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The radial probability function for a 2 s orbital is shown here. Classify the following statements as either true or false: (a) There are two maxima in this function because one electron spends most of its time at an approximate distance of 0.5 A from the nucleus and the other electron spends most of its time at an approximate distance of 3 A from the nucleus. (b) The radial probability function shown here and the probability density \([\psi(r)]^{2}\) both go to zero at the same distance from the nucleus, approximately 1 A. (c) For an \(s\) orbital, the number of radial nodes is equal to the principal quantum number, \(n .\)

Short Answer

Expert verified
The classification of the statements is as follows: (a) False - The radial probability function shows probability density, not the positions of individual electrons. Two maxima indicate likely regions, not specific electron locations. (b) True - Both the radial probability function and the probability density \(\psi^2(r)\) go to zero at the same distance (around 1 A) for s orbitals. (c) False - The number of radial nodes for an s orbital is one less than the principal quantum number n, determined by the formula: radial nodes = n - l - 1 (where l = 0 for s orbitals).

Step by step solution

01

Statement (a): One electron around 0.5 A and the other around 3 A

The statement is false. The radial probability function shows the probability density of finding an electron at a specific distance from the nucleus, not the positions of individual electrons in an orbital. Two maxima in the radial probability function indicate two regions where electrons are more likely to be found, but it doesn't specify individual electrons.
02

Statement (b): Radial probability function and probability density go to zero around 1 A

The statement is true. The radial probability function provides the probability of finding an electron within a spherical shell located at a distance r from the nucleus. On the other hand, probability density \(\psi^2(r)\) represents the probability of finding an electron at a particular position within the orbital (r coordinate). For s orbitals, both the radial probability function and \(\psi^2(r)\) go to zero at the same distance because they're both zero at the radial node (approximately 1 A).
03

Statement (c): Number of radial nodes equal to principal quantum number n for s orbitals

The statement is false. The number of radial nodes for an orbital is defined as the number of points where the probability of finding an electron at a specific distance from the nucleus goes to zero. It's determined by the formula: radial nodes = n - l - 1, where n is the principal quantum number and l is the angular momentum quantum number. For an s orbital, l = 0, so radial nodes = n - 1. This means that the number of radial nodes is one less than the principal quantum number n, not equal to n.

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Key Concepts

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

Quantum Chemistry
Quantum chemistry is a branch of chemistry focused on the application of quantum mechanics to chemical systems. Its primary goal is to understand the structure and behavior of atoms and molecules by interpreting the movement and interactions of subatomic particles.

At the heart of quantum chemistry lies the wave function, denoted as \( \psi \), which encapsulates all the information about a particle’s quantum state. The radial probability function, mentioned in the textbook exercise, is derived from the square of the radius times the wave function squared, \( r^2\psi^2(r) \) and it reflects the likelihood of finding an electron within a thin spherical shell at a distance \( r \) from the nucleus.

This function is crucial for predicting chemical behavior because it helps scientists visualize the most probable regions where electrons may exist around a nucleus, and by extension, elucidates the chemical bonds and reactions these electrons engage in.
Atomic Orbitals
Atomic orbitals are regions in the space around the nucleus of an atom where there is a high probability of finding an electron. Unlike the orbits of planets around the sun, atomic orbitals do not depict a definite path for electron motion but rather areas with varying likelihoods of electron presence.

Each type of orbital—designated as \( s \), \( p \), \( d \), and \( f \)—has a characteristic shape and energy level. The \( s \) orbitals are spherically symmetric and the \( p \) orbitals are dumbbell-shaped, for example. These shapes are determined by the wave function \( \psi \), and the variation in electron probability distributions gives rise to different radial probability functions for each type of orbital.

The exercise mentions a 2s orbital—one of the simplest types with a distinctive radial probability function that shows where an electron is most likely to be found relative to the nucleus, which is crucial for understanding electron configurations and chemical bonding.
Electron Configuration
Electron configuration refers to the arrangement of electrons in an atom’s orbitals according to the principles of quantum mechanics. It is a core concept in chemistry that explains the chemical properties and reactions of an element.

The configuration follows a specific order determined by the energy levels of the orbitals, commonly described using the Aufbau principle, Hund's rule, and the Pauli exclusion principle. These rules help explain why electrons fill lower energy orbitals before moving on to higher ones, and why orbitals of the same energy level are each occupied by one electron before any is occupied by a second one.

The exercise's radial probability function relates to electron configuration as it illustrates the likelihood of electron positions in different orbitals. By understanding the shapes and energies of these orbitals through their radial probability functions, scientists can predict how electrons will fill them in different atoms, influencing the atoms' chemical behavior.
Quantum Numbers
Quantum numbers are sets of numerical values that define the quantum state of an electron in an atom. There are four quantum numbers: the principal quantum number (\(n\)), the angular momentum quantum number (\(l\)), the magnetic quantum number (\(m_l\)), and the spin quantum number (\(m_s\)).

The principal quantum number \(n\) corresponds to the electron’s energy level and distance from the nucleus. The angular momentum quantum number \(l\) describes the shape of the orbital, and the magnetic quantum number \(m_l\) specifies the orientation of the orbital in space. Lastly, the spin quantum number \(m_s\) indicates the direction of the electron's spin.

This information becomes particularly relevant in the context of the exercise, where the number of radial nodes—zones where the probability of finding an electron is zero—is calculated using the formula: radial nodes = \(n - l - 1\). Understanding how quantum numbers interplay to define electron behavior is fundamental not only for interpreting the radial probability function but also for grasping the overall structure of atoms and the complex nature of chemical bonds.

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Most popular questions from this chapter

Is energy emitted or absorbed when the following electronic transitions occur in hydrogen? (a) from \(n=4\) to \(n=2\) , (b) from an orbit of radius 2.12 A to one of radius \(8.46 \hat{A},(\mathbf{c})\) an electron adds to the \(\mathrm{H}^{+}\) ion and ends up in the \(n=3\) shell?

(a) A green laser pointer emits light with a wavelength of 532 nm. What is the frequency of this light? (b) What is the energy of one of these photons? (c) The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to an upper excited state. When the electrons return to the ground state, they lose the excess energy in the form of 532 -nm photons. What is the energy gap between the ground state and excited state in the laser material?

How many unique combinations of the quantum numbers \(l\) and \(m_{l}\) are there when (a) \(n=3,\) (b) \(n=4 ?\)

Neutron diffraction is an important technique for determining the structures of molecules. Calculate the velocity of a neutron needed to achieve a wavelength of 1.25 A. (Refer to the inside cover for the mass of the neutron.)

Molybdenum metal must absorb radiation with a minimum frequency of \(1.09 \times 10^{15} \mathrm{s}^{-1}\) before it can eject an electron from its surface via the photoelectric effect. (a) What is the minimum energy needed to eject an electron? (b) What wavelength of radiation will provide a photon of this energy? (c) If molybdenum is irradiated with light of wavelength of \(120 \mathrm{nm},\) what is the maximum possible kinetic energy of the emitted electrons?

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