Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

As shown in Figure \(24.26\), 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) What is the spectrochemical series? 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 \(\mathrm{NH}_{3}\) ligands?

Short Answer

Expert verified
(a) The magnitude of Δ for \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) is approximately 160.37 kJ/mol. (b) The spectrochemical series is an arrangement of ligands in the order of their ability to cause splitting of energy levels of the central metal ion in a complex. If the H₂O ligands in \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) are replaced with NH₃ ligands, the magnitude of Δ will increase because NH₃ is a stronger field ligand than H₂O, causing larger splitting of energy levels.

Step by step solution

01

Relationship between energy, wavelength, and frequency

The relationship between energy (E), wavelength (λ), and frequency (ν) of a photon can be expressed as: \[E = h\nu\] where h is Planck's constant (\(6.626 \times 10^{-34} \mathrm{Js}\)). Since, \( c = \lambda\nu \), where c is the speed of light (\(3.00 \times 10^8 \mathrm{m/s}\)), the equation becomes: \[E = \frac{hc}{\lambda}\]
02

Calculate the energy difference \(\Delta\)

Now, we can calculate the energy difference Δ using the given λ value: \[E = \frac{hc}{\lambda}\] \[E = \frac{(6.626 \times 10^{-34} \mathrm{Js})(3.00 \times 10^8 \mathrm{m/s})}{500 \times 10^{-9} \mathrm{m}}\] \[E \approx 3.978 \times 10^{-19} \mathrm{J}\] Now, we convert this energy to kJ/mol. 1 eV = \(1.602 \times 10^{-19} \mathrm{J}\), 1 eV = 96.485 kJ/mol So, \[\Delta (\mathrm{kJ/mol}) = \frac{E (\mathrm{J})}{1.602\times 10^{-19} \mathrm{J/eV}} \times 96.485 \frac{\mathrm{kJ}}{\mathrm{mol\cdot eV}}\] \[\Delta \approx 160.37 \mathrm{kJ/mol}\] (a) The magnitude of Δ for \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) is approximately 160.37 kJ/mol.
03

Discuss spectrochemical series and the change in Δ

(b) The spectrochemical series is an arrangement of ligands in the order of their ability to cause splitting of energy levels of the central metal ion in a complex. The series is as follows: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NCS⁻ < SCN⁻ < NH₃ < en < bpy < phen < CN⁻ < CO In the given complex, water is the ligand. According to the spectrochemical series, NH₃ is a stronger field ligand than H₂O. Therefore, if the H₂O ligands in \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) are replaced with NH₃ ligands, the magnitude of Δ will increase. This is because stronger field ligands cause larger splitting of energy levels.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Ligand Field Theory
Ligand field theory is an essential part of coordination chemistry that explains the properties of metal complexes, such as their color, magnetism, and reactivity. It is an extension of crystal field theory, which takes the covalent nature of metal-ligand bonds into account. The theory operates on the principle that ligands, which are molecules or ions surrounding a central metal atom, can influence the energy of the d-orbitals of the central atom. When ligands approach the central metal ion, they interact with the d-orbitals and cause a phenomenon known as 'd-orbital splitting' or 'crystal field splitting'.

The degree of orbital splitting has a pivotal role in determining the spectral properties of complexes as it affects the absorption of light. This is critical in the analysis of absorption spectra and the prediction of the color of the complex. In simpler terms, the differences in energy between the split d-orbitals correspond to particular wavelengths of light that the complex can absorb. The non-absorbed light is transmitted or reflected, and this is what we perceive as the color of the complex. Therefore, ligand field theory serves as the foundation to understand why coordination compounds have distinctive colors and how they change with varying ligands.

For example, in the exercise provided, ligand field theory would explain the observed color in the solution by the specific d-d transitions occurring due to splitting of the d-orbitals in the presence of water ligands surrounding the titanium ion.
Spectrochemical Series
The spectrochemical series refers to a list of common ligands ordered by their ability to split the d-orbitals of a central metal ion. This series is a tool that chemists use to predict the magnitude of the splitting, denoted as \(\Delta\), which in turn affects the color and magnetic properties of the complex. Ligands at the low end of the series, such as iodide (I⁻), cause less splitting, while those at the high end, like cyanide (CN⁻) and carbon monoxide (CO), cause larger splitting.

