Question

Write the Lewis structure for each of the following species, and indicate the structure of each: (a) \(\mathrm{SeO}_{3}{ }^{2-} ;\) (b) \(\mathrm{S}_{2} \mathrm{Cl}_{2}\); (c) chlorosulfonic acid, \(\mathrm{HSO}_{3} \mathrm{Cl}\) (chlorine is bonded to sulfur).

Short answer

The Lewis structures for the given species are: (a) For \(\mathrm{SeO}_{3}{ }^{2-}\), the structure is: O || Se - O⁻ / O⁻ (b) For \(\mathrm{S}_{2} \mathrm{Cl}_{2}\), the structure is: Cl - S - S - Cl (c) For \(\mathrm{HSO}_{3} \mathrm{Cl}\), the structure is: O || H - S - O | . O | Cl

Step-by-step solution

(a) SeO3₂⁻: Step 1

Calculate total number of valence electrons. Se has 6, O has 6, and we have 3 O atoms. The ion also has a 2- charge, so we will add 2 more electrons. Total electrons = 6(Se) + 3*6(O) + 2 = 6 + 18 + 2 = 26 electrons.

(a) SeO3₂⁻: Step 2

Place the least electronegative atom at the center. In this case, Se will be at the center.

(a) SeO3₂⁻: Step 3

Create single bonds between the central Se atom and the three surrounding O atoms. This will consume 6 electrons (2 electrons per bond).

(a) SeO3₂⁻: Step 4

Distribute the remaining 20 electrons among the O atoms to complete their octet. Each O atom will receive 6 electrons (2 on each lone pair), and this will consume 18 electrons (6 electrons per O atom).

(a) SeO3₂⁻: Step 5

As the central Se atom is short by 2 electrons to complete its octet, we will remove a lone pair from one of the O atoms and form a double bond with the Se atom. The Lewis structure for \(\mathrm{SeO}_{3}{ }^{2-}\) is: O || Se - O⁻ / O⁻

(b) S₂Cl₂: Step 1

Calculate total number of valence electrons. S has 6, Cl has 7, and we have 2 S and 2 Cl atoms. Total electrons = 2 * 6(S) + 2 * 7(Cl) = 12 + 14 = 26 electrons.

(b) S₂Cl₂: Step 2

Place the least electronegative atoms at the center. In this case, two S atoms will be at the center.

(b) S₂Cl₂: Step 3

Create single bonds between the central S atoms and surrounding Cl atoms. This will consume 4 electrons (2 electrons per bond). Create a bond between both S atoms as well, which consumes 2 more electrons.

(b) S₂Cl₂: Step 4

Distribute the remaining 20 electrons among the Cl and S atoms to complete their octet. Each Cl atom will receive 6 electrons (3 lone pairs), and this will consume 12 electrons (6 electrons per Cl atom). As each S atom has 4 electrons, they should both receive 2 more electrons each.

(b) S₂Cl₂: Step 5

The octet rule is already satisfied, so no need to form double or triple bonds. The Lewis structure for \(\mathrm{S}_{2} \mathrm{Cl}_{2}\) is: Cl - S - S - Cl

(c) HSO₃Cl: Step 1

Calculate total number of valence electrons. H has 1, S has 6, O has 6, and we have 3 O atoms, and Cl has 7. Total electrons = 1(H) + 6(S) + 3*6(O) + 7(Cl) = 1 + 6 + 18 + 7 = 32 electrons.

(c) HSO₃Cl: Step 2

Place the least electronegative atom at the center. In this case, S will be at the center.

(c) HSO₃Cl: Step 3

Create single bonds between the central S atom and the surrounding H, three O atoms and Cl atom. This will consume 10 electrons (2 electrons per bond).

(c) HSO₃Cl: Step 4

Distribute the remaining 22 electrons among the O and Cl atoms to complete their octet. Each O atom will receive 6 electrons (2 on each lone pair), and this will consume 18 electrons (6 electrons per O atom). The Cl atom is already 7 (as it has one bond), and by adding one electron to it, we now used all 32 electrons.

(c) HSO₃Cl: Step 5

The octet rule is already satisfied, so no need to form double or triple bonds. The Lewis structure for \(\mathrm{HSO}_{3} \mathrm{Cl}\) is: O || H - S - O | . O | Cl

Key concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom and are critical to understanding how elements bond in molecules. They are responsible for the chemical properties of the element, as they are involved in the formation of chemical bonds. In the context of the exercise, calculating the total number of valence electrons for a species is the first step in drawing the Lewis structure.

For example, in the case of \(\mathrm{SeO}_{3}{ }^{2-}\), selenium (Se) has 6 valence electrons, each oxygen (O) atom has 6, and due to the 2- charge, we add 2 more electrons, tallying up to 26 valence electrons. Similarly, for \(\mathrm{S}_{2} \mathrm{Cl}_{2}\), sulfur (S) has 6 and chlorine (Cl) has 7 valence electrons. When you are counting valence electrons, it's crucial to remember that you must count all atoms present and any additional charges to ensure accuracy in the subsequent steps of drawing the Lewis structure.

Octet Rule

The octet rule is a fundamental concept used to predict the arrangement of electrons in molecules. It states that atoms tend to form bonds in such a way that each atom has eight electrons in its valence shell, giving it the same electron configuration as a noble gas. However, there are exceptions, and some elements are stable with fewer or more than eight electrons.

Applying the octet rule to the Lewis structures as in the exercise, we distributed remaining electrons after forming bonds to ensure each atom reaches an octet. For instance, in \(\mathrm{SeO}_{3}{ }^{2-}\), after forming single bonds, additional electrons were distributed to oxygen atoms to fulfill their octet, leaving selenium short of two, which led to the formation of a double bond. With \(\mathrm{HSO}_{3} \mathrm{Cl}\), each oxygen and the chlorine atom received electrons to complete their octet. Understanding this principle is crucial since the satisfaction of the octet rule helps stabilize the molecule.

Electronegativity

Electronegativity refers to the ability of an atom to attract and hold onto electrons within a chemical bond. It's a vital concept in determining how atoms bond and the structure of a molecule. In Lewis structures, atoms with higher electronegativity are typically placed on the outside, with less electronegative atoms occupying the center position.

In the provided solutions, for each species, the least electronegative atom is chosen as the central atom. For \(\mathrm{SeO}_{3}{ }^{2-}\), Se is less electronegative than O, so it becomes the central atom. With \(\mathrm{S}_{2} \mathrm{Cl}_{2}\), the less electronegative sulfur occupies the center, while in \(\mathrm{HSO}_{3} \mathrm{Cl}\), the sulfur again takes the central position, being less electronegative than both oxygen and chlorine. It's essential to comprehend electronegativity as it directly influences the molecular structure and properties, such as polarity and bond angles, in chemical compounds.