Question

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

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 option

• Option A ($1s^{2}2s^{1}$) is lithium ($Li$), a period 2 element.
• Option B ($1s^{2}2s^{2}2p^{2}$) is carbon ($C$), a period 2 element with a more filled shell than $Li$.
• Option C ($1s^{2}2s^{2}2p^{6}$) is neon ($Ne$), a noble gas with a stable electron configuration.
• Option D ($1s^{2}2s^{2}2p^{6}3s^{1}$) is sodium ($Na$), a period 3 element. The outermost electron in $Na$ is in the 3s orbital, which is further from the nucleus compared to the 2s and 2p electrons in period 2 elements. It also has more shielding from the inner - shell electrons. So, it is the easiest to remove.

Answer:

D. $1s^{2}2s^{2}2p^{6}3s^{1}$