Question

Draw the Lewis structure for a molecule of each of these compounds and ions. \(\begin{array}{ll}{\text { a. } \mathrm{H}_{2} \mathrm{S}} & {\text { c. } \mathrm{SO}_{2}} \\ {\text { b. } \mathrm{BF}_{4}-} & {\text { d. } \mathrm{SeCl}_{2}}\end{array}\)

Short answer

The Lewis structures for the given compounds and ions are as follows: a. H2S: \(\text{H}\overset {.}{\underset {.}{\text{S}}}\text{H}\) b. BF4-: \(\overset {.}{\underset {.}{\text{B}}}\overset {.}{\underset {.}{\text{F}}}[4\, \text{times}]\) c. SO2: \[ \begin{array}{{>{\displaystyle}l}}\underset{..}{\mathrm{O}}{}_{.}\mathrm{=}\overset{ }{\mathrm{S}}{}_{.}\mathrm{-}\underset{..}{\mathrm{O}}\\ {} \end{array} \] d. SeCl2: \(\text{Cl}\overset {.}{\underset {.}{\text{Se}}}\text{Cl}\)

Step-by-step solution

a. H2S

: 1. Find the total number of valence electrons: \(1 (\text{for H}) \times 2 + 6 (\text{for S}) = 8\) 2. Arrange atoms: H-S-H 3. Connect atoms with single bonds: H-S-H (4 valence electrons used) 4. Distribute remaining valence electrons as lone pairs: \(\text{H}\overset {.}{\underset {.}{\text{S}}}\text{H}\) (8 valence electrons used) 5. All atoms have a complete valence shell; no additional bonds are needed.

b. BF4-

: 1. Find the total number of valence electrons: \(5(\text{for B}) + 4 \times 7(\text{for F}) + 1(\text{extra electron for Charge}) = 32\) 2. Arrange atoms: B surrounded by 4 F atoms 3. Connect atoms with single bonds: B-F[4 times] (8 valence electrons used) 4. Distribute remaining valence electrons as lone pairs: \(\overset {.}{\underset {.}{\text{B}}}\overset {.}{\underset {.}{\text{F}}}[4\, \text{times}]\) (32 valence electrons used) 5. All atoms except B have a complete valence shell; no additional bonds are needed as boron can have an incomplete octet.

c. SO2

: 1. Find the total number of valence electrons: \(6 (\text{for S}) + 2 \times 6(\text{for O}) = 18\) 2. Arrange atoms: S with two oxygen atoms 3. Connect atoms with single bonds: S-O[2 times] (4 valence electrons used) 4. Distribute remaining valence electrons as lone pairs: \[ \begin{array}{{>{\displaystyle}l}}\underset{..}{\mathrm{O}}{}_{.}\mathrm{-}\overset{..}{\mathrm{S}}{}_{.}\mathrm{-}\underset{..}{\mathrm{O}} \\ {}\end{array} \] (18 valence electrons used) 5. Sulfur must be completing its octet by creating a double bond with one of the oxygen atoms: \[ \begin{array}{{>{\displaystyle}l}}\underset{..}{\mathrm{O}}{}_{.}\mathrm{=}\overset{ }{\mathrm{S}}{}_{.}\mathrm{-}\underset{..}{\mathrm{O}} \\ {} \end{array} \]

d. SeCl2

: 1. Find the total number of valence electrons: \(6 (\text{for Se}) + 2 \times 7(\text{for Cl}) = 20\) 2. Arrange atoms: Se with two chlorine atoms 3. Connect atoms with single bonds: Se-Cl[2 times] (4 valence electrons used) 4. Distribute remaining valence electrons as lone pairs: \(\text{Cl}\overset {.}{\underset {.}{\text{Se}}}\text{Cl}\) (20 valence electrons used) 5. All atoms have a complete valence shell; no additional bonds are needed.

Key concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom and are critical to understanding chemical bonding. These electrons have the highest energy level and are held least tightly to the nucleus, making them involved in bonding with other atoms. For example, hydrogen has one valence electron, sulfur has six, boron has three, fluorine has seven, and selenium has six. When drawing the Lewis structure, which visualizes the molecule's electron configuration, the first step is to calculate the number of valence electrons available. This will guide us in creating the correct arrangement of atoms and electrons.

As seen in the problem's solution, finding the total count allows us to determine the structure's foundation - like Hydrogen Sulfide (H₂S) having 8 valence electrons to be distributed. It's notable that some elements don't follow the octet rule strictly. For instance, boron often forms compounds where it has less than eight valence electrons, such as in BF₄^-, which is seen as an exception. Understanding valence electrons is essential for predicting how atoms will bond and react chemically.

Chemical Bonding

Chemical bonding is the process where atoms combine by sharing or transferring valence electrons to achieve greater stability. Bonds are often described in terms of Lewis structures, which show how electrons are distributed among atoms in a molecule. There are several types of chemical bonds: ionic, covalent, and metallic. In the examples provided, we focus on covalent bonding, where electrons are shared between atoms.

In molecules like H₂S and SeCl₂, atoms are held together by single covalent bonds. In BF₄^-, you find single covalent bonds but the molecule also carries a charge due to the presence of an extra valence electron. Meanwhile, sulfur dioxide (SO₂) includes a double bond to fulfill the octet rule, indicating a larger share of electrons between the sulfur and one oxygen atom. The details of shared electron pairs and how atoms bond together influence the physical properties and reactivity of the resulting molecules.

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of the atoms within a molecule, which can significantly affect its physical and chemical properties. The geometry is determined by the valence shell electron pair repulsion (VSEPR) theory, which states that electron pairs around a central atom will arrange themselves as far apart as possible to minimize repulsion.

In our exercises, the molecular geometries differ: H₂S is bent or angular due to two bonding pairs and two lone pairs on the sulfur, BF₄^- forms a tetrahedral shape with boron at the center, reflecting equal repulsion among the four bonds, SO₂ is also bent due to one sulfur with two bonding pairs and one lone pair, and SeCl₂, like H₂S, adopts a bent geometry due to two bonding and two lone pairs on selenium. Correctly predicting the molecular geometry is important for understanding how a molecule interacts with others, its polarity, and its overall reactivity.