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Chapter 9: Problem 92

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

Short Answer

Expert verified
The non-existence of F3- and the stability of I3- can be explained based on atomic size, electronegativity, and lone pair repulsion. Iodine atoms are larger with lesser electronegativity and lesser lone pair repulsion than Fluorine, leading to the formation of a stable I3- ion. Conversely, due to its significantly smaller atomic size, higher electronegativity, and greater lone pair repulsion, a corresponding F3- ion does not exist.

Step by step solution

01

Consider the Atomic Size

Examine the trends in atomic size in the periodic table. One can observe that iodine (I) is significantly larger than fluorine (F). The larger size of the Iodine atom allows the formation of more than one bond, while the smaller Fluorine atom does not permit the existence of multiatomic anions.
02

Consider the Electronegativity

Fluorine is the most electronegative element in the periodic table, meaning it holds on to its electrons very tightly. As a result, it is reluctant to share electrons with another fluorine atom for them to form F3- ion. Contrarily, Iodine is less electronegative, and thus forms the I3- anion easily.
03

Understand the Role of Lone Pair Repulsion

Due to the smaller size of Fluorine, there is a high electron density, causing large repulsions between the fluorine lone pairs. This makes the structure very unstable compared to I3-, where the repulsions between lone pairs are smaller due to the larger size of the Iodine atom. This makes I3- more stable compared to F3- being non-existent.

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