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 as follows: (a) \(\mathrm{SeO}_{3}^{2-}\): The central atom is Se, surrounded by three O atoms. Se has a lone pair, and each O atom has two lone pairs, with single bonds between Se and O. (b) \(\mathrm{S}_{2}\mathrm{Cl}_{2}\): The structure is Cl-S-S-Cl, with single bonds between Cl and S atoms and between the two S atoms. Each Cl and S atom has three lone pairs. (c) chlorosulfonic acid, \(\mathrm{HSO}_{3}\mathrm{Cl}\): The central atom is S, surrounded by one H, one Cl, and three O atoms, all bonded to S. Cl and O atoms have three lone pairs, and H forms a single bond with S.

Step-by-step solution

Count the Valence Electrons

To begin drawing Lewis structures for each given compound, first, count the total number of valence electrons in the molecule/ion. Compound (a) \(\mathrm{SeO}_{3}^{2-}\): - Selenium (Se) has 6 valence electrons (it is in group 16). - Oxygen (O) has 6 valence electrons (it is in group 16). - The 2- charge indicates an additional 2 electrons. Total valence electrons: 6 (Se) + 3 * 6 (O) + 2 (charge) = 26 Compound (b) \(\mathrm{S}_{2}\mathrm{Cl}_{2}\): - Sulfur (S) has 6 valence electrons (it is in group 16). - Chlorine (Cl) has 7 valence electrons (it is in group 17). Total valence electrons: 2 * 6 (S) + 2 * 7 (Cl) = 26 Compound (c) chlorosulfonic acid, \(\mathrm{HSO}_{3}\mathrm{Cl}\): - Hydrogen (H) has 1 valence electron (it is in group 1). - Sulfur (S) has 6 valence electrons (it is in group 16). - Oxygen (O) has 6 valence electrons (it is in group 16). - Chlorine (Cl) has 7 valence electrons (it is in group 17). Total valence electrons: 1 (H) + 6 (S) + 3 * 6 (O) + 7 (Cl) = 32

Determine the Central Atom and Arrange Remaining Atoms

The central atom is typically the one with the lowest electronegativity or the one that can form the highest number of bonds. Compound (a) \(\mathrm{SeO}_{3}^{2-}\): - Central atom: Se - Arrangement: Se surrounded by three O atoms Compound (b) \(\mathrm{S}_{2}\mathrm{Cl}_{2}\): - Central atoms: both S atoms - Arrangement: Cl-S-S-Cl Compound (c) \(\mathrm{HSO}_{3}\mathrm{Cl}\): - Central atom: S - Arrangement: S surrounded by one H, three O atoms, and one Cl atom

Distribute Electrons and Complete Octets

Place electrons (in the form of bonds or lone pairs) around the atoms to fulfill the octet rule for all atoms except hydrogen, which only needs a duet (2 electrons). Compound (a) \(\mathrm{SeO}_{3}^{2-}\) (26 electrons): 1. Place single bonds between the central Se atom and the three O atoms. This uses 6 electrons (3 bonds * 2 electrons/bond), leaving 20 electrons. 2. Complete the octets for the three O atoms with lone pairs (6 electrons each). This uses 18 electrons (3 octets * 6 electrons/octet), leaving 2 electrons. 3. Place the remaining 2 electrons as a lone pair on the Se atom. Compound (b) \(\mathrm{S}_{2}\mathrm{Cl}_{2}\) (26 electrons): 1. Place single bonds between the Cl and S atoms and between the two S atoms. This uses 6 electrons (3 bonds * 2 electrons/bond), leaving 20 electrons. 2. Complete the octets for the Cl atoms with lone pairs (6 electrons each). This uses 12 electrons (2 octets * 6 electrons/octet), leaving 8 electrons. 3. Complete the octets for the S atoms with lone pairs (4 electrons each). This uses the remaining 8 electrons (2 octets * 4 electrons/octet). Compound (c) \(\mathrm{HSO}_{3}\mathrm{Cl}\) (32 electrons): 1. Place single bonds between H and S, S and Cl, and between S and the three O atoms. This uses 8 electrons (4 bonds * 2 electrons/bond), leaving 24 electrons. 2. Complete the octets for the Cl and three O atoms with lone pairs (6 electrons each). This uses 24 electrons (4 octets * 6 electrons/octet), leaving no electrons. 3. Hydrogen does not need an octet since it can only have a duet (2 electrons). The Lewis structures for compounds (a), (b), and (c) are now complete.

