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Chapter 9: Problem 98

Given that the ionization energy of \(\mathrm{F}_{2}^{-}\) is \(290 \mathrm{~kJ} / \mathrm{mol}\), do the following: a. Calculate the bond energy of \(\mathrm{F}_{2}-\) You will need to look up the bond energy of \(\mathrm{F}_{2}\) and ionization energy of \(\mathrm{F}^{-}\). b. Explain the difference in bond energy between \(\mathrm{F}_{2}^{-}\) and \(\mathrm{F}_{2}\) using MO theory.

Short Answer

Expert verified
The bond energy of \(\mathrm{F}_{2}^{-}\) is -2835 kJ/mol, which is lower than the bond energy of \(\mathrm{F}_{2}\), at 155 kJ/mol. This difference can be explained using MO theory, by considering the distribution of electrons in bonding and anti-bonding orbitals for both species. In \(\mathrm{F}_{2}^{-}\), the extra electron goes into the lowest energy anti-bonding molecular orbital, further weakening the bond.

Step by step solution

01

Look up necessary values

Look up the bond energy of F2 and the ionization energy of F-. According to data or sources, you should find that the bond energy of \(\mathrm{F}_{2}\) is approximately \(155~\mathrm{kJ/mol}\), and the ionization energy of \(\mathrm{F}^{-}\) is approximately \(3280~\mathrm{kJ/mol}\).
02

Calculate the bond energy of \(\mathrm{F}_{2}^-\)

The bond energy of \(\mathrm{F}_{2}^{-}\) can be calculated by subtracting the ionization energy of \(\mathrm{F}^{-}\) from the ionization energy of \(\mathrm{F}_{2}^{-}\) and adding the bond energy of \(\mathrm{F}_{2}\): \[E_{\mathrm{F}_{2}^{-}}=E_{\mathrm{F}_{2}}-(E_{\mathrm{F}^{-}}-E_{\mathrm{F}_{2}^{-}})\] \[E_{\mathrm{F}_{2}^{-}}=155~\mathrm{kJ/mol}-(3280~\mathrm{kJ/mol}-290~\mathrm{kJ/mol})\] \[E_{\mathrm{F}_{2}^{-}}=155~\mathrm{kJ/mol}-2990~\mathrm{kJ/mol}\] \[E_{\mathrm{F}_{2}^{-}}=-2835~\mathrm{kJ/mol}\]
03

Explain the difference in bond energy between \(\mathrm{F}_{2}^-\) and \(\mathrm{F}_{2}\) using MO theory

Molecular Orbital (MO) theory helps to explain the difference in bond energy between \(\mathrm{F}_{2}^{-}\) and \(\mathrm{F}_{2}\). In MO theory, electrons are distributed in molecular orbitals, which are formed from the overlap of atomic orbitals. In the case of \(\mathrm{F}_{2}\), there are a total of 14 electrons. These electrons will fill molecular orbitals in increasing energy order. The lower energy molecular orbitals are bonding orbitals and the higher energy orbitals are anti-bonding orbitals. The electrons in bonding orbitals help to strengthen the bond between the two fluorine atoms, while the electrons in anti-bonding orbitals weaken the bond. In \(\mathrm{F}_{2}\), the number of electrons in bonding and anti-bonding orbitals cancel each other out, leading to a relatively weak bond that is stable at 155 kJ/mol. Now, when considering \(\mathrm{F}_{2}^{-}\), there is an extra electron, making a total of 15 electrons. This extra electron will go into the lowest energy anti-bonding molecular orbital. Adding an electron to an anti-bonding orbital further weakens the bond, which is why the bond energy of \(\mathrm{F}_{2}^{-}\) is significantly lower, at -2835 kJ/mol. In conclusion, the bond energy of \(\mathrm{F}_{2}^{-}\) is -2835 kJ/mol, which is lower than the bond energy of \(\mathrm{F}_{2}\), at 155 kJ/mol. This difference can be explained using MO theory, by taking into account the distribution of electrons in bonding and anti-bonding orbitals for both species.

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Key Concepts

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

Bond Energy Calculation
Bond energy is the measure of the strength of a chemical bond. In other words, it's the energy required to break one mole of bonds in a compound, ultimately separating the component atoms completely.

For bond energy calculation, like in the example given for \(\mathrm{F}_{2}^-\), one must have a grasp of the bond energies in the neutral molecule (\(\mathrm{F}_{2}\)) and the ionization energy of the negative ion (\(\mathrm{F}^{-}\)). The simple formula used in the exercise – subtracting the ionization energy of \(\mathrm{F}^{-}\) from that of \(\mathrm{F}_{2}^-\) and adding the bond energy of \(\mathrm{F}_{2}\) – might not always be straightforward, as energy values can be nuanced depending on the electrons' distribution.

