\(\mathrm{F}\) and \(\mathrm{F}^{-}\) have identical core charges and the valence shell is \(\mathrm{n}=2\) in both cases. Explain why the radius of \(\mathrm{F}^{-}\) is larger than the radius of \(\mathrm{F}\).

Short Answer

Expert verified
The radius of F- is larger than that of F because the extra electron in F- increases the electron-electron repulsion, causing the electron cloud to expand and thus, the atomic radius to increase. The core charge (number of protons), which remains constant, does not increase its pull on the electrons to counteract this expansion.

Step by step solution

01

Establish Basic Concepts of Atomic Radius and Ionization

The atomic radius is a measure of the size of an atom -- usually the mean or typical distance from the center of the nucleus to the boundary of the electron cloud. Ionization, on the other hand, refers to the process when an atom loses or gains electrons, thus forming an ion.
02

Understanding the Structure of F and F-

The fluorine atom (F) has 9 protons and 9 electrons when it is neutral, meaning the valence shell (n = 2) has 7 electrons. Fluorine ion (F-), however, has gained an additional electron, making the total number of electrons now 10, still with the 9 protons.
03

Analyze the Effect of Additional Electron on Atomic Radius

An additional electron in F- increases the electron-electron repulsion in the atom. This repulsion in the electron cloud causes it to expand, subsequently leading to an increase in the atomic radius. Importantly, the core charge (number of protons) remains the same, so its pull on the electrons does not increase.

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