Question

name e ifidro mejia 9/18/25 which of the following ground - state electron configurations represents the atom that has the lowest first - ionization energy? a 1s²2s¹ b 1s²2s²2p² c 1s²2s²2p⁶ d 1s²2s²2p⁶3s¹ which of the following is the ground - state electron configuration of the f⁻ ion? a 1s²2s²2p⁴ b 1s²2s²2p⁵ c 1s²2s²2p⁶

Explanation:

Step1: Recall ionization - energy trend

Ionization energy generally increases across a period and decreases down a group. Outer - shell electrons that are further from the nucleus and less shielded are easier to remove.

Step2: Analyze each electron configuration for first question

• For $1s^{2}2s^{1}$ (lithium), it has a single valence electron in the 2s orbital.
• For $1s^{2}2s^{2}2p^{2}$ (carbon), the electrons are in a more stable configuration compared to lithium and are closer to the nucleus on average.
• For $1s^{2}2s^{2}2p^{6}$ (neon), it has a full outer - shell (noble gas configuration), which is very stable and has a high ionization energy.
• For $1s^{2}2s^{2}2p^{6}3s^{1}$ (sodium), although it has a single valence electron like lithium, the 3s electron is further from the nucleus than the 2s electron in lithium, but the effective nuclear charge and shielding also play a role. Lithium has the lowest first - ionization energy among them as it has a relatively small effective nuclear charge and its single valence electron is in the 2s orbital.

Step3: Recall electron configuration of fluoride ion for second question

Fluorine has an atomic number of 9, with a ground - state electron configuration of $1s^{2}2s^{2}2p^{5}$. When it gains an electron to form $F^{-}$, it has a total of 10 electrons, and the electron configuration becomes $1s^{2}2s^{2}2p^{6}$.

Answer:

1. A. $1s^{2}2s^{1}$
2. C. $1s^{2}2s^{2}2p^{6}$