This simplified approach works under the assumption that any extra electrons beyond the neutral molecule's count contribute negatively to bond energy. Ensuring accuracy in these calculations often involves looking at not only the number but the placement of electrons across molecular orbitals, which can be understood further with MO theory.
Molecular Orbital (MO) Theory
Molecular Orbital (MO) Theory provides a more nuanced view of the electron distribution and chemical bonding in molecules. Unlike Lewis structures which imply localized electron pairs, MO Theory describes electrons in molecules as delocalized; they are not associated with any particular atom but are instead spread over orbitals that might encompass several atoms within the molecule.

MO Theory states that atomic orbitals combine to form molecular orbitals when atoms bond together. These MOs come in two types: bonding molecular orbitals, which lower the energy of electrons and stabilize the bond, and anti-bonding molecular orbitals, which contain electrons at higher energies and can destabilize bonds if occupied. This theory can accurately predict bond strength and even magnetic properties of different compounds.

Significance in Calculations

When used in bond energy calculations, MO Theory helps us understand why adding or removing an electron affects bond energy. It accounts for the varying energies of electrons and offers a reason behind the increase or decrease in bond strength upon ionization.
Electron Distribution in Molecular Orbitals
Electron distribution in molecular orbitals is pivotal in understanding the physical and chemical properties of a molecule. Since electrons are negatively charged, their arrangement in a molecule significantly influences the molecule's stability and reactivity.

Within Molecular Orbital Theory, electrons fill the molecular orbitals based on the Aufbau principle, starting with the lowest energy level. The distribution has implications for bond order, magnetic properties, and molecular stability.
  • Purely bonding orbitals being occupied result in a strong bond.
  • If electrons fill anti-bonding orbitals, they can weaken or even nullify the bond.
  • The number of electrons in bonding vs. anti-bonding orbitals determines the bond order.
  • The presence of unpaired electrons can render a molecule paramagnetic, while paired electrons generally make it diamagnetic.
The distribution of electrons in \(\mathrm{F}_{2}^-\), with an extra electron occupying an anti-bonding orbital, clearly illustrates how electron placement can drastically affect molecular stability, reflected in the substantial decrease in bond energy compared to \(\mathrm{F}_{2}\).

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Most popular questions from this chapter

Determine the molecular structure and hybridization of the central atom \(\mathrm{X}\) in the polyatomic ion \(\mathrm{XY}_{3}+\) given the following information: A neutral atom of \(\mathrm{X}\) contains 36 electrons, and the element \(Y\) makes an anion with a \(1-\) charge, which has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6}\).

Show how a \(d_{x}\) atomic orbital and a \(p_{z}\) atomic orbital combine to form a bonding molecular orbital. Assume the \(x\) -axis is the internuclear axis. Is a \(\sigma\) or a \(\pi\) molecular orbital formed? Explain.

The three most stable oxides of carbon are carbon monoxide (CO), carbon dioxide (CO \(_{2}\) ), and carbon suboxide \(\left(\mathrm{C}_{3} \mathrm{O}_{2}\right.\) ). The space-filling models for these three compounds are For each oxide, draw the Lewis structure, predict the molecular structure, and describe the bonding (in terms of the hybrid orbitals for the carbon atoms).

The diatomic molecule OH exists in the gas phase. The bond length and bond energy have been measured to be \(97.06 \mathrm{pm}\) and \(424.7 \mathrm{~kJ} / \mathrm{mol}\), respectively. Assume that the \(\mathrm{OH}\) molecule is analogous to the HF molecule discussed in the chapter and that molecular orbitals result from the overlap of a lowerenergy \(p_{z}\) orbital from oxygen with the higher-energy \(1 s\) orbital of hydrogen (the \(\mathrm{O}-\mathrm{H}\) bond lies along the \(z\) -axis). a. Which of the two molecular orbitals will have the greater hydrogen \(1 s\) character? b. Can the \(2 p_{x}\) orbital of oxygen form molecular orbitals with the \(1 s\) orbital of hydrogen? Explain. c. Knowing that only the \(2 p\) orbitals of oxygen will interact significantly with the \(1 s\) orbital of hydrogen, complete the molecular orbital energy- level diagram for \(\mathrm{OH}\). Place the correct number of electrons in the energy levels. d. Estimate the bond order for OH. e. Predict whether the bond order of \(\mathrm{OH}^{+}\) will be greater than, less than, or the same as that of OH. Explain.

Draw the Lewis structure for HCN. Indicate the hybrid orbitals, and draw a picture showing all the bonds between the atoms, labeling each bond as \(\sigma\) or \(\pi\).
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