Question

Which of the four atoms \(\mathrm{Rb}\), \(\mathrm{Sr}\), \(\mathrm{Sb}\), or \(\mathrm{Cs}\) (a) has the smallest atomic radius? (b) has the lowest ionization energy? (c) is the least electronegative?

Short answer

Answer: Antimony (Sb) has the smallest atomic radius, Cesium (Cs) has the lowest ionization energy, and Cesium (Cs) is the least electronegative element among the given elements.

Step-by-step solution

Determine the positions of given elements in the periodic table

First, we need to determine the positions of these elements in the periodic table. This will help us to understand the trends in their properties. Rubidium (Rb) - Group 1, Period 5 Strontium (Sr) - Group 2, Period 5 Antimony (Sb) - Group 15, Period 5 Cesium (Cs) - Group 1, Period 6

Periodic trends in atomic radius

Atomic radius generally decreases across a period from left to right due to an increase in effective nuclear charge and increases down a group due to the addition of a new energy level (shell) to the electron cloud. (a) To find the element with the smallest atomic radius, we will apply the periodic trend in atomic radius: - Rb and Cs are in Group 1, and Cs is below Rb in the same group, so Cs will have a larger atomic radius than Rb. - Sr is in Group 2 and in the same period as Rb, which means Sr will have a smaller atomic radius than Rb. - Sb is in Group 15 and in the same period as Rb and Sr, which means Sb will have an even smaller atomic radius. Therefore, \(\boxed{\text{Antimony (Sb)}}\) has the smallest atomic radius among the given elements.

Periodic trends in ionization energy

Ionization energy is the energy required to remove an electron from an atom. It generally increases across a period from left to right and decreases down a group. This is due to the increased effective nuclear charge experienced by electrons across a period and the increased distance from the nucleus and shielding down a group. (b) To find the element with the lowest ionization energy, we will apply the periodic trend in ionization energy: - Rb and Cs are in Group 1. Cs is below Rb in the same group, so Cs will have a lower ionization energy than Rb. - Sr is in Group 2 and in the same period as Rb, which means Sr will have a higher ionization energy than Rb. - Sb is in Group 15 and in the same period as Rb and Sr, which means Sb will have a much higher ionization energy. Therefore, \(\boxed{\text{Cesium (Cs)}}\) has the lowest ionization energy among the given elements.

Periodic trends in electronegativity

Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. It generally increases across a period from left to right and decreases down a group. This is primarily due to the increased effective nuclear charge experienced by electrons across a period and the increased distance from the nucleus and shielding down a group. (c) To find the element with the least electronegativity, we will apply the periodic trend in electronegativity: - Rb and Cs are in Group 1. Cs is below Rb in the same group, so Cs will have a lower electronegativity than Rb. - Sr is in Group 2 and in the same period as Rb, which means Sr will have a higher electronegativity than Rb. - Sb is in Group 15 and in the same period as Rb and Sr, which means Sb will have a much higher electronegativity. Therefore, \(\boxed{\text{Cesium (Cs)}}\) is the least electronegative element among the given elements.