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Chapter 9: Problem 96

Describe how the ionization energies of the ions \(\mathrm{Be}^{+}, \mathrm{B}^{+}, \mathrm{C}^{+}, \mathrm{N}^{+}, \mathrm{O}^{+}, \mathrm{F}^{+}, \mathrm{Ne}^{+},\) and \(\mathrm{Na}^{+}\) vary with atomic number.

Short Answer

Expert verified
As atomic number increases, ionization energy generally increases due to the increased nuclear charge. However, ions with half-filled or fully filled sub-shells have higher ionization energies due to increased stability.

Step by step solution

01

Analyze the electron configuration

First, let's start by writing down the electron configuration for each ion. Note that each ion is a positively charged ion, meaning that it has lost one electron compared to the neutral atom. For the first ion, \(\mathrm{Be}^{+}\), the electron configuration of the neutral atom \(Be\) is \(1s^22s^2\). So, for \(\mathrm{Be}^{+}\), which has lost an electron, the configuration would be \(1s^22s^1\). Similarly, figure out the electron configuration for each of the other ions.
02

Analyze the trend in ionization energies based on electron configuration

Next, let's analyze how the electron configuration affects the ionization energy. Atoms that have half-filled or fully filled sub-shells have more stability, hence they have higher ionization energies. This is because these stable configurations are more tightly held by the nucleus, and more energy is needed to remove an electron. Also, as atomic number (which relates to nuclear charge) increases, the attractive force on the electrons increases. This means the ionization energy generally increases with an increase in atomic number.
03

Predict the variation in ionization energies

Comparing the electron configurations and knowing the factors that affect ionization energy, we predict that as atomic number increases, ionization energy will generally increase - provided that the increase in atomic number does not result in an ion with a more stable half-filled or fully filled sub-shell configuration. In those cases, the ion with the more stable configuration will have a higher ionization energy.

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Most popular questions from this chapter

An ion that is isoelectronic with \(\mathrm{Se}^{2-}\) is (a) \(\mathrm{S}^{2-} ;\) (b) I \(^{-}\) (c) \(\mathrm{Xe} ;\) (d) \(\mathrm{Sr}^{2+}.\)

Describe how the ionization energies of the ions \(\mathrm{He}^{-}, \mathrm{Li}^{-}, \mathrm{Be}^{-}, \mathrm{B}^{-}, \mathrm{C}^{-}, \mathrm{N}^{-}, \mathrm{O}^{-},\) and \(\mathrm{F}^{-}\) vary with atomic number.

Which of the following species has the greatest number of unpaired electrons (a) \(\mathrm{Ge} ;\) (b) \(\mathrm{Cl} ;\) (c) \(\mathrm{Cr}^{3+}\) (d) Br -?

In estimating the boiling point and melting point of bromine in Example \(9-5,\) could we have used Celsius or Fahrenheit instead of Kelvin temperature? Explain.

All the isoelectronic species illustrated in the text had the electron configurations of noble gases. Can two ions be isoelectronic without having noble-gas electron configurations? Explain.
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