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(a) Why does the quantum mechanical description of manyelectron atoms make it difficult to define a precise atomic radius? (b) When nonbonded atoms come up against one another, what determines how closely the nuclear centers can approach?

Short Answer

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In quantum mechanics, the probabilistic nature of electron positions makes it difficult to define a precise atomic radius, as there is no well-defined boundary where electron density drops to zero. For nonbonded atoms, the closest approach of their nuclear centers is determined by the balance between attractive forces (van der Waals interactions) and repulsive forces (resulting from the Pauli exclusion principle). This distance between the nuclear centers when attractive and repulsive forces are equal is called the van der Waals radius.

Step by step solution

01

Part (a): Understanding Quantum Mechanical Description

In quantum mechanics, electrons around an atom are described by wave functions, which represent the probability distribution of finding an electron in a certain region around the nucleus. It means that electrons do not have precise positions in an atom, making it hard to determine a specific boundary for the atom. An atomic radius is the distance from the nucleus to the outermost electron shell. However, due to the probabilistic nature of the electron positions, it becomes challenging to define the precise atomic radius.
02

Part (a): Electron Distribution and Precise Atomic Radius

The electrons in many-electron atoms are distributed in different orbitals with various shapes and energies. The distribution depends on the principle quantum number, azimuthal quantum number, and magnetic quantum number. Due to the distinct electron distribution, there is no well-defined boundary where the electron density drops to zero. Thus, defining a precise atomic radius becomes difficult as it relies heavily on the electron distribution in atomic orbitals.
03

Part (b): Factors Determining Closest Approach of Nuclear Centers

When nonbonded atoms come close to each other, the closest approach of their nuclear centers is determined by the balance between attractive and repulsive forces: 1. Attractive Forces: These forces primarily come from the van der Waals interactions. These interactions arise due to the temporary fluctuations in the electron distribution around the atom, causing temporary regions of positive and negative charge. The induced dipoles attract neighboring atoms, leading to weak attractive forces. 2. Repulsive Forces: When atoms get very close, the outer electron clouds start to overlap, which leads to repulsive forces as a result of the Pauli exclusion principle. This principle states that no two electrons can have the same set of quantum numbers. Thus, when electron clouds overlap, the electrons in the overlapping region experience repulsion due to the requirement that they occupy different quantum states. The closest approach of nuclear centers depends on the balance between these two forces. When the attractive and repulsive forces are equal, the atoms are in a stable position, and the distance between their nuclear centers is called the van der Waals radius.

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Most popular questions from this chapter

Hydrogen is an unusual element because it behaves in some ways like the alkali metal elements and in other ways like nonmetals. Its properties can be explained in part by its electron configuration and by the values for its ionization energy and electron affinity. (a) Explain why the electron affinity of hydrogen is much closer to the values for the alkali elements than for the halogens. (b) Is the following statement true? "Hydrogen has the smallest bonding atomic radius of any element that forms chemical compounds. If not, correct it. If it is, explain in terms of electron configurations. (c) Explain why the ionization energy of hydrogen is closer to the values for the halogens than for the alkali metals. (d) The hydride ion is \(\mathrm{H}\). Write out the process corresponding to the first ionization energy of the hydride ion. (e) How does the process in part (d) compare to the process for the electron affinity of a neutral hydrogen atom?

Using only the periodic table, arrange each set of atoms in order of increasing radius: (a) \(\mathrm{Ba}, \mathrm{Ca}, \mathrm{Na}\); (b) In, \(\mathrm{Sn}, \mathrm{As}\); (c) \(\mathrm{Al}, \mathrm{Be}, \mathrm{Si}\).

In Table \(7.8\), the bonding atomic radius of neon is listed as \(0.58\) \(\AA\), whereas that for xenon is listed as \(1.40 \AA\). A classmate of yours states that the value for Xe is more realistic than the one for Ne. Is she correct? If so, what is the basis for her statement?

Which will experience the greater effective nuclear charge, the electrons in the \(n=3\) shell in Ar or the \(n=3\) shell in \(\mathrm{Kr}\) ? Which will be closer to the nucleus?

It is possible to define metallic character as we do in this book and base it on the reactivity of the element and the ease with which it loses electrons. Alternatively, one could measure how well electricity is conducted by each of the elements to determine how "metallic" the elements are. On the basis of conductivity, there is not much of a trend in the periodic table: Silver is the most conductive metal, and manganese the least. Look up the first ionization energies of silver and manganese; which of these two elements would you call more metallic based on the way we define it in this book?

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