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The value of \(\Delta\) for the \(\left[\mathrm{CrF}_{6}\right]^{3-}\) complex is \(182 \mathrm{k} / / \mathrm{mol}\). Calculate the expected wavelength of the absorption corresponding to promotion of an electron from the lower-energy to the higher-energy d-orbital set in this complex. Should the complex absorb in the visible range?

Short Answer

Expert verified
The expected wavelength of the absorption corresponding to the electron promotion in the \(\left[\mathrm{CrF}_{6}\right]^{3-}\) complex is found using the energy-wavelength relationship, \(\lambda = \frac{hc}{\Delta E}\). First, convert the given energy \(\Delta = 182\ \mathrm{k}/\mathrm{mol}\) to Joules, resulting in \(\Delta E\). Then, substitute the values for \(h\), \(c\), and \(\Delta E\) to solve for \(\lambda\). Finally, compare the calculated wavelength to the visible range (\(400-700\ \text{nm}\)) to determine if the complex absorbs in the visible range.

Step by step solution

01

1. Understand the energy-wavelength relationship

To find the wavelength associated with a given energy difference, we can use the Planck's equation: \[E = h\nu\] Where \(E\) is the energy, \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \ \text{J s}\)), and \(\nu\) is the frequency. Additionally, we know that the speed of light, \(c\) (\(3.0 \times 10^{8} \ \text{m/s}\)), is related to the frequency and wavelength \(\lambda\) as: \[c = \lambda\nu\] Combining the two equations, we have: \[E = h\frac{c}{\lambda}\]
02

2. Convert the given energy to Joules

The given energy is in \(\mathrm{k}/\mathrm{mol}\). We need to convert it to Joules to use it in our calculations. Given that 1 electronic volt (eV) is equal to \(96.485\ \mathrm{k}\ \mathrm{mol}^{-1}\), and 1 eV equals \(1.602 \times 10^{-19}\) J, we have: \[\Delta E = \left(\frac{182\ \text{k}}{96.485\ \mathrm{kmol}^{-1}}\right) \times \left(\frac{1.602 \times 10^{-19}\text{J}}{1\ \text{eV}}\right)\]
03

3. Calculate the wavelength

Now that we have the energy in Joules, we can use the Planck's equation to find the wavelength: \[\lambda = \frac{hc}{\Delta E}\] Substitute the values for \(h\), \(c\), and \(\Delta E\) to solve for \(\lambda\).
04

4. Determine if the complex absorbs in the visible range

The visible range of the electromagnetic spectrum is approximately \(400-700\ \text{nm}\). Compare the calculated wavelength to this range to determine if the complex absorbs in the visible range.
05

5. Summary

To solve this exercise: 1. Understand the energy-wavelength relationship. 2. Convert the given energy to Joules. 3. Calculate the wavelength. 4. Determine if the complex absorbs in the visible range.

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Key Concepts

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

d-orbitals
The term "d-orbitals" refers to a set of five orbitals in the d-subshell of an atom's electron cloud. These orbitals are crucial in chemistry, especially when discussing transition metals like chromium, present in the \(\left[\text{CrF}_6\right]^{3-}\) complex.

In transition metal complexes, d-orbitals split into two energy levels due to interactions with ligands. This is known as crystal field splitting. The arrangement and energy difference between these orbitals are influenced by the ligands' nature and placement around the metal.

Understanding the splitting of d-orbitals helps explain various phenomena in coordination chemistry, such as color and magnetism. The energy difference between the split levels is represented by \(\Delta\), which is key in determining the wavelength of absorbed light.
wavelength calculation
Wavelength calculation is essential in determining the color of light absorbed by a complex in an absorption spectroscopy experiment. We begin with the energy, \(\Delta E\), which represents the energy difference between split d-orbitals.

This energy can be converted into a wavelength using Planck's equation:
  • \(E = hu\) where \(E\) is energy, \(h\) is Planck's constant, and \(u\) is frequency.
  • The relationship between wavelength \(\lambda\) and frequency \(u\) is given by \(c = \lambdau\), where \(c\) is the speed of light.
Combining these, we find:
\[\lambda = \frac{hc}{E}\]
By converting the given \(\Delta E\) from \(\text{kJ/mol}\) to Joules and plugging it into this formula, we determine the exact wavelength of light absorbed by the complex.
absorption spectroscopy
Absorption spectroscopy is a technique used to study the absorption characteristics of substances, often revealing valuable information about molecular structure and behavior.

