Question

1. explore: examine other groups and periods in the periodic table to see if the same trends exist. what trends do you see in ionization energy down a group and across a period?7. think and discuss: as you move down a group, you will recall that the radius increases. why do you think an increase in atomic radius would result in a lower ionization energy?8. think and discuss: as you move across a period, you will recall that the radius decreases. why do you think a decrease in atomic radius would result in a greater ionization energy?

Explanation:

• Question 6: Ionization energy is the energy required to remove an electron from an atom. As we move down a group in the periodic table, the number of electron shells increases. This means that the outermost electrons are further away from the nucleus and are shielded by the inner - shell electrons. So, the ionization energy decreases down a group. As we move across a period (from left to right), the atomic number increases, and the electrons are added to the same shell. The effective nuclear charge (the net positive charge experienced by an electron) increases across a period. This stronger attraction between the nucleus and the electrons makes it more difficult to remove an electron, so ionization energy increases across a period.
• Question 7: The force of attraction \(F\) between a nucleus (charge \(+Ze\), where \(Z\) is the atomic number and \(e\) is the elementary charge) and an electron (charge \(-e\)) is given by Coulomb's law \(F=\frac{kZe\times e}{r^{2}}\) (where \(k\) is a constant and \(r\) is the distance between the nucleus and the electron). When the atomic radius \(r\) increases (as we move down a group), the distance between the nucleus and the outermost electron increases. The force of attraction between the nucleus and the outermost electron decreases. So, less energy is required to remove the electron (lower ionization energy).
• Question 8: As we move across a period, the atomic radius \(r\) decreases. Using Coulomb's law \(F = \frac{kZe\times e}{r^{2}}\), a smaller \(r\) (with an increase in \(Z\) - the number of protons in the nucleus) leads to a stronger force of attraction \(F\) between the nucleus and the electrons. Since ionization energy is related to overcoming this attractive force, a stronger force means more energy is required to remove an electron (greater ionization energy).

Answer:

• Question 6: Ionization energy decreases down a group and increases across a period.
• Question 7: A larger atomic radius means the out - ermost electron is farther from the nucleus. The electrostatic attraction between the nucleus and the electron is weaker (by Coulomb's law), so less energy is needed to remove the electron (lower ionization energy).
• Question 8: A smaller atomic radius means the outermost electron is closer to the nucleus. The electrostatic attraction (by Coulomb's law) between the nucleus and the electron is stronger. So, more energy is required to remove the electron (greater ionization energy).