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 increases across a period (left to right) and decreases down a group (top to bottom) in the periodic table. This is because as we move across a period, the effective nuclear charge increases, pulling electrons closer and making it harder to remove them. Down a group, the principal quantum number (\(n\)) increases, so electrons are in higher energy levels, farther from the nucleus, and easier to remove.

Step2: Classify elements by groups and periods

• Li (Lithium) and Cs (Cesium): Both are alkali metals (Group 1). Cs is below Li in Group 1 (Cs is in Period 6, Li in Period 2). So down the group, IE decreases, so Cs has lower IE than Li.
• N (Nitrogen), F (Fluorine): Both are in Period 2 (N in Group 15, F in Group 17). Across Period 2, IE increases (N to F), so N has lower IE than F.
• I (Iodine): Halogen (Group 17), Period 5.

Now compare the alkali metals (Li, Cs) with non - metals (N, F, I). Alkali metals have only one valence electron, which is easier to remove compared to non - metals (which have more valence electrons and higher effective nuclear charge for their valence electrons). Among Li and Cs, since Cs is lower in Group 1, it has a larger atomic radius, less effective nuclear charge on the valence electron, so it's easier to remove the valence electron from Cs than from Li.

Now check the relative positions: Cs is in Group 1, Period 6; Li (Group 1, Period 2); N (Group 15, Period 2); F (Group 17, Period 2); I (Group 17, Period 5).

The general trend for first ionization energy: Group 1 metals have lower IE than non - metals in the same period. And within Group 1, lower IE as we go down. So Cs, being the lowest in Group 1 among these, will have the smallest first ionization energy.

Answer:

Cs (Cesium) has the smallest first ionization energy among Li, Cs, N, F, I.