For complexes like \(\left[\text{CrF}_6\right]^{3-}\), absorption spectroscopy measures the wavelengths of light absorbed as electrons transition between energy levels. When light of a particular wavelength is absorbed, it causes electrons to move from lower-energy d-orbitals to higher-energy d-orbitals.
  • This transition can be used to determine electronic configurations and the energy difference \(\Delta\) between orbital sets.
  • The data gathered help predict the compound's color based on the wavelengths in the visible light range.
Understanding these transitions is pivotal for interpreting how substances interact with light.
energy-wavelength relationship
The energy-wavelength relationship is fundamental in understanding how complexes absorb light. The key to this concept is realizing that light energy and wavelength are inversely related.

According to Planck's equation \(E = hu\), energy \(E\) is directly proportional to the frequency \(u\) of the absorbed light. As frequency increases, energy increases, but wavelength \(\lambda\), which is inversely proportional to frequency, decreases:
\[\lambda = \frac{hc}{E}\]
This equation highlights that smaller wavelengths correspond to higher energies.
  • Knowing the energy difference \(\Delta\) in a complex helps us calculate the light wavelength absorbed.
  • This wavelength determines if the absorption happens within the visible spectrum (approximately 400-700 nm).
This relationship allows us to link an energy difference in d-orbitals to the colorful appearance of many transition metal complexes.

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Most popular questions from this chapter

(a) A complex absorbs photons with an energy of \(4.51 \times 10^{-19}\). What is the wavelength of these photons? (b) If this is the only place in the visible spectrum where the complex absorbs light, what color would you expect the complex to be?

(a) What is the difference between a monodentate ligand and a bidentate ligand? (b) How many bidentate ligands are necessary to fill the coordination sphere of a six-coordinate complex? (c) You are told that a certain molecule can serve as a tridentate ligand. Based on this statement, what do you know about the molecule? mathrm{Br}$

(c) When the coordinated water to the \(\mathrm{Zn}(\mathrm{II})\) center in carbonic anhydrase is deprotonated, what ligands are bound to the Zn(II) center? Assume the three nitrogen ligands are unaffected. (d) The \(\mathrm{F} K_{a}\) of \(\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{d}\right]^{2+}\) is 10 . Suggest an explanation for the difference between this \(\mathrm{pK} \mathrm{K}_{\text {and }}\) that of carbonic anhydrase. (e) Would you expect carbonic anhydrase to have a decp color, like hemoglobin and other metalion containing proteins do? Explain. Two different compounds have the formulation \(\mathrm{CoBr}\left(\mathrm{SO}_{4}\right) \cdot 5 \mathrm{NH}_{3}\). Compound \(\mathrm{A}\) is dark violet, and compound B is red-violet. When compound \(A\) is treated with \(\mathrm{AgNO}_{3}(\mathrm{Gq})\), no reaction occurs, whereas compound \(\mathrm{B}\)

Which of the complexes shown here are chiral? [Section 23.4] $$ \mathrm{Cr} \quad \mathrm{Cr}=\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2} \quad \Theta=\mathrm{Cl} \quad \theta=\mathrm{NH}_{3} $$ (1) ( (3) (4)

The total concentration of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) in a sample of hard water was determined by titrating a \(0.100\) - L sample of the water with a solution of EDTA \({ }^{4-}\). The EDTA \({ }^{4-}\) chelates the two cations: $$ \begin{aligned} \mathrm{Mg}^{2+}+[\mathrm{EDTA}]^{4-} & \longrightarrow[\mathrm{Mg}(\mathrm{EDTA})]^{2-} \\ \mathrm{Ca}^{2+}+\left[\mathrm{EDTA}^{4-}\right.& \longrightarrow[\mathrm{Ca}(\mathrm{EDTA})]^{2-} \end{aligned} $$ It requires \(31.5 \mathrm{~mL}\) of \(0.0104 \mathrm{M}[\mathrm{EDTA}]^{4-}\) solution to reach the end point in the titration. A second \(0.100-L\) sample was then treated with sulfate ion to precipitate \(\mathrm{Ca}^{2+}\) as calcium sulfate. The \(\mathrm{Mg}^{2+}\) was then titrated with \(18.7 \mathrm{~mL}\) of \(0.0104 \mathrm{M}\) [EDTA] ]- Calculate the concentrations of \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) in the hard water in mg/I.

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