Final answer:
The atomic radius of a fluorine atom is 64 pm, based on the internuclear distance in a diatomic fluorine molecule. When fluorine forms an anion (F-), its ionic radius increases to 119 pm due to increased electron repulsion. The ionic radius is different from the atomic radius and can vary significantly when atoms gain or lose electrons.
Step-by-step explanation:
The atomic and ionic radii are important concepts in chemistry to understand the size of atoms and ions. To measure the atomic radius, one typically looks at the distance between the nuclei of diatomic molecules. For example, the internuclear distance in a diatomic molecule of fluorine is 128 pm. So, the atomic radius of a fluorine atom is calculated by dividing this distance by two, yielding an atomic radius of 64 pm. However, this is different from the ionic radius, which is the size of an ion.
When a fluorine atom gains an electron to become a fluoride ion (F-), it increases in size due to increased electron-electron repulsion, which outweighs the nucleus's effective charge on the outer electrons. This results in an expansion of the electron probability region, and thus the ionic radius increases. The ionic radius of the fluoride ion is 119 pm, indicating that the addition of an electron has nearly doubled the size of the atom's radius.
Different ions have different sizes. For example, a calcium ion (Ca2+) typically has an ionic radius of around 126 pm. The similarity between the ionic radii of fluoride and calcium ions allows for them to form a certain type of crystalline structure.