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For both atoms and molecules, ionization energies (Section 7.4) are related to the energies of orbitals: The lower the energy of the orbital, the greater the ionization energy. The first ionization energy of a molecule is therefore a measure of the energy of the highest occupied molecular orbital (HOMO). See the "Chemistry Put to Work" box on Orbitals and Energy. The first ionization energies of several diatomic molecules are given in electron-volts in the following table: \begin{tabular}{ll} \hline Molecule & \(I_{1}(\mathrm{eV})\) \\ \hline \(\mathrm{H}_{2}\) & \(15.4\) \\ \(\mathrm{~N}_{2}\) & \(15.6\) \\ \(\mathrm{O}_{2}\) & \(12.1\) \\ \(\mathrm{~F}_{2}\) & \(15.7\) \\ \hline \end{tabular} (a) Convert these ionization energies to \(\mathrm{kJ} / \mathrm{mol} .(\mathrm{b})\) On the same plot, graph \(I_{1}\) for the \(\mathrm{H}, \mathrm{N}, \mathrm{O}\), and \(\mathrm{F}\) atoms (Figure 7.11) and \(I_{1}\) for the molecules listed. (c) Do the ionization energies of the molecules follow the same periodic trends as the ionization energies of the atoms? (d) Use molecular orbital energy-level diagrams to explain the trends in the ionization energies of the molecules.

Short Answer

Expert verified
(a) The ionization energies in kJ/mol are: H2: 1485.86 kJ/mol, N2: 1504.37 kJ/mol, O2: 1167.46 kJ/mol, and F2: 1513.61 kJ/mol. (b) To graph the ionization energies, plot the atomic and molecular values on the y-axis against the element symbols on the x-axis, using different markers or colors for atoms and molecules. (c) From the graph, the ionization energies of the molecules do not follow the same periodic trends as the atoms. For example, O2 has a lower ionization energy compared to the other molecules, whereas the opposite is true for the corresponding atom. (d) The trends in the ionization energies can be explained by the molecular orbital energy-level diagrams. The presence of bonding and anti-bonding orbitals affects the ionization energies, as electrons are more likely to be removed from higher energy (anti-bonding) orbitals. For example, O2 has a lower ionization energy due to its unpaired electrons in the anti-bonding orbitals.

Step by step solution

01

(a) Converting electron-volts to kilojoules per mole

For this step, we will use the conversion 1 eV = 96.485 kJ/mol. To convert the ionization energies from electron-volts to kilojoules per mole, multiply the given values by the conversion factor. For H2: \(15.4\,\text{eV}*\dfrac{96.485\,\text{kJ}}{1\,\text{eV}\,\text{mol}} = 1485.86\,\text{kJ/mol}\) For N2: \(15.6\,\text{eV}*\dfrac{96.485\,\text{kJ}}{1\,\text{eV}\,\text{mol}} = 1504.37\,\text{kJ/mol}\) For O2: \(12.1\,\text{eV}*\dfrac{96.485\,\text{kJ}}{1\,\text{eV}\,\text{mol}} = 1167.46\,\text{kJ/mol}\) For F2: \(15.7\,\text{eV}*\dfrac{96.485\,\text{kJ}}{1\,\text{eV}\,\text{mol}} = 1513.61\,\text{kJ/mol}\) Now we have the ionization energies in kJ/mol.
02

(b) Graphing Ionization Energies of Atoms and Molecules

To create this graph, plot the ionization energy values against the element symbols on the x-axis. Plot both the atomic ionization energies from Figure 7.11 and the molecular ionization energies calculated above. Use two different markers or colors for atoms and molecules to distinguish them in the graph.
03

(c) Discussing Periodic Trends in Molecular Ionization Energies

Observe the graph created in the previous step and compare the trends of the atomic and molecular ionization energies. Do the ionization energies of the molecules follow the same periodic trends as the ionization energies of the atoms? Discuss your observations regarding these trends.
04

