Question

1. carbon has an electron configuration of 1s² 2s² 2p². which electron configuration has a higher electronegativity than carbon?

o: 1s² 2s² 2p⁴
cl: 1s² 2s² 2p⁶ 3s² 3p⁵
p: 1s² 2s² 2p⁶ 3s² 3p³
mg: 1s² 2s² 2p⁶ 3s²

Explanation:

Step1: Recall electronegativity trend

Electronegativity increases across a period and decreases down a group in the periodic table.

Step2: Analyze each option

• O: Oxygen (\(O\)) has electron configuration \(1s^{2}2s^{2}2p^{4}\). Oxygen is to the right of carbon (\(C\)) in period 2. Since electronegativity increases across a period, \(O\) has higher electronegativity than \(C\).
• Cl: Chlorine (\(Cl\)) has electron configuration \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}\). Although \(Cl\) is more electronegative than many elements, comparing to \(O\) (smaller atomic radius in period 2 vs \(Cl\) in period 3), \(O\) is more electronegative than \(C\) (relevant to the question of which has higher electronegativity than \(C\), but we need to check all).
• P: Phosphorus (\(P\)) has electron configuration \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}\). \(P\) is below \(N\) (and to the right of \(Si\) etc.), but in lower period than \(O\) (relevant period - group trend). \(P\) is less electronegative than \(O\) and we are comparing to \(C\). Since \(O\) is better (as per step 1 - 2 analysis for \(O\) vs \(C\) based on period trend).
• Mg: Magnesium (\(Mg\)) has electron configuration \(1s^{2}2s^{2}2p^{6}3s^{2}\). \(Mg\) is a metal, has lower electronegativity than non - metals like \(C\) (metals generally have lower electronegativity than non - metals in relevant contexts).

Answer:

O. \(1s^{2}2s^{2}2p^{4}\)