Question

1. element x has an electron configuration of 1s²2s²2p⁶3s¹, while element z has an electron configuration of 1s²2s²2p⁵. x = sodium and z = fluorine

a) which element would have greater first ionization energy?
b) which element would have a larger radius?
c) which element would have higher electronegativity?
d) which element would form an ion that has a larger radius?

Explanation:

To solve these, we analyze Element X (Sodium, \( \text{Na} \), \( 1s^22s^22p^63s^1 \)) and Element Z (Fluorine, \( \text{F} \), \( 1s^22s^22p^5 \)):

Part (a): First Ionization Energy

Ionization energy (IE) is the energy to remove an electron.

• \( \text{Na} \) (Group 1) has 1 valence electron (e⁻) in \( 3s \), easy to remove (low IE).
• \( \text{F} \) (Group 17) has 7 valence e⁻, close to octet; removing an e⁻ is hard (high IE).

Thus, Fluorine (Z) has greater first IE.

Part (b): Atomic Radius

Atomic radius increases down a group, decreases across a period (left→right).

• \( \text{Na} \) is in Period 3, \( \text{F} \) in Period 2.
• Larger period = larger radius. Thus, Sodium (X) has a larger radius.

Part (c): Electronegativity

Electronegativity (EN) increases across a period (left→right) and up a group.

• \( \text{F} \) is right of \( \text{Na} \) in Period 2, so \( \text{F} \) has higher EN.

Thus, Fluorine (Z) has higher electronegativity.

Part (d): Anion Radius

• \( \text{Na} \) forms \( \text{Na}^+ \) (loses 1 e⁻, smaller radius).
• \( \text{F} \) forms \( \text{F}^- \) (gains 1 e⁻, larger radius due to electron-electron repulsion).

Thus, the anion of Fluorine (Z) (\( \text{F}^- \)) has a larger radius.

Answer:

s:
a) Fluorine (Z)
b) Sodium (X)
c) Fluorine (Z)
d) Fluorine (Z) (anion \( \text{F}^- \))