You learned in this chapter that ionization generally increases as you move from left to right across the periodic table. However, consider the following data, which shows the ionization energies of the period 2 and 3 elements: $$ \begin{array}{ccccc} \text { Group } & \begin{array}{c} \text { Period 2 } \\ \text { Elements } \end{array} & \begin{array}{c} \text { lonization } \\ \text { Energy } \\ \text { (kJ/mol) } \end{array} & \begin{array}{c} \text { Period 3 } \\ \text { Elements } \end{array} & \begin{array}{c} \text { Ionization } \\ \text { Energy } \\ \text { (kJ/mol) } \end{array} \\ \text { 1A } & \text { Li } & 520 & \text { Na } & 496 \\ \text { 2A } & \text { Be } & 899 & \text { Mg } & 738 \\ \text { 3A } & \text { B } & 801 & \text { Al } & 578 \\ \text { 4A } & \text { C } & 1086 & \text { Si } & 786 \\ \text { 5A } & \text { N } & 1402 & \text { P } & 1012 \\ \text { 6A } & \text { 0 } & 1314 & \text { S } & 1000 \\ \text { 7A } & \text { F } & 1681 & \text { Cl } & 1251 \\ \text { 8A } & \text { Ne } & 2081 & \text { Ar } & 1521 \\ \hline \end{array} $$ Notice that the increase is not uniform. In fact, ionization energy actually decreases a bit in going from elements in group \(2 \mathrm{~A}\) to \(3 \mathrm{~A}\) and then again from \(5 \mathrm{~A}\) to \(6 \mathrm{~A}\). Use what you know about electron configurations to explain why these dips in ionization energy exist.

Step by step solution

01

Review of Electron Configuration and Ionization Energy

Ionization energy is the energy required to remove an electron from an atom in its gaseous state. Generally, ionization energy increases across a period because elements have more protons as you move across the period, thus there is a greater pull on the electrons (effective nuclear charge). However, the pattern is not always consistent due to the electronic structure of the atoms. The arrangement of electrons in different energy levels and subshells can cause variations in ionization energy.

02

Analyze Group 2A to 3A Decrease

The decrease in ionization energy from Group 2A to Group 3A can be explained by the subshell configuration of these elements. Be (Group 2A) has a full 's' subshell, which is more stable than the partially filled 'p' subshell in B (Group 3A). The full subshell in beryllium results in a higher ionization energy because electrons in a filled subshell are more difficult to remove.

03

Analyze Group 5A to 6A Decrease

The decrease in ionization energy from Group 5A to Group 6A can be attributed to electron-electron repulsion in the 'p' subshell. N (Group 5A) has half-filled 'p' subshell, which is relatively stable due to the symmetrical distribution of electrons. When moving to O (Group 6A), an additional electron is placed in a 'p' orbital, resulting in increased electron-electron repulsion within the 'p' subshell, which slightly reduces the ionization energy.

04

Conclusion

The dips in ionization energy for certain groups can be explained by electron configurations and the stability associated with filled and half-filled subshells. The stability of a full or half-full subshell (as seen in Group 2A and 5A) requires more energy to remove an electron, while additional electron-electron repulsion in less stable configurations (as seen in Group 3A and 6A) slightly decreases the ionization energy.

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Key Concepts

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

Electron Configuration

Understanding electron configuration is fundamental when analyzing ionization energy trends.
Ionization energy refers to the energy required to remove the most loosely bound electron from an atom in its gaseous state. Atoms have electrons distributed in various orbitals around the nucleus, and these fill up according to the Aufbau principle - with electrons occupying lower energy levels first.

Each electron configuration follows a distinct pattern of 'shells' and 'subshells'. For example, the configuration of beryllium (Be) in Group 2A is represented as 1s² 2s², indicating filled 's' orbital in the second energy level. A filled subshell like this is more stable and holds its electrons more tightly, leading to higher ionization energies. In contrast, boron (B) in Group 3A starts filling the 'p' subshell (2p¹), and since 'p' orbitals are higher in energy than 's', they are easier to remove, resulting in a lower ionization energy.

Periodic Table

The periodic table is not just a list of elements but a mosaic of atomic properties, revealing numerous trends including those of ionization energy.
As you move from left to right across a period on the periodic table, the ionization energy generally increases. This is because the number of protons in the nucleus increases, leading to a stronger attraction between the positively charged nucleus and the negatively charged electrons (effective nuclear charge), and thus a greater amount of energy is required to detach one electron from the atom.

Exceptions to Ionization Energy Trends

However, this pattern is not uniform. There are subtle nuances, such as the aforementioned dips in energy levels, that provide insights into the complexity of electron behavior. Elements in the same group share similar electron configurations, which can help explain why these trends occur and why certain groups exhibit decreases in ionization energy.

Effective Nuclear Charge

The concept of effective nuclear charge is pivotal in understanding why electrons are held with varying degrees of attraction.
Effective nuclear charge is the net positive charge experienced by an electron in a multi-electron atom. Although the total charge of the nucleus is equal to the number of protons it contains, the actual charge that an electron feels is less because of the shielding effect of other electrons. This effect is particularly evident in transition from Group 5A to Group 6A, where nitrogen (N) has a half-filled p subshell, which exhibits a relatively stable and symmetrical electron distribution, thereby increasing the ionization energy.

When oxygen (O) adds an extra electron, it increases repulsion within the same subshell due to its relatively higher electron density. This repulsion diminishes the effective nuclear charge felt by each electron, making it easier to remove one, and this is manifest in a lower ionization energy compared to nitrogen.