Question

1. based on their positions in the periodic table, predict which has the smallest first ionization energy: li, cs, n, f, i.

Explanation:

Step1: Recall ionization energy trends

Ionization energy (IE) generally decreases down a group and increases across a period. Down a group, atomic size increases, so valence electrons are farther from the nucleus, requiring less energy to remove. Across a period, effective nuclear charge increases, so IE increases.

Step2: Classify elements by group/period

• Li (lithium) and Cs (cesium) are in Group 1 (alkali metals). Cs is below Li in the group.
• N (nitrogen), F (fluorine) are in Period 2, Group 15 and 17, respectively. I (iodine) is in Group 17, below F.

Step3: Compare IE for Group 1

In Group 1, as we go down (Li → Na → K → Rb → Cs), atomic radius increases, so first IE decreases. Thus, Cs has lower IE than Li.

Step4: Compare with non - metals

N, F, and I are non - metals. Non - metals generally have higher IE than alkali metals because they have more valence electrons and smaller atomic radii (for same period) or, for halogens, they are closer to achieving a stable octet (but down the group, IE decreases for halogens too, but still, alkali metals have much lower IE than non - metals). So among Li, Cs, N, F, I, Cs is in the group where IE decreases down the group and is an alkali metal (which has low IE compared to non - metals), and it is lower than Li in the group.

Answer:

Cs (Cesium) has the smallest first ionization energy.