a) i) A: Beryllium has a higher first ionization energy (IE) than lithium. Beryllium has a higher nuclear charge than lithium, and its valence electrons are in the same principal energy level (n=2). The increased nuclear charge pulls the electrons closer, requiring more energy to remove the first electron. Additionally, beryllium has a filled 2s subshell, which is more stable. B: The first ionization energy of boron is less than that of beryllium. Beryllium has a filled 2s subshell ($1s^2 2s^2$), which is very stable. Boron has its outermost electron in a 2p orbital ($1s^2 2s^2 2p^1$). The 2p electron in boron is at a slightly higher energy level and experiences more shielding from the 2s electrons, making it easier to remove than a 2s electron from beryllium. ii) Step 1: Determine the common hydride formula for each element based on its group and typical valency. • Lithium (Group 1): Forms $\text{LiH}$ • Beryllium (Group 2): Forms $\text{BeH}_2$ • Boron (Group 13): Forms $\text{BH}_3$ (often dimerizes to $\text{B}_2\text{H}_6$) • Carbon (Group 14): Forms $\text{CH}_4$ • Nitrogen (Group 15): Forms $\text{NH}_3$ • Oxygen (Group 16): Forms $\text{H}_2\text{O}$ • Fluorine (Group 17): Forms $\text{HF}$ • Neon (Group 18): Does not typically form hydrides. | Element | Li | Be | B | C | N | O | F | Ne | | :------ | :-- | :--- | :--- | :--- | :--- | :---- | :-- | :-- | | Hydride | $\text{LiH}$ | $\text{BeH}_2$ | $\text{BH}_3$ | $\text{CH}_4$ | $\text{NH}_3$ | $\text{H}_2\text{O}$ | $\text{HF}$ | None | iii) Step 1: Identify the hydrides that are neutral. • $\text{CH}_4$ (methane) is a neutral hydride. • $\text{BH}_3$ (borane) is also considered neutral, though it is electron-deficient and often dimerizes. The two elements that form neutral hydrides are Carbon and Boron. That's 2 down. 3 left today — send the next one.