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Chapter 8: Problem 70

Why are Group \(1 \mathrm{~B}\) elements more stable than Group 1A elements even though they seem to have the same outer electron configuration, \(n s^{1},\) where \(n\) is the principal quantum number of the outermost shell?

Short Answer

Expert verified
Group 1B elements are more stable than Group 1A elements even though they have the same outer electron configuration, \(n s^{1}\), due to the higher nuclear charge in Group 1B elements. This increased nuclear charge results in greater stability.

Step by step solution

01

Understanding Electron Configuration of Group 1A and Group 1B Elements

Group 1A elements have the general electron configuration of \(n s^{1}\), where \(n\) is the principal quantum number of the outermost shell. This means that in their outermost shell, there is one electron. Similarly, Group 1B elements also have the same outermost electron configuration of \(n s^{1}\). This suggests that both groups should have similar chemical behaviors.
02

Understanding The Principal Quantum Number

The principal quantum number, \(n\), indicates the energy level of the electron in an atom. The larger the number, the higher the energy and the farther the electron is from the nucleus, which indicates a greater degree of instability.
03

Explaining the Stability Differences

Even though both Group 1A and Group 1B elements have similar outer electron configurations, the main difference lies in their nuclear charge. Group 1B elements have more protons in their nucleus compared to Group 1A elements. More protons mean greater nuclear attraction force, pulling the electrons closer to the nucleus, thus reducing the energy of the electron and increasing stability. Therefore, Group 1B elements are generally more stable than Group 1A elements.

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Most popular questions from this chapter

As a group, the noble gases are very stable chemically (only \(\mathrm{Kr}\) and Xe are known to form compounds). Use the concepts of shielding and the effective nuclear charge to explain why the noble gases tend to neither give up electrons nor accept additional electrons.

How do the chemical properties of oxides change from left to right across a period? From top to bottom within a particular group?

Explain, in terms of their electron configurations, why \(\mathrm{Fe}^{2+}\) is more easily oxidized to \(\mathrm{Fe}^{3+}\) than \(\mathrm{Mn}^{2+}\) is to \(\mathrm{Mn}^{3+}\)

The first and second ionization energies of \(\mathrm{K}\) are 419 kJ/mol and \(3052 \mathrm{~kJ} / \mathrm{mol}\), and those of Ca are \(590 \mathrm{~kJ} / \mathrm{mol}\) and \(1145 \mathrm{~kJ} / \mathrm{mol},\) respectively. Compare their values and comment on the differences.

A technique called photoelectron spectroscopy is used to measure the ionization energy of atoms. A sample is irradiated with UV light, and electrons are ejected from the valence shell. The kinetic energies of the ejected electrons are measured. Because the energy of the UV photon and the kinetic energy of the ejected electron are known, we can write $$ h \nu=I E+\frac{1}{2} m u^{2} $$ where \(\nu\) is the frequency of the UV light, and \(m\) and \(u\) are the mass and velocity of the electron, respectively. In one experiment the kinetic energy of the ejected electron from potassium is found to be \(5.34 \times 10^{-19} \mathrm{~J}\) using a UV source of wavelength \(162 \mathrm{nm} .\) Calculate the ionization energy of potassium. How can you be sure that this ionization energy corresponds to the electron in the valence shell (that is, the most loosely held electron)?
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