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Molecules that are brightly colored have a small energy gap between filled and empty electronic states (the HOMOLUMO gap; see Exercise 9.104 ). Suppose you have two samples, one is lycopene which is responsible for the red color in tomato, and the other is curcumin which is responsible for the yellow color in turmeric. Which one has the larger HOMO-LUMO gap?

Short Answer

Expert verified
Lycopene, responsible for the red color in tomatoes, has a larger HOMO-LUMO gap than curcumin, responsible for the yellow color in turmeric. This is because red light is less energetic than yellow light, and a less energetic color corresponds to a larger HOMO-LUMO gap.

Step by step solution

01

Identify the colors and wavelengths of lycopene and curcumin

Lycopene is responsible for the red color in tomato, and curcumin is responsible for the yellow color in turmeric. According to the visible light spectrum, red light has a longer wavelength (approx. 620-750 nm) and lower energy than yellow light (approx. 570-590 nm).
02

Relate the energy of colors to the HOMO-LUMO gap

According to the relationship between color and energy gap mentioned in the analysis, a less energetic color has a larger HOMO-LUMO gap and a more energetic color has a smaller HOMO-LUMO gap. Since red light (lycopene) is less energetic than yellow light (curcumin), the HOMO-LUMO gap is expected to be larger for lycopene than for curcumin.
03

Conclusion

Comparing their colors, lycopene (red) is less energetic than curcumin (yellow). Hence, lycopene has a larger HOMO-LUMO gap than curcumin.

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Key Concepts

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

The Visible Light Spectrum
The visible light spectrum is the portion of the electromagnetic spectrum that humans can see. This spectrum spans wavelengths from approximately 400 nm to 700 nm. Different wavelengths within this range are perceived as various colors. For instance:
  • Violet: 400-450 nm
  • Blue: 450-495 nm
  • Green: 495-570 nm
  • Yellow: 570-590 nm
  • Orange: 590-620 nm
  • Red: 620-750 nm
Each color has a different wavelength, with violet having the shortest wavelength and red the longest. Shorter wavelengths (like violet and blue) correspond to higher energy light, while longer wavelengths (like red) represent lower energy. This spectrum plays a key role when studying chemical compounds that interact with light to produce color.
Color and Energy Relationship
The relationship between color and energy in the context of light is crucial to understanding molecular behavior. Every color of light in the visible spectrum possesses a specific energy level, inversely related to its wavelength. The equation that connects these properties is given by Planck's relation: \[ E = \frac{hc}{\lambda} \]where:
  • E is the energy of the photon
  • h is Planck's constant
  • c is the speed of light
  • \( \lambda \) is the wavelength of light
This means that:
  • Longer wavelengths (like red) have lower energy
  • Shorter wavelengths (like blue) have higher energy
In terms of the HOMO-LUMO gap, a lower energy (or longer wavelength) corresponds to a larger gap between the Highest Occupied Molecular Orbital (HOMO) and the Lowest Unoccupied Molecular Orbital (LUMO). Thus, substances that absorb light with longer wavelengths tend to have larger HOMO-LUMO gaps.
Lycopene and Curcumin
Lycopene and curcumin are two naturally occurring compounds known for their vibrant colors - lycopene is red, and curcumin is yellow. These colors arise from the distinct ways in which these molecules interact with light, specifically through their HOMO-LUMO gaps.
Lycopene, found in tomatoes, reflects red light. As mentioned, red light corresponds to longer wavelengths and thus lower energy in the visible spectrum. This implies that the HOMO-LUMO gap in lycopene is relatively large.
Curcumin, the pigment in turmeric, has a yellow coloration. Yellow light has a slightly shorter wavelength than red, leading to a higher energy absorption. Consequently, curcumin's HOMO-LUMO gap is smaller than that of lycopene. Understanding how these compounds interact with visible light can provide insights into their stability and reactivity. In this case, the larger HOMO-LUMO gap of lycopene suggests it is less reactive than curcumin, as it requires more energy to transition electrons from the HOMO to the LUMO.

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Most popular questions from this chapter

Butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6},\) is a planar molecule that has the following carbon-carbon bond lengths: $$ \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{134 \mathrm{pm}} \mathrm{CH}=\mathrm{CH}_{2} $$ (a) Predict the bond angles around each of the carbon atoms and sketch the molecule. (b) From left to right, what is the hybridization of each carbon atom in butadiene? (c) The middle \(\mathrm{C}-\mathrm{C}\) bond length in butadiene \((148 \mathrm{pm})\) is a little shorter than the average \(\mathrm{C}-\mathrm{C}\) single bond length (154 pm). Does this imply that the middle \(\mathrm{C}-\mathrm{C}\) bond in butadiene is weaker or stronger than the average \(\mathrm{C}-\mathrm{C}\) single bond? (d) Based on your answer for part (c), discuss what additional aspects of bonding in butadiene might support the shorter middle \(\mathrm{C}-\mathrm{C}\) bond.

A compound composed of \(6.7 \% \mathrm{H}, 40.0 \% \mathrm{C},\) and \(53.3 \% \mathrm{O}\) has a molar mass of approximately \(60 \mathrm{~g} / \mathrm{mol}\). (a) What is the molecular formula of the compound? (b) What is its Lewis structure if the two \(\mathrm{O}\) are bonded to \(\mathrm{C} ?\) (c) What is the geometry and hybridization of the \(\mathrm{C}\) atom that is bonded to \(2 \mathrm{O}\) atoms? (d) How many \(\sigma\) and how many \(\pi\) bonds are there in the molecule?

Indicate whether each statement is true or false. (a) \(p\) orbitals can only make \(\sigma\) or \(\sigma^{*}\) molecular orbitals. (b) The probability is always \(0 \%\) for finding an electron in an antibonding orbital. (c) Molecules containing electrons that occupy antibonding orbitals must be unstable. (d) Electrons cannot occupy a nonbonding orbital.

From their Lewis structures, determine the number of \(\sigma\) and \(\pi\) bonds in each of the following molecules or ions: (a) hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\) (b) hydrogen cyanide, HCN; (c) sulphur trioxide, \(\mathrm{SO}_{3} ;\) (d) ozone, \(\mathrm{O}_{3}\).

For each statement, indicate whether it is true or false. (a) \(\ln\) order to make a covalent bond, the orbitals on each atom in the bond must overlap. (b) A \(p\) orbital on one atom cannot make a bond to an \(s\) orbital on another atom. \((\mathbf{c})\) Lone pairs of electrons on an atom in a molecule influence the shape of a molecule. (d) The 1 s orbital has a nodal plane. \((\mathbf{e})\) The \(2 p\) orbital has a nodal plane.

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