Question

Bromine forms a number of oxides of varying stability. (a) One oxide has \(90.90 \%\) Br and \(9.10 \%\) O. Assuming its empirical and molecular formulas are the same, draw a Lewis structure of the molecule and specify the hybridization of the central atom (O). (b) Another oxide is unstable BrO. Assuming the molecular orbital diagram in Figure 9.18 applies to BrO, write its electron configuration (where Br uses \(4 s\) and \(4 p\) orbitals). What is the HOMO for the molecule?

Short answer

The Br2O structure has O as \(sp^3\) hybridized; the HOMO for BrO is \(\pi^*(4p)\).

Step-by-step solution

Determine Formula of Oxide from Percent Composition

For part (a), we have the composition of the oxide as 90.90% Br and 9.10% O. Calculating the moles: for Br, assuming 100 g of compound, \(\frac{90.90}{79.90}\) mol; for O, \(\frac{9.10}{16.00}\) mol. The smallest mole ratio gives approximately 2 Br to 1 O, suggesting the formula is \(\text{Br}_2\text{O}\).

Draw Lewis Structure of Br2O

For \(\text{Br}_2\text{O}\), oxygen is the central atom. Connect each Br with a single bond to O. Then, assign lone pairs to satisfy the octet rule. Both Br atoms will be singly bonded to O, and O will have two lone pairs.

Determine Oxygen Hybridization in Br2O

In the \(\text{Br}_2\text{O}\) molecule, oxygen forms two sigma bonds and has two lone pairs, implying a steric number of 4. Therefore, the hybridization of the central atom (O) is \(sp^3\).

Write Electron Configuration of BrO

For part (b), assuming BrO uses the same molecular orbital arrangement as diatomic molecules, Br uses 4s and 4p orbitals. The electron configuration is similar to that of a heteronuclear diatomic molecule: \([\sigma(4s)]^2[\sigma^*(4s)]^2[\pi(4p)]^4[\sigma(4p)]^2[\pi^*(4p)]^1\).

Identify HOMO for BrO

The Highest Occupied Molecular Orbital (HOMO) for BrO, following our configuration, is \(\pi^*(4p)\).

Key concepts

Lewis Structure

Understanding the Lewis structure is crucial in chemistry as it helps us visualize the arrangement of atoms in a molecule. For the bromine oxide \(\text{Br}_2\text{O}\), the oxygen is the central atom. This atom is bonded to two bromine atoms, each with a single bond.

To complete the Lewis structure, assign lone pairs of electrons to achieve the stable octet rule. Since oxygen typically forms two bonds, it will need two lone pairs of electrons. Each Br atom will remain satisfied with one bond and three remaining lone pairs, making them chemically stable.

In summary, in the Lewis structure of \(\text{Br}_2\text{O}\):

• Oxygen has two single bonds and two lone pairs.
• Each bromine atom is connected to the oxygen atom and has three lone pairs.

This arrangement provides a visual representation of the electron distribution and bonds within the molecule.

Molecular Formula

The molecular formula gives insight into the exact makeup of a compound, specifying the number of each type of atom present. To determine the molecular formula of a compound like bromine oxide (\(\text{Br}_2\text{O}\)) from percent composition, a step-by-step calculation is necessary.

First, assume you have 100 grams of the substance:

• Calculate the number of moles for bromine: \(\frac{90.90}{79.90}\).
• Do the same for oxygen: \(\frac{9.10}{16.00}\).

The ratio of these moles gives a clearer picture of the molecular composition. Rounding yields a 2:1 ratio, suggesting that two Br atoms bond with one O atom.

The empirical and molecular formulas are similar, giving us \(\text{Br}_2\text{O}\), accurately representing the composition and structure of the molecule.

Hybridization

Hybridization offers a deeper understanding of how atomic orbitals mix to form molecule geometries and bonds. Within \(\text{Br}_2\text{O}\), the central oxygen atom's hybridization provides insight into its electron geometry.

Here's what you need to consider:

• Oxygen is primarily involved with two sigma bonds with Br atoms.
• It also has two lone pairs.

The steric number for oxygen in \(\text{Br}_2\text{O}\) is 4, which results from the sum of sigma bonds and lone pairs.

A steric number of 4 typically indicates \(sp^3\) hybridization. This means the oxygen's orbitals have rearranged into four equivalent \(sp^3\) hybrid orbitals, allowing the formation of the observed bonds and electron pairings.

Electron Configuration

Understanding electron configuration aids in grasping the behavior of electrons within a molecule. In the bromine monoxide (BrO) molecule, electron arrangement can be likened to that of typical diatomic molecules using molecular orbitals.

Here's a breakdown of the electron configuration for BrO:

• \([\sigma(4s)]^2\) and \([\sigma^*(4s)]^2\): Electrons in the 4s orbitals.
• \([\pi(4p)]^4\): The presence of functional orbitals, showing filled bonding states.
• \([\sigma(4p)]^2\): Additional electrons within orbitals.
• \([\pi^*(4p)]^1\): Describes singly occupied antibonding orbitals.

The highest occupied molecular orbital (HOMO) for BrO is \(\pi^*(4p)\).

Understanding this configuration unveils the reactivity and bond formation proficiency of BrO, lending insight into its molecular behavior.