The series informs us about the 'field strength' of ligands, with 'weak field' ligands causing lesser splitting and 'strong field' ligands causing more significant splitting. In the context of the original exercise, the spectrochemical series indicates that by replacing H2O with NH3, a ligand that appears further down the series and is known as a stronger field ligand, the value of \(\Delta\) would increase. This increase means that the d-d transitions would require more energy, and the complex may absorb light at shorter wavelengths, potentially changing the color observed.
Energy Level Splitting
Energy level splitting is a fundamental concept to understand d-d transitions in coordination compounds. When a metal ion is surrounded by ligands, the degenerate (equally energized) d-orbitals split into two groups - a lower-energy set and a higher-energy set. The energy difference between these sets of d-orbitals is symbolized as \(\Delta\), and it varies depending on the nature of the metal ion and the ligands involved.

This splitting is crucial because it determines the energy required for an electron to 'jump' from a lower-energy d-orbital to a higher-energy d-orbital. When such a d-d transition occurs, the complex absorbs a specific quantum of light, corresponding to the energy difference \(\Delta\). As a result, the absorbed light's wavelength is directly related to the \(\Delta\) value. If the value of \(\Delta\) is small, the complex tends to absorb longer wavelengths (redder light), and if it is large, it absorbs shorter wavelengths (bluer light).

In the exercise, the given wavelength of 500 nm and the calculated \(\Delta\) of approximately 160.37 kJ/mol, are intrinsically connected through the formula \[E = \frac{hc}{\lambda}\]. This formula, which bridges the energy of light with its wavelength, allows for the calculation of the splitting energy that determines the color of the complex seen in the laboratory.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\), abbreviated py, is the following molecule: (a) Why is pyridine referred to as a monodentate ligand? (b) Consider the following equilibrium reaction: \(\left[\mathrm{Ru}(\mathrm{py})_{4}(\mathrm{bipy})\right]^{2+}+2 \mathrm{py} \rightleftharpoons\left[\mathrm{Ru}(\mathrm{py})_{6}\right]^{2+}+\mathrm{bipy}\) What would you predict for the magnitude of the equilibrium constant for this equilibrium? Explain the basis for your answer.

Write balanced chemical equations to represent the following observations. (In some instances the complex involved has been discussed previously in the text.) (a) Solid silver chloride dissolves in an excess of aqueous ammonia. (b) The green complex \(\left[\mathrm{Cr}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}\), on treatment with water over a long time, converts to a brown-orange complex. Reaction of \(\mathrm{AgNO}_{3}\) with a solution of the product precipitates \(3 \mathrm{~mol}\) of \(\mathrm{AgCl}\) per mole of Cr present. (Write two chemical equations.) (c) When an \(\mathrm{NaOH}\) solution is added to a solution of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{a}\) precipitate forms. Addition of excess \(\mathrm{NaOH}\) solution causes the precipitate to dissolve. (Write two chemical equations.) (d) A pink solution of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) turns deep blue on addition of concentrated hydrochloric acid.

Write the formula for each of the following compounds, being sure to use brackets to indicate the coordination sphere: (a) hexaamminechromium(III) nitrate (b) tetraamminecarbonatocobalt(III) sulfate (c) dichlorobis(ethylenediamine)platinum(IV) bromide (d) potassium diaquatetrabromovanadate(III) (e) bis(ethylenediamine) zinc(II) tetraiodomercurate(II)

Based on the molar conductance values listed here for the series of platinum(IV) complexes, write the formula for each complex so as to show which ligands are in the coordination sphere of the metal. By way of example, the molar conductances of \(0.050 \mathrm{M} \mathrm{NaCl}\) and \(\mathrm{BaCl}_{2}\) are \(107 \mathrm{ohm}^{-1}\) and 197 ohm \(^{-1}\), respectively. $$ \begin{array}{ll} \hline \text { Complex } & \begin{array}{l} \text { Molar Conductance (ohm }^{-1} \text { ) }^{*} \\ \text { of } \mathbf{0 . 0 5 0 M \text { Solution }} \end{array} \\ \hline \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{4} & 523 \\ \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{4} & 228 \\ \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{4} & 97 \\ \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{4} & 0 \\ \mathrm{KPt}\left(\mathrm{NH}_{3}\right) \mathrm{Cl}_{5} & 108 \\ \hline \end{array} $$

For each of the following polydentate ligands, determine (i) the maximum number of coordination sites that the ligand can occupy on a single metal ion and (ii) the number and type of donor atoms in the ligand: (a) ethylenediamine (en), (b) bipyridine (bipy), (c) the oxalate anion \(\left(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right)\), (d) the \(2-\) ion of the porphine molecule (Figure 24.8); (e) [EDTA] \(^{4-}\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free