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Which type of radiation would be most useful in examining the dimensionality of molecules? (A) Ultraviolet (B) Visible (C) Infrared (D) Microwave

Short Answer

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(C) Infrared

Step by step solution

01

Understand radiation types

The first step is to understand what each type of radiation does. (A) Ultraviolet radiation is absorbed by molecules when electrons are excited to higher energy levels. Typically, UV radiation is used to study electronic transitions. (B) Visible radiation, similar to UV, is absorbed during electronic transitions and is usually used for studying visible-light spectroscopy. (C) Infrared radiation is absorbed by molecules when bonds vibrate. It's typically used to study vibrational transitions. (D) Microwave radiation is absorbed when molecules rotate, which means it is used to study rotational transitions.
02

Analyze the appropriate radiation

Dimensionality primarily concerns the shape, bonding, and movement of molecules. In general, the vibrational and rotational transitions provide the most information about a molecule's size, shape, and structure because they involve changes in molecular bonds and shapes. Thus, Infrared radiation and Microwave radiation are likely to be the most useful.
03

Decision based on the Information

Between Infrared and Microwave radiations, Infrared radiation is absorbed by bonds when they vibrate, marking changes in the shape of a molecule. This gives a more detailed picture of the molecule's dimensionality than rotations provided by Microwave radiation. Therefore, Infrared radiation is considered to be most useful in examining the dimensionality of molecules.

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Key Concepts

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

Molecular Dimensionality
When we talk about the dimensionality of molecules, we are discussing their geometric configuration or how they look in three-dimensional space. This involves the arrangement of atoms within the molecule and how these atoms bond and interact with each other. Understanding molecular dimensionality is crucial for determining the physical and chemical properties of substances.
Molecular dimensionality can influence the color, reactivity, phase of matter, and even the biological activity of a molecule.
  • The shape of a molecule can dictate how it interacts with other molecules, including how it binds in biological systems.
  • The dimensionality determines how a molecule absorbs energy in the form of electromagnetic radiation.
This is why certain types of spectroscopic methods are better suited for examining molecular dimensionality. Infrared and microwave radiations, which are crucial in studying vibrational and rotational transitions, provide valuable insights into these three-dimensional structures.
Vibrational Transitions
Vibrational transitions occur when the energy absorbed by a molecule causes its bonds to vibrate. Picture the molecule's bonds as tiny springs holding the atoms together. When infrared radiation hits these bonds, it causes them to stretch, bend, and twist, depending on the energy and nature of the bond.
This vibrational movement is what infrared spectroscopy primarily examines. By measuring how different molecular bonds absorb infrared radiation, scientists can identify various functional groups within a molecule and often "see" the shape and structure of the molecule without having to visualize it directly.
  • Vibrations help in discerning details about bond strength and length.
  • Infrared spectroscopy provides fingerprints that can identify molecules.
Ultimately, vibrational transitions enable chemists to piece together a highly detailed map of a molecule's geometry, which is invaluable when contemplating its dimensionality and overall chemical nature.
Rotational Transitions
Rotational transitions involve changes in a molecule’s energy as it rotates in space. Think of molecules as spinning on axes through their centers of mass. When microwave radiation is absorbed by molecules, it can cause these rotations to change.
This absorption and associated change in rotational energy levels provide information about the molecule’s moment of inertia and structure.
  • These transitions are ideal for determining the relative orientation of atoms.
  • They help understand the mass distribution within a molecule.
Microwave spectroscopy, which studies these transitions, is very useful for examining small molecules and gaining insights into their precise shapes. However, compared to vibrational transitions, rotational transitions give a more general picture of molecular shapes, often used for smaller and simpler molecules. Understanding these transitions lays a foundation for comprehending how molecules behave when perturbed and how they interact with light to reveal their dimensions in space.

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Most popular questions from this chapter

Use the following information to answer questions 14-16 The radius of atoms and ions is typically measured in Angstroms \((A),\) which is equivalent to \(1 * 10^{-10} \mathrm{m} .\) Below is a table of information for three different elements. TABLE NOT AVAILABLE The phosphorus ion is larger than a neutral phosphorus atom, yet a zinc ion is smaller than a neutral zinc atom. Which of the following statements best explains why? (A) The zinc atom has more protons than the phosphorus atom. (B) The phosphorus atom is paramagnetic, but the zinc atom is diamagnetic. (C) Phosphorus gains electrons when forming an ion, but zinc loses them. (D) The valence electrons in zinc are further from the nucleus than those in phosphorus.

$$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Cl}_{2}(g) \mapsto 2 \mathrm{NOCl}(g) \Delta G^{\circ}=132.6 \mathrm{kJ} / \mathrm{mol}$$ For the equilibrium above, what would happen to the value of \(\Delta G^{\circ}\) if the concentration of \(\mathrm{N}_{2}\) were to increase and why? (A) It would increase because the reaction would become more themodynamically favored. (B) It would increase because the reaction would shift right and create more products. (C) It would decrease because there are more reactants present. (D) It would stay the same because the value of \(K_{e q}\) would not change.

$$\mathrm{Br}_{2}(g)+\mathrm{I}_{2}(g) \leftrightarrow 2 \mathrm{IBr}(g)$$ At \(150^{\circ} \mathrm{C},\) the equilibrium constant, \(K_{c},\) for the reaction shown above has a value of \(300 .\) This reaction was allowed to reach equilibrium in a sealed container and the partial pressure due to IBr(g) was found to be 3 atm. Which of the following could be the partial pressures due to \(\operatorname{Br}_{2}(g)\) and \(I_{2}(g)\) in the container? \(\begin{array}{lll}{} & {\operatorname{Br}_{2}(g)} & {\mathrm{I}_{2}(g)} \\\ {\text { (A) }} & {0.1 \mathrm{atm}} & {0.3 \mathrm{atm}} \\ {\text { (B) }} & {0.3 \mathrm{atm}} & {1 \mathrm{atm}} \\ {\text { (C) }} & {1 \mathrm{atm}} & {1 \mathrm{atm}} \\ {\text { (D) }} & {1 \mathrm{atm}} & {3 \mathrm{atm}}\end{array}\)

A 2.0 L flask holds 0.40 g of helium gas. If the helium is evacuated into a larger container while the temperature is held constant, what will the effect on the entropy of the helium be? (A) It will remain constant because the number of helium molecules does not change. (B) It will decrease because the gas will be more ordered in the larger flask. (C) It will decrease because the molecules will collide with the sides of the larger flask less often than they did in the smaller flask. (D) It will increase because the gas molecules will be more dispersed in the larger flask.

If the solubility of \(\mathrm{BaF}_{2}\), is equal to \(x,\) which of the following expressions is equal to the solubility product, \(K_{s p}\) , for \(\operatorname{BaF}_{2} ?\) (A) \(x^{2}\) (B) 2\(x^{2}\) (C) 2\(x^{3}\) (D) 4\(x^{3}\)

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