Understanding the Atomic Size of Li? and Al3?: A Comprehensive Guide

The atomic size of ions is influenced by several factors including effective nuclear charge, electron configuration, and the number of electron shells. This article aims to explain why the atomic size of Li? is greater than that of Al3?, despite both ions having the same electron configuration. Let's delve into the key factors:

Key Factors

Effective Nuclear Charge (Z_eff)

The effective nuclear charge is the net positive charge experienced by an electron in a multi-electron atom. It accounts for the shielding effect of inner-shell electrons.

For Li?:

Nuclear Charge Z: 3 from 3 protons Electron Configuration: 1s2 Effective Nuclear Charge: The 2 electrons in 1s experience a high Z_eff because there are no other electrons to shield them effectively. Hence, they feel a strong attraction to the nucleus.

For Al3?:

Nuclear Charge Z: 13 from 13 protons Electron Configuration: 1s2 2s2 2p6 Effective Nuclear Charge: The electrons in Al3? also experience a strong attraction but due to more protons, they are closer to the nucleus and experience a higher effective nuclear charge. The 10 inner electrons are effective at shielding the outer electrons.

Size and Electron Configuration

Both ions have the same electron configuration of 1s2 but the effective nuclear charge is significantly different. Li? has fewer protons, leading to a lower Z_eff compared to Al3?. As a result, the electrons in Li? are not pulled in as tightly as those in Al3? in the atomic structure.

Ionic Radius

The ionic radius of an ion is influenced by the number of protons. Since Al3? has more protons than Li?, it exerts a stronger attractive force on its electrons, resulting in a smaller ionic size. Therefore, Li? has a larger ionic radius compared to Al3?.

Conclusion

In summary, while both Li? and Al3? have the same electron configuration of 1s2, the greater effective nuclear charge in Al3? due to its larger number of protons leads to a stronger attraction on the electrons. This results in a smaller ionic radius compared to Li?. Consequently, Li? has a larger atomic size than Al3?.