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Chapter 7: Problem 6

Explain why a graph of ionization energy versus atomic number (across a row) is not linear. Where are the exceptions? Why are there exceptions?

Short Answer

Expert verified
The graph of ionization energy versus atomic number across a row is not linear because of the increasing nuclear charge, shielding effects, and stability of certain electron configurations such as half-filled and fully-filled subshells. Exceptions to the general trend are observed between groups 2 and 3 (s and p orbital interaction) due to the shift from an s^2 to an s^1p^1 configuration and between groups 5 and 6 (half-filled p orbitals) because of the greater stability provided by the p^3 configuration compared to the p^4 configuration.

Step by step solution

01

Understand Ionization Energy

Ionization energy is the energy required to remove the most loosely bound electron (valence electron) from a neutral atom in the gaseous state. It is an important property of elements because it indicates the behavior of an atom in a chemical reaction. The ionization energy of an element depends on the atomic number (number of protons) and the electronic configuration of the element.
02

Describe Ionization Energy Trend Across a Row (Period)

As we move across a row (or period) in the periodic table from left to right, the ionization energy generally increases. This is because the atomic number (i.e., the number of protons) increases, which results in a greater force of attraction between the positively charged nucleus and the negatively charged electrons. The increasing number of protons also means that the electrons are pulled closer to the nucleus. Therefore, more energy is required to overcome the stronger electrostatic force and to remove an electron from the atom.
03

Understand Exceptions in Ionization Energy Trend

Although the ionization energy generally increases across a row, there are some exceptions. These exceptions occur when an atom has a half-filled or fully-filled subshell of electrons, which provide extra stability to the electronic configuration. Particularly, exceptions are observed between groups 2 and 3 (s and p orbital interaction) as well as between groups 5 and 6 (half-filled p orbitals).
04

Explain Exception Between Groups 2 and 3

At the boundary between groups 2 and 3, there is a drop in ionization energy. This is due to the change in the subshell from an s^2 configuration to an s^1p^1 configuration. The electron added to the new p orbital is further away from the nucleus and experiences greater shielding by the inner (core) electrons, which makes it easier to remove the electron and thus leads to a decrease in ionization energy.
05

Explain Exception Between Groups 5 and 6

Another notable exception occurs between groups 5 and 6, where the ionization energy should typically increase but a decrease is observed instead. This is due to the presence of half-filled p orbitals in elements of group 5 (p^3 configuration). Half-filled orbitals offer greater stability than a p^4 configuration (one additional electron), which can be explained by electron-electron repulsion and exchange energy in the half-filled configuration. Thus, elements in group 5 have higher ionization energy compared to elements in group 6.
06

Conclusion

In conclusion, the graph of ionization energy versus atomic number (across a row) is not linear because of factors such as increasing nuclear charge, shielding effects, and stability of certain electron configurations (half-filled and fully-filled subshells). Exceptions to the general trend are observed between groups 2 and 3, and groups 5 and 6, which can be explained by the unique electronic configurations and the stability they confer.

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Most popular questions from this chapter

Give a possible set of values of the four quantum numbers for all the electrons in a boron atom and a nitrogen atom if each is in the ground state.

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Consider the following ionization energies for aluminum: $$\begin{array}{c}{\operatorname{Al}(g) \longrightarrow \mathrm{Al}^{+}(g)+\mathrm{e}^{-} \quad I_{1}=580 \mathrm{kJ} / \mathrm{mol}} \\\ {\mathrm{Al}^{+}(g) \longrightarrow \mathrm{Al}^{2+}(g)+\mathrm{e}^{-} \quad I_{2}=1815 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{Al}^{2+}(g) \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} \quad I_{3}=2740 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{Al}^{3+}(g) \longrightarrow \mathrm{Al}^{4+}(g)+\mathrm{e}^{-} \quad I_{4}=11,600 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ a. Account for the trend in the values of the ionization energies. b. Explain the large increase between \(I_{3}\) and \(I_{4}\)
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