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Molecules that are brightly colored have a small energy gap between filled and empty electronic states (the HOMO-LUMO gap; see Exercise 9.104 ). Sometimes you can visually tell which HOMO-LUMO gap is larger for one molecule than another. Suppose you have samples of two crystalline powders-one is white, and one is green. Which one has the larger HOMO-LUMO gap?

Short Answer

Expert verified
The white crystalline powder has a larger HOMO-LUMO gap than the green crystalline powder, as the green powder is absorbing some wavelength of visible light, implying a smaller gap, whereas the white powder reflects most visible light, indicating a larger gap.

Step by step solution

01

Identify the relationship between color and HOMO-LUMO gap

Molecules with bright colors have a small HOMO-LUMO gap. This gap represents the energy difference between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO). The smaller this gap, the more likely it is for a molecule to absorb visible light and appear colored.
02

Analyze the color of the two powders

We have two crystalline powders - one is white and the other is green. This suggests that the green powder is absorbing certain parts of the visible light spectrum, while the white powder is reflecting most of the visible light.
03

Comparing HOMO-LUMO gaps using the colors of the powders

Since the green powder is brightly colored (green), it is absorbing some wavelength of visible light and therefore implies a smaller HOMO-LUMO gap. In contrast, the white powder reflects most of the visible light and does not appear colored, indicating a larger HOMO-LUMO gap.
04

Conclusion

The white crystalline powder has a larger HOMO-LUMO gap than the green crystalline powder.

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Key Concepts

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

Understanding Molecular Orbitals
Molecular orbitals are a cornerstone of chemistry, especially when it comes to understanding the colors of compounds. These orbitals are formed when atomic orbitals combine as atoms bond together to form molecules.

In simplest terms, the HOMO, or 'Highest Occupied Molecular Orbital,' is the most energetically favorable orbital that contains electrons. On the other end, the LUMO, or 'Lowest Unoccupied Molecular Orbital,' is the next orbital available for an electron to move into but it is currently empty. The energy gap between these two orbitals, known as the HOMO-LUMO gap, is critical as it determines a molecule's electronic properties and its interaction with light.

To visualize this concept, imagine a ladder where the HOMO represents the highest step on which someone is standing (occupied) and the LUMO represents the next available higher step (unoccupied). The energy required to move an individual from the HOMO to the LUMO is closely related to the color those compounds may exhibit.
The Role of Visible Light Absorption
Visible light absorption is inherently connected to the molecular structure of substances, particularly the HOMO-LUMO gap. In the context of visible light, which is part of the electromagnetic spectrum, only certain wavelengths are absorbed by a substance based on the energy difference between its molecular orbitals.

Each color we observe corresponds to a certain wavelength of visible light. When white light, which contains all colors, shines on a substance, the electrons in the HOMO can absorb energy and jump to the LUMO if the energy of a certain light color matches the HOMO-LUMO gap. This energy absorption causes certain colors to 'disappear' from the spectrum of reflected light, leaving us with the complementary color of the absorbed one. Therefore, a green powder appears green because it absorbs the light that is complementary to green, which in most cases is red.
Color and Energy Relationship
The color and energy relationship is fundamental in predicting the appearance of materials. It's a direct application of the aforementioned concepts. A smaller HOMO-LUMO gap corresponds to the absorption of light with lower energy, which is found in the red to yellow part of the visible spectrum. Conversely, a larger HOMO-LUMO gap means that the molecule can absorb higher energy light, towards the violet part of the spectrum.

A substance that appears white reflects all visible wavelengths, implying that its HOMO-LUMO gap is too large for visible light to be absorbed. Consequently, it does not interact significantly with visible light energies and remains colorless in appearance. On the other hand, a substance that appears green absorbs a specific part of the spectrum. This absorption is due to a smaller HOMO-LUMO gap, fitting the energy of red light which is the complementary color to green.

Therefore, when we see colored objects, we are actually observing the 'aftereffects' of a molecular tug-of-war, where certain wavelengths of light are captured and others are released for our eyes to perceive. The green powder's ability to absorb light of a certain wavelength is evidence of a smaller HOMO-LUMO gap compared to the larger gap of the white powder, which doesn't absorb in the visible range.

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Most popular questions from this chapter

The structure of borazine, \(\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6},\) is a six-membered ring of alternating \(\mathrm{B}\) and \(\mathrm{N}\) atoms. There is one \(\mathrm{H}\) atom bonded to each \(\mathrm{B}\) and to each \(\mathrm{N}\) atom. The molecule is planar. (a) Write a Lewis structure for borazine in which the formal charge on every atom is zero. (b) Write a Lewis structure for borazine in which the octet rule is satisfied for every atom. (c) What are the formal charges on the atoms in the Lewis structure from part (b)? Given the electronegativities of \(\mathrm{B}\) and \(\mathrm{N},\) do the formal charges seem favorable or unfavorable? (d)Do either of the Lewis structures in parts (a) and (b) have multiple resonance structures? (e) What are the hybridizations at the B and N atoms in the Lewis structures from parts (a) and (b)? Would you expect the molecule to be planar for both Lewis structures? (f) The six \(B-N\) bonds in the borazine molecule are all identical in length at 1.44 A. Typical values for the bond lengths of \(\mathrm{B}-\mathrm{N}\) single and double bonds are 1.51 \(\mathrm{A}\) and \(1.31 \mathrm{A},\) respectively. Does the value of the \(\mathrm{B}-\mathrm{N}\) bond length seem to favor one Lewis structure over the other? (g) How many electrons are in the \(\pi\) system of borazine?

How would you expect the extent of overlap of the bonding atomic orbitals to vary in the series IF, ICl, IBr, and \(I_{2} ?\) Explain your answer.

Indicate whether each statement is true or false. (a) \(s\) orbitals can only make \(\sigma\) or \(\sigma^{*}\) molecular orbitals. (b) The probability is 100\(\%\) for finding an electron at the nucleus in a \(\pi^{*}\) orbital. (c) Antibonding orbitals are higher in energy than bonding orbitals (if all orbitals are created from the same atomic orbitals). (d) Electrons cannot occupy an antibonding orbital.

(a) An AB \(_{2}\) molecule is linear. How many non bonding electron pairs are around the A atom from this information? (b) How many non bonding electrons surround the Xe in \(\mathrm{XeF}_{2} ?(\mathbf{c})\) Is XeF \(_{2}\) linear?

Consider the molecule \(\mathrm{BF}_{3}\). (a) What is the electron configuration of an isolated B atom? (b) What is the electron configuration of an isolated F atom? (c) What hybrid orbitals should be constructed on the B atom to make the B–F bonds in \(\mathrm{B} \mathrm{F}_{3}\)?(d) What valence orbitals, if any, remain unhybridized on the B atom in \(\mathrm{BF}_{3} ?\)

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