Ionization energy is the minimum energy required to remove an electron from a gaseous atom or ion in its ground state.

Ionization energy is the minimum energy required to remove an electron from a gaseous atom or ion in its ground state.

• Trends in Ionization Energy:

  • Across a Period (e.g., Li to Be): Ionization energy generally increases across a period. This is because the nuclear charge increases, pulling the valence electrons closer to the nucleus and making them harder to remove.
    • From the table, the first ionization energy ($I_1$) of Li is 520.2 kJ/mol, while for Be it is 899.5 kJ/mol. Be has a higher $I_1$ than Li, consistent with this trend.
  • Down a Group: Ionization energy generally decreases down a group. This is because the atomic radius increases, and the outermost electrons are further from the nucleus and experience greater shielding from inner electrons, making them easier to remove. (The provided table does not contain data to illustrate this trend directly).

• Formation of Ions by Alkali Metals (e.g., Lithium, Li): Lithium (an alkali metal) has the electron configuration $1s^2 2s^1$.

  • The first ionization energy ($I_1$) for Li is 520.2 kJ/mol: $$\text{Li(g)} \to \text{Li}^+\text{(g)} + \text{e}^- \quad (I_1 = 520.2 \text{ kJ/mol})$$ Removing this $2s^1$ electron results in a stable $\text{Li}^+$ ion with a $1s^2$ electron configuration, which is isoelectronic with Helium (a noble gas).
  • The second ionization energy ($I_2$) for Li is 7298.2 kJ/mol: $$\text{Li}^+\text{(g)} \to \text{Li}^{2+}\text{(g)} + \text{e}^- \quad (I_2 = 7298.2 \text{ kJ/mol})$$
  • The third ionization energy ($I_3$) for Li is 11815.0 kJ/mol: $$\text{Li}^{2+}\text{(g)} \to \text{Li}^{3+}\text{(g)} + \text{e}^- \quad (I_3 = 11815.0 \text{ kJ/mol})$$

  There is a very large jump in energy from $I_1$ to $I_2$ (approximately 14 times greater). This significant increase indicates that removing the second electron requires much more energy because it is being removed from a stable, full inner electron shell ($1s^2$). Therefore, alkali metals readily form compounds with a +1 ion but not +2 or +3 ions.

• Formation of Ions by Alkaline Earth Metals (e.g., Beryllium, Be): Beryllium (an alkaline earth metal) has the electron configuration $1s^2 2s^2$.

  • The first ionization energy ($I_1$) for Be is 899.5 kJ/mol: $$\text{Be(g)} \to \text{Be}^+\text{(g)} + \text{e}^- \quad (I_1 = 899.5 \text{ kJ/mol})$$
  • The second ionization energy ($I_2$) for Be is 1757.1 kJ/mol: $$\text{Be}^+\text{(g)} \to \text{Be}^{2+}\text{(g)} + \text{e}^- \quad (I_2 = 1757.1 \text{ kJ/mol})$$ Removing both $2s^2$ electrons results in a stable $\text{Be}^{2+}$ ion with a $1s^2$ electron configuration, which is isoelectronic with Helium. The increase from $I_1$ to $I_2$ is relatively small (less than double), indicating that removing both valence electrons is energetically favorable.
  • The third ionization energy ($I_3$) for Be is 14848.8 kJ/mol: $$\text{Be}^{2+}\text{(g)} \to \text{Be}^{3+}\text{(g)} + \text{e}^- \quad (I_3 = 14848.8 \text{ kJ/mol})$$
  • The fourth ionization energy ($I_4$) for Be is 21006.6 kJ/mol: $$\text{Be}^{3+}\text{(g)} \to \text{Be}^{4+}\text{(g)} + \text{e}^- \quad (I_4 = 21006.6 \text{ kJ/mol})$$

  There is a very large jump in energy from $I_2$ to $I_3$ (approximately 8.5 times greater). This significant increase indicates that removing the third electron requires much more energy because it is being removed from a stable, full inner electron shell ($1s^2$). Therefore, alkaline earth metals readily form compounds with a +2 ion but not +3 or +4 ions.