Question

Write Lewis structures for each molecule or ion. Use expanded octets as necessary. \begin{equation}\mathrm{ a. }{ClF}_{5} \quad \text { b. } \mathrm{AsF}_{6}-\quad \text { c. } \mathrm{Cl}_{3} \mathrm{PO} \quad \text { d. IF }_{5}\end{equation}

Short answer

ClF5: Central Cl atom with 5 single bonds to F, expanded octet with 2 lone pairs. AsF6-: Central As atom with 6 bonds to F, one extra electron for charge. Cl3PO: Central P atom with a double bond to O, three single bonds to Cl. IF5: Central I atom with 5 single bonds to F, expanded octet with one lone pair.

Step-by-step solution

Lewis Structure for ClF5

For ClF5, start by determining the total number of valence electrons. Chlorine has 7 and each fluorine has 7, so the total is 7 + 5*7 = 42. Place Cl in the center and arrange 5 F atoms around it. Give each F atom a single bond with Cl and complete the octets for F atoms. Chlorine can have an expanded octet; add the remaining electrons to Cl to fulfill the expanded octet rule.

Lewis Structure for AsF6-

For AsF6-, start by calculating the total number of valence electrons. Arsenic has 5, each fluorine has 7, and there is an extra electron due to the negative charge, making a total of 5 + 6*7 + 1 = 48. Place As in the center and arrange 6 F atoms around. Give each F atom a single bond with As and complete the octets for F atoms. As is the central atom and can have an expanded octet; assign the remaining electrons to complete the octet expansion.

Lewis Structure for Cl3PO

For Cl3PO, start with the total count of valence electrons: phosphorus has 5, oxygen has 6, and each chlorine has 7, giving a total of 5 + 6 + 3*7 = 32. Place P in the center, O and the 3 Cl atoms around it. Connect P with a double bond to O and single bonds to Cl atoms. Complete the octets for O and Cl atoms. Add any leftover electrons to Cl or P to satisfy the octet rule (expanded for P if necessary).

Lewis Structure for IF5

For IF5, begin by finding the total number of valence electrons: iodine has 7 and each fluorine has 7, hence the total is 7 + 5*7 = 42. Place I in the center with 5 F atoms around. Connect I with a single bond to each F atom and fill the octets for F atoms. Since I can have an expanded octet, place any remaining electrons on it to fulfill the expansion.

Key concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom and are instrumental in chemical bonding. Understanding the role of valence electrons is the first step in predicting how atoms will interact with each other to form molecules. For instance, in our exercise with the molecule ClF₅, chlorine (Cl) has seven valence electrons, and fluorine (F) also has seven. To draft a Lewis structure, it’s essential to calculate the total number of valence electrons; here, ClF₅ has 42.
Valence electrons are represented by dots surrounding the atomic symbols in the Lewis structure. These electrons can participate in the forming of bonds with adjacent atoms or can be non-bonding. When two atoms share valence electrons, a covalent bond is formed. In our solutions, we distributed these electrons to form single bonds between chlorine and fluorine, and then used the remaining valence electrons to complete the octets. Notably, these valence electrons do more than just form bonds; they establish the reactivity, phase, and physical properties of the substance.

Expanded Octets

While the octet rule states that atoms tend to form bonds until they are surrounded by eight electrons, certain elements in the third periodic row and beyond can exceed this boundary. This ability to have more than eight electrons around the central atom is known as an 'expanded octet.'
Elements such as phosphorus (P), sulfur (S), chlorine (Cl), and others can host more than eight valence electrons, typically due to their empty d-orbitals which can be used to accommodate extra electrons. Take AsF₆^- and IF₅ in our exercise for example — both arsenic (As) and iodine (I) extend beyond the octet rule, holding more than eight electrons. This is shown in the Lewis structures by adding lone pairs of electrons until all are accounted for. The ability to have expanded octets is crucial when drawing accurate Lewis structures for these types of molecules and offers insight into the geometry and bonding behavior of such compounds.

Chemical Bonding

Chemical bonding describes the force that holds atoms together in compounds. The Lewis structures employed in our solutions illustrate the bonds between atoms—lines represent shared electron pairs forming covalent bonds. Chemical bonds are crucial for understanding molecule stability and how compounds interact in chemical reactions.
In a molecule like ClF₅, each single line between Cl and F indicates a pair of electrons shared between the two atoms, making five single bonds and fulfilling the valence requirement of the fluorine atoms. Meanwhile, elements that can have an expanded octet, like those shown in our exercise—ClF₅, AsF₆^-, Cl₃PO, and IF₅—form additional bonds or retain lone pairs to accommodate the extra electrons. Each bond type, be it single, double, or triple, is tied closely to the theory of valence shell electron pair repulsion (VSEPR), which can predict molecular shapes. By understanding chemical bonding, students can gain insights into the physical properties of substances, such as melting points, boiling points, and reactivity.