The triiodide ion \(\left(\mathrm{I}_{3}^{-}\right)\) in which the I atoms are arranged in a straight line is stable, but the corresponding \(\mathrm{F}_{3}^{-}\) ion does not exist. Explain.

Short Answer

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The \(I_3^-\) ion exists due to the larger size of iodine atoms which allows for the formation of weak inter-atomic iodine-iodine bonds. However, the \(F_3^-\) ion does not exist because the small size of fluorine atoms causes strong electron-electron repulsion, destabilizing the ion.

Step by step solution

01

Understand the nature of Iodine and Fluorine Atoms

Firstly, iodine atoms are much larger than fluorine atoms. Larger atoms have more electron shells, so their outer electrons are less tightly held and can be more easily involved in bond formation. The larger size of the iodine atom allows the formation of weak iodine-iodine bonds.
02

Discuss the Electron Repulsion

In a \(F_3^-\) ion, due to small size of fluorine atoms, the outer shell electrons of the three fluorine atoms would be very close. This closeness would result in very strong repulsion among the outer shell electrons of the three fluorine atoms. This high electron-electron repulsion would make the \(F_3^-\) ion unstable.
03

Combine the Understanding of Both Factors

Because iodine atoms are larger than fluorine atoms, their outer shell electrons are less tightly held. This allows for the formation of weak iodine-iodine bonds, allowing the \(I_3^-\) ion to exist. However, in the case of the \(F_3^-\) ion, the small size of the fluorine atoms combined with strong repulsion among outer shell electrons makes this ion unstable and therefore it doesn't exist.

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