Question

Although \(\mathrm{I}_{3}^{-}\)is known, \(\mathrm{F}_{3}^{-}\)is not. Which statement is the most correct explanation? (a) Iodine is more likely to be electron-deficient. (b) Fluorine is too small to accommodate three nonbonding electron pairs and two bonding electron pairs. (c) Fluorine is too electronegative to form anions. (d) \(\mathrm{I}_{2}\) is known but \(\mathrm{F}_{2}\) is not. (e) Iodine has a larger electron affinity than fluorine.

Short answer

The most correct explanation for the fact that \(\mathrm{I}_{3}^{-}\) is known while \(\mathrm{F}_{3}^{-}\) is not is statement (b): Fluorine is too small to accommodate three nonbonding electron pairs and two bonding electron pairs.

Step-by-step solution

Analyze statement (a)

Iodine is more likely to be electron-deficient. It is true that iodine is less electronegative than fluorine, and may be more likely to accept additional electrons. Since we are considering the ions where these elements gain electrons, this option might be relevant. However, we can't conclude yet if it's the most correct explanation. Let's analyze the other statements.

Analyze statement (b)

Fluorine is too small to accommodate three nonbonding electron pairs and two bonding electron pairs. Fluorine is indeed smaller than iodine, so this option may have some relevance in terms of size effects on the formation of anions. The larger size of iodine might provide more space for the electron pairs to disperse, which could make the formation of \(\mathrm{I}_{3}^{-}\) more favorable. This seems to be a potentially relevant explanation.

Analyze statement (c)

Fluorine is too electronegative to form anions. Fluorine is the most electronegative element, which means it has a higher affinity for electrons than all other elements. This implies that fluorine actually forms anions quite readily, so this statement is not a correct explanation for the lack of \(\mathrm{F}_{3}^{-}\). We can rule out this option.

Analyze statement (d)

\(\mathrm{I}_{2}\) is known but \(\mathrm{F}_{2}\) is not. This statement is incorrect, as both \(\mathrm{I}_{2}\) and \(\mathrm{F}_{2}\) are known diatomic molecules. This option does not provide an explanation for why \(\mathrm{I}_{3}^{-}\) is known and \(\mathrm{F}_{3}^{-}\) is not. We can rule out this option.

Analyze statement (e)

Iodine has a larger electron affinity than fluorine. This statement is incorrect. Fluorine has the largest electron affinity among all elements, meaning it is more favorable for fluorine to gain an electron than iodine. This option doesn't provide an explanation for the given facts. We can rule out this option.

Identify the most correct explanation

Having analyzed all five statements, the most correct explanation for the fact that \(\mathrm{I}_{3}^{-}\) is known while \(\mathrm{F}_{3}^{-}\) is not is statement (b): Fluorine is too small to accommodate three nonbonding electron pairs and two bonding electron pairs. The smaller size of fluorine compared to iodine can cause repulsion between electron pairs, making the formation of \(\mathrm{F}_{3}^{-}\) less favorable.

Key concepts

Chemical Bonding

Chemical bonding is the process where atoms combine to form molecules. This fundamental action is what allows the formation of the diverse array of substances we see around us. There are several types of chemical bonds, including:

• Covalent Bonds: Atoms share electrons to fill their outer shells.
• Ionic Bonds: Electrons are transferred from one atom to another, creating oppositely charged ions that attract each other.
• Metallic Bonds: Electrons are shared across a lattice of metal atoms.

When we look at why certain molecules form, like \( I_3^- \) compared to \( F_3^- \), we're seeing how the rules of chemical bonding play out in real world scenarios.
In the case of \( I_3^- \), iodine can stretch bonds and handle electron repulsion well due to its larger size. Fluorine, on the other hand, is much smaller. This small size makes it difficult to accommodate extra bonding, especially when multiple electrons and bonds are involved. The repulsion between bonding and non-bonding electrons in \( F_3^- \) would lead to instability, preventing this molecule from forming.

Electron Pair Repulsion

Electron pair repulsion is a core concept in understanding molecular geometry. It revolves around the idea that electron pairs around an atom will position themselves as far apart as possible to minimize repulsion. This principle is a cornerstone of the VSEPR (Valence Shell Electron Pair Repulsion) theory. Here's how it works:

• Lone Pairs: Non-bonding pairs that occupy space and create repulsion forces.
• Bonding Pairs: Shared pairs between atoms that also repel each other.

In much the same way, when analyzing molecular combinations like \( I_3^- \) and \( F_3^- \), you consider electron pair repulsion. For \( I_3^- \), the larger size of iodine allows the electron pairs to be arranged without strong repulsive forces pushing them apart excessively. In contrast, the compact structure of fluorine means there's insufficient space for electron pairs to reduce repulsion adequately. As a result, any attempt to form \( F_3^- \) would result in significant electron pair repulsion, making the molecule unstable and unfeasible.

Electronegativity

Electronegativity is the measure of an atom's ability to attract and hold onto electrons in a chemical bond. It's a crucial concept in predicting the behavior of atoms in molecules. Generally, the higher the electronegativity, the stronger an atom will pull electrons towards itself in a bond. Key aspects of this include:

• Periodic Trend: Electronegativity tends to increase across a period and decrease down a group in the periodic table.
• Comparison of Elements: Fluorine, being the most electronegative element, has the greatest ability to attract electrons, while iodine has a much lower electronegativity.

In the context of \( I_3^- \) and \( F_3^- \), electronegativity differences play a role in explaining why one exists and the other does not. Fluorine's high electronegativity means it is less inclined to form bonds that could lead to an overall charge imbalance, like in \( F_3^- \). This is because it wants to retain its electrons rather than share or extend its electron cloud excessively. Meanwhile, iodine's lower electronegativity and larger size mean it's more willing and able to accommodate extra electrons, forming stable anions like \( I_3^- \).