The electron affinities of the elements from aluminum to chlorine are \(-44,-120,-74,-200.4\), and \(-384.7 \mathrm{~kJ} / \mathrm{mol}\), respectively. Rationalize the trend in these values.

Short Answer

Expert verified
The electron affinity trend from aluminum to chlorine can be rationalized by considering the effective nuclear charge (Z_eff) and the stability of half-filled orbitals. As we move across the period, Z_eff generally increases, leading to more negative electron affinity values. However, at phosphorus, the 3p orbital is already half-filled, providing extra stability and making it less energetically favorable to add an electron. This results in a less negative electron affinity for phosphorus compared to silicon and sulfur. Overall, the trend in electron affinity is: more negative from Al to Si, less negative from Si to P, and more negative from P to Cl.

Step by step solution

01

Identify the elements and their positions on the periodic table

The elements with the given electron affinities are aluminum (Al), silicon (Si), phosphorus (P), sulfur (S), and chlorine (Cl). These elements are in Period 3 of the periodic table. Their group numbers are as follows: - Aluminum (Al): Group 13 - Silicon (Si): Group 14 - Phosphorus (P): Group 15 - Sulfur (S): Group 16 - Chlorine (Cl): Group 17
02

Examine the trend in electron affinity values

We are given the electron affinity values as: Al (-44 kJ/mol), Si (-120 kJ/mol), P (-74 kJ/mol), S (-200.4 kJ/mol), and Cl (-384.7 kJ/mol). The overall trend in these values is: - First, the electron affinity becomes more negative from Al to Si; - Then, it becomes less negative from Si to P; - Finally, it becomes more negative from P to Cl.
03

Understand electron affinity's relationship with atomic structure and the periodic table

Electron affinity is influenced by the effective nuclear charge (Z_eff) and the distance between the nucleus and the added electron (represented by the principal quantum number n). As we move from left to right across a period, Z_eff generally increases, which means the attractive force between the nucleus and the outermost electrons becomes stronger. However, the distance (n) remains roughly the same within the same period.
04

Rationalize the trend in electron affinities

The Z_eff increases as we move across the period from aluminum to chlorine, which explains why the electron affinities generally become more negative. This indicates that the elements on the right side of the periodic table (such as chlorine) will have a stronger attraction for the added electrons than those on the left side (such as aluminum). However, there is an exception at the phosphorus element. This can be explained by the fact that, in phosphorus, an electron will be added to the 3p orbital, which is already half-filled (electron configuration: [Ne] 3s² 3p³) in its neutral state. A half-filled orbital is considered as extra stable, so adding an extra electron to a half-filled p orbital is less energetically favorable than adding an electron to a different element with a not half-filled orbital (like silicon or sulfur). So, considering both the increase in Z_eff and the half-filled orbital stability, the given trend in the electron affinity values can be rationalized as: more negative from Al to Si, less negative from Si to P (due to half-filled 3p orbital stability), and continuing more negative from P to Cl.

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Most popular questions from this chapter

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