Problem 102 - Chemistry Matter and Change

Question

Draw the Lewis structures for these molecules, each of which has a central atom that does not obey the octet rule. \(\begin{array}{ll}{\text { a. } P C l_{5}} & {\text { c. } C l F_{5}} \\\ {\text { b. BF }_{3}} & {\text { d. BeH }_{2}}\end{array}\)

Step-by-Step Solution

Answer

The Lewis structures for the given molecules that do not obey the octet rule are: 1. \(PCl_5\): Cl | Cl - P - Cl | Cl 2. \(BF_3\): F | F - B - F 3. \(ClF_5\): F | F - Cl - F - F | F 4. \(BeH_2\): H - Be - H

Step 1: Draw the Lewis structure for \(PCl_5\)

First, find the total number of valence electrons for \(PCl_5\). Phosphorus (P) has 5 valence electrons, and Chlorine (Cl) has 7 valence electrons. Since there are 5 Chlorine atoms, the total valence electrons are \(5 + 7(5) = 40\). Place the central atom (P) in the middle and arrange the surrounding atoms (Cl) around it. Then, place the single bonds between the central atom and the surrounding atoms and distribute the remaining valence electrons in pairs: Cl | Cl - P - Cl | Cl

Step 2: Draw the Lewis structure for \(BF_3\)

Calculate the total number of valence electrons for \(BF_3\). Boron (B) has 3 valence electrons, and Fluorine (F) has 7 valence electrons. Since there are 3 Fluorine atoms, total valence electrons are \(3 + 7(3) = 24\). Place the central atom (B) in the middle and arrange the surrounding atoms (F) around it. Then, place the single bonds between the central atom and the surrounding atoms: F | F - B - F

Step 3: Draw the Lewis structure for \(ClF_5\)

Determine the total number of valence electrons for \(ClF_5\). Chlorine (Cl) has 7 valence electrons, and Fluorine (F) has 7 valence electrons. Since there are 5 Fluorine atoms, total valence electrons are \(7 + 7(5) = 42\). Place the central atom (Cl) in the middle and arrange the surrounding atoms (F) around it. Then, place the single bonds between the central atom and the surrounding atoms and distribute the remaining valence electrons in pairs: F | F - Cl - F - F | F

Step 4: Draw the Lewis structure for \(BeH_2\)

Calculate the total number of valence electrons for \(BeH_2\). Beryllium (Be) has 2 valence electrons, and Hydrogen (H) has 1 valence electron. Since there are 2 Hydrogen atoms, total valence electrons are \(2 + 1(2) = 4\). Place the central atom (Be) in the middle and arrange the surrounding atoms (H) around it. Then, place the single bonds between the central atom and the surrounding atoms: H - Be - H The resulting Lewis structures for these molecules are as follows: 1. \(PCl_5\): Cl | Cl - P - Cl | Cl 2. \(BF_3\): F | F - B - F 3. \(ClF_5\): F | F - Cl - F - F | F 4. \(BeH_2\): H - Be - H

Key Concepts

Octet Rule Exceptions

Understanding the octet rule is fundamental when studying chemical bonding, but equally important is recognizing its exceptions. The octet rule states that atoms are at their most stable when surrounded by eight electrons in their valence shell. Yet, certain elements frequently break this rule. One class of exceptions includes elements such as boron and beryllium, which are perfectly stable with fewer than eight electrons. In the case of BF₃ and BeH₂, Boron and Beryllium, respectively, have fewer than eight electrons; Boron has six, while Beryllium has four.

Another notable exception involves molecules with elements from the third period or below on the periodic table, such as phosphorus in PCl₅ and sulfur in ClF₅. These elements can have more than eight electrons because they have access to an additional orbital (d-orbital) allowing them to accommodate more electrons. Understanding these exceptions is crucial when predicting molecule structure and reactivity.

Valence Electrons

Valence electrons are the electrons in the outermost shell of an atom and play a pivotal role in chemical bonding and molecular structure. These electrons determine how an atom will interact with others to form compounds. For instance, when creating Lewis structures, it is vital to count the number of valence electrons to predict molecular bonding patterns correctly.

To illustrate, using the molecules given in the exercise, Phosphorus has five valence electrons, while Chlorine has seven. By understanding the count of valence electrons, one can distribute them to form bonds and lone pairs properly, which is a critical step in creating an accurate Lewis structure.

Counting Valence Electrons

Simply refer to the group number for main-group elements. Phosphorus, for instance, belongs to group 15 and thus has five valence electrons. Chlorine, in group 17, has seven valence electrons. Hydrogen, despite being in group 1, has only one valence electron it can share.

Molecular Geometry

Once the Lewis structures are drawn using the appropriate amount of valence electrons, the next step is understanding the molecular geometry—the three-dimensional arrangement of atoms around a central atom in a molecule. Even if elements violate the octet rule, they still adopt specific shapes that minimize repulsion between electrons.

PCl₅ and ClF₅ will have an expanded octet and exhibit trigonal bipyramidal and square pyramidal geometries, respectively. Conversely, molecules like BF₃, which obey the octet rule but have less than eight electrons around the central atom, will typically form a trigonal planar shape to minimize repulsion. In the case of BeH₂, with only two attachments to the central atom, the molecule adopts a linear geometry.

Angular Arrangements

For each molecule, the angles between the bonds are determined by the repulsion between the valence electrons. In a model like BF₃, for example, the 120-degree angles are a result of the electrons arranging themselves equally around the central boron atom to be as far apart as possible.