Question

Although \(\mathrm{I}_{3}\) is a known ion, \(\mathrm{F}_{3}^{-}\) is not. (a) Draw the Lewis structure for \(\mathrm{I}_{3}^{-}\) (it is linear, not a triangle). (b) One of your classmates says that \(\mathrm{F}_{3}^{-}\) does not exist because \(\mathrm{Fis}\) too electronegative to make bonds with another atom. Give an example that proves your classmate is wrong. (c) Another classmate says \(\mathrm{F}_{3}^{-}\) does not exist because it would violate the octet rule. Is this classmate possibly correct? (d) Yet another classmate says \(\mathrm{F}_{3}^{-}\) does not exist because \(\mathrm{F}\) is too small to make bonds to more than one atom. Is this classmate possibly correct?

Short answer

The Lewis structure for \(I_{3}^{-}\) has a linear arrangement of three Iodine atoms, with three lone pairs on the central atom and two lone pairs on each outer atom. The non-existence of \(F_{3}^{-}\) is likely due to the violation of the octet rule and the small size of the Fluorine atom, making it difficult to bond with more than one atom. However, Fluorine's high electronegativity alone is not a sufficient reason, as it does form bonds in compounds like HF and PF6-.

Step-by-step solution

To draw the Lewis structure for I3-, first, arrange the three Iodine (I) atoms linearly because the question states that I3- is linear. Each Iodine atom has 7 valence electrons. Since the ion carries a negative charge, add one extra electron to the total, making it a total of 22 valence electrons to distribute. Place a single bond between the central atom and the two Iodine atoms on each side. Use the remaining electrons to complete the octets of all three Iodine atoms by placing lone pairs on them. This will give you three lone pairs on the central atom and two lone pairs on each outer atom. #Step 2: Discuss F3- ion existence or not using electronegativity#

One of the claims made by a classmate is that F3- ion does not exist because Fluorine (F) is too electronegative to make bonds with another atom. This is not entirely true, as Fluorine does form bonds with other atoms, often in the form of binary compounds like HF or in polyatomic anions such as PF6-. Thus, this claim cannot be the sole reason for the non-existence of F3- ion. #Step 3: Analyze F3- ion in terms of the octet rule#

Another classmate claims that F3- ion does not exist because it would violate the octet rule. If we attempt to draw a Lewis structure for F3-, we will find that Fluorine being the central atom, forms bonds with the other two Fluorine atoms. But, since each Fluorine atom already has 7 valence electrons and only needs one to complete its octet, it is impossible for the central Fluorine atom to form two bonds with two other Fluorine atoms without exceeding its octet. Therefore, this claim could be a valid reason for the non-existence of the F3- ion. #Step 4: Analyze F3- ion with respect to the size of Fluorine atom#

Another classmate claims that F3- ion does not exist because Fluorine is too small to make bonds with more than one atom. Small atoms like Fluorine, having high electron density in their outermost shell, may not be able to form multiple covalent bonds due to the repulsion between their negatively charged electron clouds (lone pairs). Thus, this claim can also be a possible factor contributing to the non-existence of F3- ion. In conclusion, the reasons for the non-existence of F3- ion could be the violation of the octet rule and the small size of the Fluorine atom which causes difficulty in bonding with more than one atom. The high electronegativity of Fluorine does not automatically disqualify it from forming bonds with other atoms, as evidenced by compounds such as HF and PF6-.

Key concepts

Octet Rule

The octet rule is a fundamental principle in chemistry that helps us understand how atoms bond. It states that atoms are most stable when they have eight electrons in their outer valence shell, resembling the electron configuration of noble gases.
This rule is particularly important when drawing Lewis structures, as it guides us in allotting the correct number of electrons around each atom.
In the case of the \(\mathrm{I}_{3}^{-}\) ion mentioned in the exercise, each iodine atom aims to complete its octet. Iodine naturally has seven valence electrons, and when it forms bonds, these interactions aim to satisfy the octet rule.
Conversely, for the hypothetical \(\mathrm{F}_{3}^{-}\) ion, the central fluorine would need to form bonds that extend beyond its usual octet, creating what is termed a hypervalent molecule, which isn't typical for small atoms like fluorine.
Understanding the octet rule helps in predicting why certain compounds like \(\mathrm{F}_{3}^{-}\) are unstable. Fluorine's inability to exceed the octet makes the formation of such a structure less feasible.

Electronegativity

Electronegativity is a measure of an atom's ability to attract and hold onto electrons within a bond. It plays a critical role in determining the type of bond that will form between atoms.
The electronegativity scale, with fluorine as the most electronegative element, explains why it forms strong bonds when it does so, often forming polar covalent bonds.
In the original exercise, a classmate suggests that the \(\mathrm{F}_{3}^{-}\) ion does not exist due to the high electronegativity of fluorine. While it's true that fluorine strongly attracts electrons, resulting in the formation of strong bonds in molecules like HF or the hexafluorophosphate ion (\(\mathrm{PF}_{6}^{-}\)), it does not automatically mean \(\mathrm{F}_{3}^{-}\) can't form.
The issue arises more from the structure not complying with the octet rule, and less so from electronegativity alone. Therefore, although electronegativity is a significant factor in bonding, it is not the only reason for the non-existence of a specific ion, as higher electronegativity doesn't prevent bond formation by itself.

Covalent Bonding

Covalent bonding occurs when atoms share electrons to achieve stability. This sharing can happen between similar or different atoms, and the nature of this sharing determines the bond's characteristics.
In the case of the exercise, the \(\mathrm{I}_{3}^{-}\) ion is a good example of covalent bonding. The iodine atoms share electrons to complete their octets, resulting in a stable linear triatomic ion.
For \(\mathrm{F}_{3}^{-}\), covalent bonding becomes complicated due to fluorine's inability to comfortably share more electrons than needed to satisfy the octet rule. Unlike iodine, which can share its electrons more freely to participate in multi-atom covalent bonding, fluorine's small size and electron configuration inhibit such bonding.
Overall, covalent bonding is a versatile way for atoms to adhere to the octet rule by sharing electrons. However, certain atoms like fluorine have limitations based on size and electron configuration that make specific structures like \(\mathrm{F}_{3}^{-}\) less feasible.