Periodic Trend: Successive Ionization Energies: Videos & Practice Problems

Ionization energy, denoted as $E_1$ for the first ionization energy, refers to the energy required to remove the first electron from a gaseous atom, resulting in a positively charged ion. Successive ionization energies involve removing additional electrons one at a time, with each subsequent ionization requiring more energy as the ion becomes more positively charged. This increase in energy is significant when inner core electrons are removed after the valence electrons have been stripped away, as seen in the large jumps in ionization energy values. These jumps can help predict the number of valence electrons in an element, as elements tend to lose electrons to achieve a noble gas electron configuration. For instance, lithium (in Group 1A) has a large increase in ionization energy after the first electron is removed, reflecting its desire to lose only one electron to resemble a noble gas. Similarly, oxygen would experience a significant increase after losing six electrons, corresponding to the number needed to reach a noble gas-like state.

Successive Ionizations

The first ionization energy, denoted as \( I_{e1} \), refers to the energy required to remove the first electron from a gaseous atom, resulting in a positively charged ion. For nitrogen, this energy is approximately 1402 kJ/mol. When an electron is removed, the atom transitions from a neutral state to a positively charged state, specifically \( N^+ \) in this case.

Successive ionizations involve the removal of additional electrons one at a time, rather than all at once. For instance, after removing the first electron, nitrogen becomes \( N^+ \). The second electron can then be removed, leading to the formation of \( N^{2+} \). This process continues, with each removal increasing the positive charge of the ion. Thus, the second ionization energy, \( I_{e2} \), is the energy needed to remove the second electron from \( N^+ \), and the third ionization energy, \( I_{e3} \), is the energy required to remove the third electron from \( N^{2+} \).

In summary, the ionization process for nitrogen can be represented as follows:

• First ionization: \( N \rightarrow N^+ + e^- \) (with \( I_{e1} = 1402 \, \text{kJ/mol} \))
• Second ionization: \( N^+ \rightarrow N^{2+} + e^- \) (with \( I_{e2} \))
• Third ionization: \( N^{2+} \rightarrow N^{3+} + e^- \) (with \( I_{e3} \))

Each successive ionization requires more energy due to the increased positive charge of the ion, which leads to a stronger attraction between the remaining electrons and the nucleus.

Successive Ionizations Example

The 4th ionization energy of a manganese atom can be represented by the following equation, which reflects the process of removing an electron from a manganese ion that has already lost three electrons. In this scenario, manganese is in a gaseous state and has a +3 charge before the removal of the fourth electron.

The equation for the 4th ionization energy can be expressed as:

\[\text{Mn}^{3+} (g) \rightarrow \text{Mn}^{4+} (g) + e^-\]

In this equation, \(\text{Mn}^{3+}\) represents the manganese ion with a +3 charge, and upon the removal of the fourth electron, it becomes \(\text{Mn}^{4+}\), which has a +4 charge. The process involves the absorption of energy to overcome the attraction between the positively charged ion and the negatively charged electron being removed. This energy required to remove the fourth electron is known as the 4th ionization energy.

Successive Ionization Energies

Successive ionization energies refer to the increasing amount of energy required to remove electrons from an atom, with each subsequent electron removal demanding more energy. This trend is evident when examining the ionization energies of elements, where the first ionization energy is lower than the second, and so on. For instance, lithium's first ionization energy is 520.2 kJ/mol, but the second ionization energy skyrockets to 7,298 kJ/mol, illustrating the significant increase in energy needed to remove additional electrons.

Elements typically lose valence electrons to achieve an electron configuration similar to that of noble gases, a state known as being isoelectronic. A notable increase in ionization energy occurs when an electron is removed from an inner core, as opposed to a valence electron. For example, lithium, located in group 1A, aims to lose just one electron to resemble a noble gas. However, attempting to remove a second electron results in a dramatic increase in required energy, exceeding a tenfold rise.

Beryllium, found in group 2A, seeks to lose two electrons to attain noble gas configuration. The transition from the first to the second ionization energy shows an increase that is less than double, but removing a third electron leads to a substantial jump in energy requirements. This pattern allows for predictions regarding where significant increases in ionization energy will occur, typically after an element has achieved noble gas status.

