})} $$
02

Calculate the energy of the photon in Joules

Now we can calculate the energy of the photon, which is: $$ E = \frac{(6.63 \times 10^{-34}\mathrm{Js}) \times (3 \times 10^{8}\mathrm{m/s})}{(500 \times 10^{-9}\mathrm{~m})} = 3.978 \times 10^{-19}\mathrm{~J} $$
03

Convert the energy in Joules to energy in kJ/mol

To convert the energy of a single photon into energy per mole of photons, we need to multiply by Avogadro's number (\(6.022 \times 10^{23}\mathrm{mol^{-1}}\)) and convert Joules to kJ by dividing by \(10^3\): $$ \Delta = E \times \text{Avogadro's number} = (3.978 \times 10^{-19}\mathrm{~J}) \times (6.022 \times 10^{23}\mathrm{mol^{-1}}) \times \frac{1\mathrm{~kJ}}{10^3\mathrm{~J}} = 239.34\mathrm{~kJ/mol} $$ (a) The magnitude of \(\Delta\) for \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) in \(\mathrm{kJ/mol}\) is approximately \(239.34 \mathrm{~kJ/mol}\).
04

Predict the change in \(\Delta\) for replacing \(\mathrm{H}_{2}\mathrm{O}\) ligands with \(\mathrm{NH}_{2}\) ligands

When the \(\mathrm{H}_{2} \mathrm{O}\) ligands in \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) are replaced with \(\mathrm{NH}_{2}\) ligands, the new compound will be \(\left[\mathrm{Ti}\left(\mathrm{NH}_{2}\right)_{6}\right]^{3+}\). The \(\mathrm{NH}_{2}\) ligand is a stronger field ligand than \(\mathrm{H}_{2}\mathrm{O}\), which means it will cause the energy difference between the d orbitals to increase. Thus, the value of Δ will be greater for \(\left[\mathrm{Ti}\left(\mathrm{NH}_{2}\right)_{6}\right]^{3+}\) than for \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\). (b) The magnitude of \(\Delta\) will increase if the \(\mathrm{H}_{2} \mathrm{O}\) ligands in \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) were replaced with \(\mathrm{NH}_{2}\) ligands.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding d-d Transitions
When electrons in a transition metal complex move between different energy levels within the 'd' orbitals, the process is known as a d-d transition. These transitions are responsible for the vibrant colors often exhibited by coordination compounds, as they absorb particular wavelengths of light that correspond to the energy gap between two 'd' orbitals.

In the exercise, the titanium(III) complex with water ligands absorbs light with a wavelength of 500 nm. This absorption is indicative of a d-d transition occurring within the complex, and it serves as a fingerprint for the energy difference between the two involved orbitals. Using Planck's equation, which relates energy with the frequency of light, we calculate the energy associated with this particular wavelength to determine the crystal field splitting energy, also noted as \( \Delta \).
Ligand Field Stabilization Energy (LFSE)
The ligand field stabilization energy (LFSE) refers to the energy difference that arises due to the splitting of the d orbitals in a transition metal complex when ligands are present. Ligands create an electrostatic field that splits the degenerate d orbitals into two or more energy levels. The LFSE is the amount of energy released when d electrons are distributed among these newly created levels according to the Aufbau principle and Hund's rule.

The actual LFSE depends on several factors, including the number of d electrons, the arrangement of the d orbitals, and the strength of the ligands, which is often described by the spectrochemical series. A larger LFSE means that the complex is more stable, hence, its formation is energetically favored.
Spectrochemical Series
The spectrochemical series is a list that ranks ligands based on their ability to split the d orbital energies in a transition metal complex, from weakest to strongest field. This has significant implications for properties such as color, magnetic behavior, and reactivity. Ligands like \( \mathrm{H}_2\mathrm{O} \) are considered weak-field ligands, causing less splitting, while others like \( \mathrm{NH}_2 \) are strong-field ligands, leading to greater splitting.

As shown in the exercise solution, replacing \( \mathrm{H}_2\mathrm{O} \) with a stronger field ligand like \( \mathrm{NH}_2 \) increases \( \Delta \), indicating a larger energy gap due to more significant d orbital splitting. This directly affects the d-d transitions, which in turn alters observable properties such as the color of the complex.
Planck's Equation
Planck's equation is a fundamental concept in quantum mechanics that describes the quantization of energy. The equation \( E = \frac{h \times c}{\lambda} \) relates the energy of a photon (E) to its wavelength (\( \lambda \)) by multiplying Planck's constant (h) with the speed of light (c), then dividing by the wavelength of light. This relationship is key to solving a variety of problems related to the interaction of light with matter, including the analysis of d-d transitions in coordination chemistry.

When a transition metal complex absorbs light, we can use Planck's equation to calculate the energy of the absorbed photons, which corresponds to the crystal field splitting energy, \( \Delta \). This equation was applied in the exercise to find the energy of the titanium(III) complex absorption, allowing us to describe the behavior of the complex with different ligands.

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Most popular questions from this chapter

One of the more famous species in coordination chemistry is the Creutz-Taube complex: It is named for the two scientists who discovered it and initially studied its properties. The central ligand is pyrazine, a planar six-membered ring with nitrogens at opposite sides. (a) How can you account for the fact that the complex, which has only neutral ligands, has an odd overall charge? (b) The metal is in a low-spin configuration in both cases. Assuming octahedral coordination, draw the d-orbital energy-level diagram for each metal. (c) In many experiments the two metal ions appear to be in exactly equivalent states. Can you think of a reason that this might appear to be so, recognizing that electrons move very rapidly compared to nuclei?

Draw the crystal-field energy-level diagrams and show the placement of electrons for the following complexes: (a) \(\left[\mathrm{VCl}_{6}\right]^{3-}\), (b) \(\left[\mathrm{FeF}_{6}\right]^{3-}\) (a high-spin complex), (c) \([\text { Ru(bipy) }]^{3+}\) (a low-spin complex), (d) \(\left[\mathrm{NiCl}_{4}\right]^{2-}\) (tetrahedral), (e) \(\left[\mathrm{PtBr}_{6}\right]^{2-},(f)\left[\mathrm{Ti}(\mathrm{en})_{3}\right]^{2+}\). S in the spectrochemical series?

(a) What is the meaning of the term coordination number as it applies to metal complexes? (b) Give an example of a ligand that is neutral and one that is negatively charged. (c) Would you expect ligands that are positively charged to be common? Explain. (d) What type of chemical bonding is characteristic of coordination compounds? Illustrate with the compound \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}\). (e) What are the most common coordination numbers for metal complexes?

The red color of ruby is due to the presence of \(\mathrm{Cr}(\mathrm{III})\) ions at octahedral sites in the dose-packed exide lattice of \(\mathrm{Al}_{2} \mathrm{O}_{2}\). Draw the crystal-field splitting diagram for \(\mathrm{Cr}\) (III) in this environment. Suppose that the ruby crystal is subjected to high pressure. What do you predict for the variation in the wavelength of absorption of the ruby as a function of pressure? Explain.

Solutions of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+},\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (both octahedral), and \(\left[\mathrm{CoCl}_{4}\right]^{2-}\) (tetrahedral) are colored. One is pink, one is blue, and one is yellow. Based on the spectrochemical series and remembering that the energy splitting in tetrahedral complexes is normally much less than that in ectahedral ones, assign a color to each complex.

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