(d) Explaining Trends using Molecular Orbital Energy-Level Diagrams

To explain the trends in the ionization energies of these molecules, refer to the molecular orbital energy-level diagrams for H2, N2, O2, and F2. Discuss how the energy levels and the presence of bonding and anti-bonding orbitals can help explain the observed trends in the ionization energies of the molecules.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Orbitals
Molecular orbitals are vital to understanding the behavior and properties of molecules. These orbitals are formed when atomic orbitals combine as atoms join to form molecules.
When two atoms approach each other, their atomic orbitals overlap, leading to the creation of new molecular orbitals. These can be bonding, anti-bonding, or non-bonding orbitals.
  • **Bonding Orbitals:** Formed when atomic orbitals constructively combine, lowering the energy and increasing stability of the molecule.
  • **Anti-bonding Orbitals:** Created through destructive interference, leading to higher energy and decreasing stability.
  • **Non-bonding Orbitals:** Orbitals that do not participate in bonding but hold electrons, typically seen in larger or more complex molecules.
Understanding these orbitals helps in predicting the molecule’s stability, reactivity, and other related phenomena, such as ionization energy.
Periodic Trends
Periodic trends refer to the predictable changes in the properties of elements across the periodic table. These trends are observed because of the regular variations in atomic structure as one moves across or down the table.
Here are some key trends:
  • Atomic size: Increases down a group and decreases across a period.
  • Ionization energy: Generally increases across a period and decreases down a group.
  • Electronegativity: Increases across a period and decreases down a group.
Ionization energy, a core aspect discussed here, measures the energy required to remove an electron from an atom or molecule. Periodic trends provide insights into why some elements have higher or lower ionization energies than others, based on their position on the periodic table.
Electron-volts to Kilojoules per Mole Conversion
Conversion between electron-volts (eV) and kilojoules per mole (kJ/mol) is crucial in chemistry, especially when dealing with ionization energies. The unit eV is commonly used at the atomic scale, while kJ/mol is practical when discussing or calculating energies in mole quantities.
  • The conversion factor is: 1 eV = 96.485 kJ/mol.
To convert ionization energies from eV to kJ/mol, you multiply the value in eV by this conversion factor. For example, if a molecule's ionization energy is expressed in eV, converting it allows for easier comparisons in broader chemical contexts, such as reaction energetics.
Molecular Orbital Diagrams
Molecular orbital diagrams visually represent the relative energy levels of molecular orbitals within a molecule. These diagrams help chemists understand the distribution of electrons in a molecule and the overall molecular stability.
  • Filling Molecular Orbitals: Electrons are placed in these orbitals starting from the lowest energy levels, following Hund’s rule and Pauli’s exclusion principle.
  • Bond Order Calculation: This is determined by the difference between the number of electrons in bonding and anti-bonding orbitals, which indicates the stability and strength of the bond.
For example, in diatomic molecules like H₂, N₂, O₂, and F₂,
these diagrams can explain trends in ionization energies by showing how electron placements in these orbitals influence overall energies.
Understanding these diagrams aids in predicting molecular behavior, including response to energy absorption or electron removal.

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Most popular questions from this chapter

From their Lewis structures, determine the number of \(\sigma\) and \(\pi\) bonds in each of the following molecules or ions: (a) \(\mathrm{CO}_{2} ;\) (b) thiocyanate ion, \(\mathrm{NCS}^{-}\) : (c) formaldehyde, \(\mathrm{H}_{2} \mathrm{CO} ;\) (d) formic acid, \(\mathrm{HCOOH}\), which has one \(\mathrm{H}\) and two \(\mathrm{O}\) atoms attached to \(\mathrm{C}\).

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide \(\left(\mathrm{OSF}_{4}\right)\) according to the following unbalanced reaction: $$ \mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g) $$ The \(\mathrm{O}\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central \(S\) atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.4) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electrondomain geometry of \(\mathrm{OSF}_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) Which of the molecular geometries in part (d) is more likely to be observed for the molecule? Explain.

(a) What is the physical basis for the VSEPR model? (b) When applying the VSEPR model, we count a double or triple bond as a single electron domain. Why is this justified?

Azo dyes are organic dyes that are used for many applications, such as the coloring of fabrics. Many azo dyes are derivatives of the organic substance azobenzene, \(\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{~N}_{2}\). A closely related substance is hydrazobenzene, \(\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{~N}_{2} .\) The Lewis structures of these two substances are c1ccc(N=Nc2ccccc2)cc1 \(\begin{array}{ll}\text { Azobenzene } & \text { Hydrazobenzene }\end{array}\) (Recall the shorthand notation used for benzene.) (a) What is the hybridization at the \(\mathrm{N}\) atom in each of the substances? (b) How many unhybridized atomic orbitals are there on the \(\mathrm{N}\) and the \(\mathrm{C}\) atoms in each of the substances? (c) Predict the \(\mathrm{N}-\mathrm{N}-\mathrm{C}\) angles in each of the substances. (d) Azobenzene is said to have greater delocalization of its \(\pi\) electrons than hydrazobenzene. Discuss this statement in light of your answers to (a) and (b). (e) All the atoms of azobenzene lie in one plane, whereas those of hydrazobenzene do not. Is this observation consistent with the statement in part (d)? (f) Azobenzene is an intense red-orange color, whereas hydrazobenzene is nearly colorless. Which molecule would be a better one to use in a solar energy conversion device? (See the "Chemistry Put to Work" box for more information about solar cells.)

(a) Starting with the orbital diagram of a sulfur atom, describe the steps needed to construct hybrid orbitals appropriate to describe the bonding in \(S F_{2}\). (b) What is the name given to the hybrid orbitals constructed in (a)? (c) Sketch the large lobes of the hybrid orbitals constructed in part (a). (d) Would the hybridization scheme in part (a) be appropriate for \(\mathrm{SF}_{4} ?\) Explain.

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