Key concepts

Valence Electrons

Valence electrons play a crucial role in the formation of chemical bonds. These are the outermost electrons of an atom and are involved in bonding with other atoms. Understanding how to count valence electrons is essential for drawing Lewis structures, which visually represent molecular structures.

• Selenium (Se) has 6 valence electrons because it is in group 16.
• Oxygen (O) also has 6 valence electrons and is in the same group as selenium.
• Sulfur (S), another group 16 element, has the same number of valence electrons.
• Chlorine (Cl) has 7 valence electrons as it is in group 17.
• Hydrogen (H), with only 1 valence electron, is an exception being in group 1.

Counting valence electrons allows us to predict how atoms will bond and comply with the octet rule. The total number of valence electrons provides the starting point for creating Lewis structures, as seen in compounds like a (Selenium oxide, a a a SO_3^{2-} a a ) having 26 electrons, or a SO_3^{2-} a a ) having 32 electrons. This also helps us determine the arrangement of atoms in the molecule.

Octet Rule

The octet rule is a fundamental concept in chemistry that helps predict and explain the bonding behavior of atoms. It states that atoms are most stable when they have eight electrons in their valence shell. This results from their strive to mimic the electron configuration of noble gases, which are inherently stable.
The octet rule guides how electrons are distributed around atoms in a molecule. For instance, when drawing Lewis structures for compounds:

• Each oxygen atom in a (SeO_3 ^{2-} a ) aims to have 8 electrons, forming bonds or using lone pairs.
• Sulfur and chlorine in a (HSO_3 Cl a ) also achieve complete octets, stabilizing the molecule.
• However, hydrogen is the exception, requiring only 2 electrons to be stable. This is sometimes referred to as the "duet rule."

Failing to satisfy the octet rule can lead to less stable molecules, which are less common in nature. By ensuring elements achieve a full octet (or duet in hydrogen's case), we can draw Lewis structures that accurately reflect stable chemical species.

Central Atom

The central atom is a pivotal concept in understanding how Lewis structures are constructed. Typically, the central atom is the one with the lowest electronegativity, which allows it to share electrons more easily. For a (SeO_3 ^{2-} a ), selenium (Se) acts as the central atom because it can form multiple bonds with the surrounding oxygen atoms. Similarly, in a (HSO_3 Cl a ), sulfur (S) takes the central position due to its ability to support multiple connections with hydrogen (H), oxygen, and chlorine (Cl).
Choosing the correct central atom is crucial because it affects the molecule's shape and electron distribution. For molecules like a (S_2 Cl_2 a ), both sulfur atoms act as central atoms, bonded directly to each other, which reflects their similar electronegativities. This understanding is essential for drawing accurate Lewis structures as it determines how other atoms are arranged around the central atom, influencing the overall geometry of the molecule.

Electron Distribution

Electron distribution in Lewis structures involves placing electrons to satisfy the octet rule for each atom. The distribution process requires that all valence electrons are accounted for, including those contributed by any charges on the molecule.
For example, in a (SeO_3 ^{2-} a ), the distribution needs to fulfill the octet for each oxygen atom and possibly the selenium by.

• Forming bonds and allocating lone pairs on oxygen.
• Placing any remaining electrons on the more substantial atoms like Se.

In the case of a (S_2 Cl_2 a ), the distribution focuses on ensuring both S and Cl atoms achieve full octets. Electrons are spread across single bonds, and the remaining electrons form lone pairs on Cl atoms first and then S.
Electron distribution determines the molecular shape and stability energetically. Properly understanding this allows us to predict molecular behavior effectively, as not all atoms will always form exactly eight electrons, particularly when resonance or expanded octets are involved.