Question

In the Lewis structure shown below, \(\mathrm{A}, \mathrm{D}, \mathrm{E}, \mathrm{Q}, \mathrm{X}\), and \(\mathrm{Z}\) represent elements in the first two rows of the periodic table (H-Ne). Identify all six elements so that the formal charges of all atoms are zero. [Section 8.3]

Short answer

The elements in the given Lewis structure that have zero formal charges are: A - Hydrogen (H) D - Oxygen (O) E - Carbon (C) Q - Fluorine (F) X - Fluorine (F) Z - Oxygen (O)

Step-by-step solution

Examine the Lewis structure

Let's first examine the given Lewis structure: A | D - E - Q - X | Z Now, we will analyze the number of bonds and lone pairs for each element in the structure.

Calculate the number of valence electrons for each element

Let's denote the number of valence electrons for each element as follows: A - Valence electrons (A), Bonds (1), Lone pairs (0) D - Valence electrons (D), Bonds (1), Lone pairs (2) E - Valence electrons (E), Bonds (4), Lone pairs (0) Q - Valence electrons (Q), Bonds (1), Lone pairs (3) X - Valence electrons (X), Bonds (1), Lone pairs (3) Z - Valence electrons (Z), Bonds (1), Lone pairs (2) In order to have a formal charge of zero for all atoms, the number of their valence electrons should be equal to the sum of half the electrons in the bond plus the number of lone pair electrons.

Match the elements to their correct valence electrons and structure

Using the information from the periodic table (first two rows: H-Ne), we can match the elements: A - Hydrogen (H), Valence electrons: 1 D - Oxygen (O), Valence electrons: 6 E - Carbon (C), Valence electrons: 4 Q - Fluorine (F), Valence electrons: 7 X - Fluorine (F), Valence electrons: 7 Z - Oxygen (O), Valence electrons: 6 Now, let's check if the formal charges are zero for all atoms: A (H): 1 = 1/2(2) + 0 => Formal charge = 0 D (O): 6 = 1/2(2) + 4 => Formal charge = 0 E (C): 4 = 1/2(8) + 0 => Formal charge = 0 Q (F): 7 = 1/2(2) + 6 => Formal charge = 0 X (F): 7 = 1/2(2) + 6 => Formal charge = 0 Z (O): 6 = 1/2(2) + 4 => Formal charge = 0 Indeed, the formal charges of all atoms are zero. So, the identified elements are: A - Hydrogen (H) D - Oxygen (O) E - Carbon (C) Q - Fluorine (F) X - Fluorine (F) Z - Oxygen (O)

Key concepts

Formal Charge

In a Lewis structure, understanding formal charge is crucial for evaluating the stability of a molecule. Formal charge is used to estimate the distribution of electrons within a molecule. It helps chemists to determine the most reasonable structure for a molecule by ensuring minimal charge distribution across atoms.

To calculate formal charge, use the formula:

• Formal Charge = Valence Electrons - (Non-Bonding Electrons + 1/2 Bonding Electrons)

The goal is usually to have a formal charge of zero for all atoms, suggesting an optimal structure where atoms are neither overburdened with too many electrons nor deficient.

For example, in the solution provided, each atom's formal charge is zero, which indicates that electrons in the bonds are equally shared, and the structure is likely stable.

Valence Electrons

Valence electrons are the outermost electrons of an atom and play a crucial role in chemical bonding and reactions. These electrons determine how atoms interact and bond with each other. In the context of the first two rows of the periodic table, understanding valence electrons is straightforward yet essential.

Every element has a specific number of valence electrons that corresponds to its group number in the periodic table. For example, hydrogen (H) has 1 valence electron, while oxygen (O) has 6 valence electrons. These valence electrons are pivotal because they are involved in forming bonds to complete the atom's electron shell, achieving a state similar to the noble gases, which are inert due to their full valence shell.

When creating Lewis structures, knowing the number of valence electrons helps you accurately depict the number and types of bonds an atom can make, as illustrated in the exercise.

First Two Rows of Periodic Table

The first two rows of the periodic table, comprising elements from hydrogen (H) to neon (Ne), represent the simplest elements with increasing atomic numbers and complexity of electron configurations. These elements are key players in basic chemistry and are often encountered in introductory chemical problems.

Elements in these rows occupy only the s and p subshells, which makes their electron configurations straightforward. Such simplicity aids in predicting their chemical behavior and bonding patterns.

• Row 1 Elements: H (Hydrogen), He (Helium)
• Row 2 Elements: Li (Lithium), Be (Beryllium), B (Boron), C (Carbon), N (Nitrogen), O (Oxygen), F (Fluorine), Ne (Neon)

The exercise presented involves identifying elements using their location in these first two rows, which ensures a focus on simple and fundamental interactions often first covered in chemistry courses. Understanding these basics lays the groundwork for grasping more complex chemical concepts later on.