Question

Write a Lewis structure for each of the following simple molecules. Show all bonding valence electron pairs as lines and all nonbonding valence electron pairs as dots. a. \(\mathrm{GeH}_{4}\) b. ICl c. \(\mathrm{NI}_{3}\) d. \(\mathrm{PF}_{3}\)

Short answer

The Lewis structures for the given molecules are as follows: a. GeH4: ``` H | H--Ge--H | H ``` b. ICl: ``` : : Cl:--I: : : : ``` c. NI3: ``` I : : : N---I---I : : : : ``` d. PF3: ``` :F: : : : : F---P---F : : ```

Step-by-step solution

a. GeH4

Step 1: Determining the valence electrons Germanium (Ge) has 4 valence electrons and Hydrogen (H) has 1 valence electron. There are 4 Hydrogen atoms, so in total, we have \(4 + 4 \times 1 = 8\) valence electrons. Step 2: Central atom and surrounding atoms Ge is the central atom and the Hydrogen atoms will be surrounding it. Step 3: Place bonds Place single bonds between Ge and each of the four H atoms. Step 4: Distribute remaining electrons There are no remaining electrons as all 8 of them are used in the 4 bonds. Step 5: Check the octet rule Ge has 4 bonds (8 electrons) and each H atom has a single bond (2 electrons). Thus, the octet rule is satisfied. The Lewis structure for GeH4 is: ``` H | H--Ge--H | H ```

b. ICl

Step 1: Determining the valence electrons Iodine (I) has 7 valence electrons and Chlorine (Cl) has 7 valence electrons. So, we have \(7 + 7 = 14\) valence electrons in total. Step 2: Central atom and surrounding atoms Iodine or Chlorine can be considered as the central atom since it is a diatomic molecule, but Iodine is less electronegative, so it will be the central atom. Step 3: Place bonds Place a single bond between Iodine (I) and Chlorine (Cl). Step 4: Distribute remaining electrons After placing the single bond, we have 12 electrons remaining. Let's distribute these electrons as lone pairs, starting with the terminal atom, Cl, and then I. Both Cl and I will have 3 lone pairs to complete their octet. Step 5: Check the octet rule Both I and Cl have a single bond (2 electrons) and 3 lone pairs (6 electrons), so they satisfy the octet rule. The Lewis structure for ICl is: ``` : : Cl:--I: : : : ```

c. NI3

Step 1: Determining the valence electrons Nitrogen (N) has 5 valence electrons and Iodine (I) has 7 valence electrons. There are 3 Iodine atoms, so in total we have \(5 + 3 \times 7 = 26\) valence electrons. Step 2: Central atom and surrounding atoms N is the central atom since there is only one N atom. Step 3: Place bonds Place single bonds between N and each of the three I atoms. Step 4: Distribute remaining electrons After placing the three single bonds, we have 20 electrons remaining. Distribute the electrons as lone pairs, starting with the terminal atoms. Each Iodine atom will have 3 lone pairs (apart from the single bond). Nitrogen already has a complete octet with its 3 bonds. Step 5: Check the octet rule N has 3 bonds (6 electrons) and each I atom has a single bond (2 electrons) and 3 lone pairs (6 electrons). The octet rule is satisfied. The Lewis structure for NI3 is: ``` I : : : N---I---I : : : : ```

d. PF3

Step 1: Determining the valence electrons Phosphorus (P) has 5 valence electrons and Fluorine (F) has 7 valence electrons. There are 3 Fluorine atoms, so we have a total of \(5 + 3 \times 7 = 26\) valence electrons. Step 2: Central atom and surrounding atoms P is the central atom since there is only one P atom. Step 3: Place bonds Place single bonds between P and each of the three F atoms. Step 4: Distribute remaining electrons After placing the three single bonds, we have 20 electrons remaining. Distribute the electrons as lone pairs, starting with the terminal atoms. Each F atom will have 3 lone pairs (apart from the single bond). Phosphorus already has a complete octet with its 3 bonds. Step 5: Check the octet rule P has 3 bonds (6 electrons) and each F atom has a single bond (2 electrons) and 3 lone pairs (6 electrons), so the octet rule is satisfied. The Lewis structure for PF3 is: ``` :F: : : : : F---P---F : : ```

Key concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom that are available for bonding with other atoms. These electrons play a pivotal role in chemical reactions and the formation of molecules. In the context of Lewis structures, valence electrons are depicted as dots or lines representing electron pairs.

For instance, in the solution for GeH4, the Germanium atom has 4 valence electrons, and each Hydrogen atom has 1. Visualizing and counting these electrons is the first crucial step in drawing a Lewis structure, allowing students to predict how the atoms will bond together. It's also essential to be aware that different elements have varying numbers of valence electrons, which determines their chemical properties and how they interact in compounds.

Understanding the distribution of valence electrons helps in building the molecular framework during Lewis structure formulation, with shared electron pairs drawn as lines (bonds) and non-bonding electrons as dots. This helps predict molecular shape and reactivity.

Octet Rule

The octet rule is a chemical rule of thumb that states atoms are most stable when they have eight electrons in their valence shell, resembling the electron configuration of a noble gas. This rule explains the tendency of atoms to bond in such a way that each atom has an 'octet' of electrons in its valence shell.

Looking at the step-by-step solutions for molecules such as GeH4 and ICl, we see that the central atoms, Ge and I respectively, follow the octet rule once the structure is complete. The exceptions to this rule, such as Hydrogen which is stable with two electrons (a 'duet'), are also depicted in the Lewis structures. While the octet rule is highly useful, it is worth noting that there are exceptions, particularly among elements in lower periods of the periodic table or those forming expanded octets. For students, mastering the octet rule is crucial in correctly drawing Lewis structures and understanding the stability and bonding of molecules.

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. The spatial configuration of molecules influences their chemical reactivity, polarity, color, biological activity, and more. Lewis structures help us predict molecular geometry by illustrating the locations of different atoms and non-bonding electron pairs.

For example, GeH4, with its four single bonds around the central Germanium atom, will adopt a tetrahedral geometry. Similarly, PF3, with three bonds around phosphorus and one lone pair, is expected to have a trigonal pyramidal shape. When drawing Lewis structures, we primarily focus on the two-dimensional aspect, but it's vital to extend this information to understand the molecule's real-world three-dimensional shape as determined by VSEPR (Valence Shell Electron Pair Repulsion) theory.

This understanding aids students significantly in visualizing molecules and their interactions, which is vital for subjects like organic chemistry, biochemistry, and molecular biology. It's always useful to connect the dots between Lewis structures, the octet rule, and molecular geometry to gain a well-rounded grasp of chemical bonding and molecule behavior.