For example, oxygen, which prefers to gain electrons, would need to lose six electrons to become isoelectronic with a noble gas. However, the energy required to remove a seventh electron would be exceedingly high, indicating a spike in ionization energy. This pattern of increasing ionization energy highlights the stability associated with noble gas configurations and the challenges of removing additional electrons once that stability is achieved.

Successive Ionization Energies Example

When analyzing the successive ionization energies of elements, it's important to understand how the position of an atom in the periodic table influences its ionization energy. Ionization energy is the energy required to remove an electron from an atom, and it generally increases as you move from left to right across a period and decreases as you move down a group.

In this context, we are tasked with identifying which atom has the smallest increase in its second ionization energy. Elements in Group 1A, such as lithium, typically have a significant jump in ionization energy when removing a second electron. This is because they only need to lose one electron to achieve a stable noble gas configuration, and removing a second electron requires significantly more energy.

After eliminating Group 1A options, we focus on aluminum, magnesium, and beryllium. These elements are in Groups 3A and 2A, respectively, and do not achieve noble gas status after losing just one electron. As we consider their positions in the periodic table, we note that ionization energy increases towards the top right corner. Beryllium and magnesium, being in Group 2A, will have lower ionization energies compared to aluminum in Group 3A.

Since magnesium is the furthest left among the remaining options, it will have the lowest initial ionization energy. Consequently, the increase in its second ionization energy will be less pronounced compared to aluminum and beryllium. Therefore, magnesium is the best choice for the atom with the smallest increase in its second ionization energy.

Which of the following represents the third ionization of Mn?

• A. Mn⁻ (g) + e⁻ → 2 Mn²⁻ (g)
• B. Mn²⁺ (g) → Mn³⁺ (g) + e⁻
• C. Mn²⁻ (g) + e⁻ → 2 Mn³⁻ (g)
• D.
  Mn (g) → Mn³⁺ (g) + 3 e⁻

Answer: (B)

Of the following atoms, which has the largest third ionization energy? 

• A. Al
• B. Ca
• C. K
• D. Ba
• E. Cs

Answer: (B)

The successive ionization energies for an unknown element are:
IE₁ = 896 kJ/molIE₂ = 1752 kJ/molIE₃ = 14,807 kJ/molIE₄ = 17,948 kJ/mol
To which family in the periodic table does the unknown element most likely belong?

• A. 1A
• B. 2A
• C. 3A
• D. 4A
• E. 5A

Answer: (B)

Here’s what students ask on this topic:

What is the first ionization energy?

The first ionization energy is the amount of energy required to remove the most loosely bound electron from a neutral atom in its gaseous state, to form a cation. It is usually expressed in electronvolts (eV) or kilojoules per mole (kJ/mol). The process can be represented by a chemical equation, where 'X' represents an element:

$X\left(g\right)\to X^{+}\left(g\right)+e-$

This energy is a measure of the force holding the electron within the atom and is indicative of how an element will react chemically. Elements with low first ionization energies tend to lose their outer electrons easily and are typically metals. In contrast, elements with high first ionization energies hold onto their electrons more tightly and are often nonmetals. The first ionization energy generally increases across a period (from left to right) in the periodic table and decreases down a group (from top to bottom).

What is the second ionization energy?

The second ionization energy refers to the amount of energy required to remove the second electron from an atom after the first electron has already been removed. This process results in the formation of a doubly charged ion from a singly charged ion. It is important to note that the second ionization energy is always higher than the first ionization energy. This increase in energy is due to the fact that, after the first electron is removed, the remaining electrons experience a greater effective nuclear charge—meaning the attraction between the nucleus and the electrons is stronger since there are fewer electrons to shield the positive charge of the nucleus. As a result, more energy is needed to overcome this attraction and remove an additional electron. Understanding ionization energies is crucial in the study of chemical reactivity and the formation of ions in various chemical processes.

Why do successive ionization energies increase?

Successive ionization energies increase because it becomes more difficult to remove an electron from an atom as you take more away. When you remove the first electron, the remaining electrons are held by the positive charge of the nucleus with a relatively weaker force since there are more electrons to spread the positive charge over. However, once you remove an electron, the positive charge of the nucleus is spread over fewer electrons, which increases the effective nuclear charge on each remaining electron. This means that each subsequent electron is held more tightly by the nucleus, and thus, more energy is required to remove it. Additionally, as you strip away electrons, the electron you are trying to remove may be closer to the nucleus and in a lower energy level, which further increases the ionization energy required to remove it.