Problem 87 - Chemistry: The Central Science

Question

Although \(\mathrm{I}_{3}^{-}\) is known, \(\mathrm{F}_{3}^{-}\) is not. Using Lewis structures, explain why \(\mathrm{F}_{3}^{-}\) does not form.

Step-by-Step Solution

Answer

The F−3 ion does not form because it would violate the octet rule, due to the lack of available orbitals in fluorine's 2p valence shell to accommodate an expanded octet. On the other hand, the I−3 ion is stable and does not violate the octet rule as its larger size allows it to accommodate an expanded octet via its d-orbitals.

Step 1: Understanding Lewis Structures

A Lewis structure is a representation of a molecule or an ion showing the arrangement of electrons around atoms, including the lone pairs and bond pairs of electrons. In a Lewis structure, each atom is represented by its element symbol and is surrounded by the electron pairs (which can be shown as dots or lines).

Step 2: Drawing the Lewis Structure for I−3 Ion

The I−3 ion consists of three iodine (I) atoms forming a singly bonded linear structure with two extra electrons on each iodine atom (as iodine has a total of 7 electrons). In I−3 ion, the central iodine atom shares one electron each with the two adjacent iodine atoms, forming single bonds. Each iodine atom now has 2 bond pairs and 3 lone pairs (3 pairs of unshared electrons). The Lewis structure of I−3 ion can be drawn as: I − I − I Where each dash (-) represents a bond pair of electrons.

Step 3: Drawing the Lewis Structure for F−3 Ion (Hypothetically)

Let's now try to draw a Lewis structure for the hypothetical F−3 ion, which consists of three fluorine (F) atoms. Fluorine has a total of 7 electrons, which can form one bond by sharing one electron. For the F−3 ion to exist, it would have three fluorine atoms sharing 1 electron each to form single bonds, and the rest of the electrons would be lone pairs. But this arrangement would lead to 10 electrons around a central fluorine atom, giving each fluorine atom an expanded octet, which is not possible due to the lack of available orbitals in fluorine's 2p valence shell.

Step 4: Comparing the Stability of I−3 and F−3 Ions

The I−3 ion is stable due to the presence of single bonds and no violation of the octet rule. Iodine atoms can extend their octet and form larger structures due to d-orbitals, which are needed for accommodating additional electrons. On the other hand, the hypothetical F−3 ion has an unstable structure, as the central fluorine atom would have to accommodate 10 electrons, violating the octet rule. In reality, such a structure does not exist because fluorine in the 2p valence shell lacks any vacant d-orbitals that are needed for an expanded octet. Thus F−3 ion does not form due to the lack of orbitals to accommodate the expanded octet. In summary, F−3 ion does not form due to a violation of the octet rule caused by the lack of available orbitals in fluorine's 2p valence shell. Whereas the I−3 ion is stable and does not violate the octet rule as its larger size allows it to accommodate an expanded octet.