   Chapter 9, Problem 71SCQ

Chapter
Section
Textbook Problem

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. 16 applies to BrO, write its electron configuration (where Br uses 45 and 4p orbitals). What is the highest occupied molecular orbital (HOMO) for the molecule?

(a)

Interpretation Introduction

Interpretation:

Lewis structure of the oxide of bromine should be drawn and the hybridization of central oxygen atom should be determined.

Concept Introduction

Valance bond (VBT) theory: This theory explain a chemical bonding theory that explains the bonding between two atoms is caused by the overlap of half-filled atomic orbitals. The two atoms share each other's unpaired electron to form a filled orbital to form a hybrid orbital and bond together.

Hybridization is the mixing of valence atomic orbitals to get equivalent hybridized orbitals that having similar characteristics and energy.

Geometry of a molecule can be predicted by knowing its hybridization.

Geometry of different types of molecule with respect to the hybridizations are mentioned are mentioned below,

TypeofmoleculeHybridaizationAtomicorbitalsusedforhybridaizationGeometryAX2sp1s+1pLinearAX3,AX2Bsp21s+2pTrigonalplanarAX4,AX3B,AX2B2sp31s+3pTetrahedralAX5,AX4B,AX3B2,AX2B3sp3d1s+3p+1dTrigonalbipyramidalAX6,AX5B,AX4B2sp3d21s+3p+2dOctahedralACentralatomXAtomsbondedtoABNonbondingelectronpairsonA

Empirical formula of a compound represents the smallest whole number relative ratio of elements in that compound.

Explanation

Empirical formula methods

1. 1. Start with the number of grams of each element, given in the problem.
2. 2. Convert the mass of each element to moles using the molar mass from the periodic table.
3. 3. Divide each mole value by the smallest number of moles calculated.
4. 4. Round to the nearest whole number. This is the mole ratio of the elements and is.

So given the statement 90.09% presented for bromine molecule, the total number of atoms two (Br-Br)

Than this statement 9.10% presented for oxygen molecule, the total number of atoms one (O)

Empirical formula: Br2O

Lewis structure obtained Br2Omolecule

(b)

Interpretation Introduction

Interpretation:

Electronic configuration of BrO in MO term should be written and the HOMO for the molecule should be determined.

Concept Introduction

Molecular orbital (MO) theory:  is a method for determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule.

According to this theory there are two types of orbitals,

1. (1) Bonding orbitals
2. (2) Antibonding orbitals

Electrons in molecules are filled in accordance with the energy; the anti-bonding orbital has more energy than the bonding orbitals.

The electronic configuration of oxygen molecule O2 can be represented as follows,

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2p)2( π2p)4( π*2p)2

The * represent the antibonding orbital

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