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Explain the following: (a) The peroxide ion, \(\mathrm{O}_{2}^{2-}\), has a longer bond length than the superoxide ion, \(\mathrm{O}_{2}^{-}\). (b) The magnetic properties of \(\mathrm{B}_{2}\) are consistent with the \(\pi_{2 \mathrm{p}}\) MOs being lower in energy than the \(\sigma_{2 p}\) MO. (c) The \(\mathrm{O}_{2}^{2+}\) ion has a stronger O- \(O\) bond than \(\mathrm{O}_{2}\) itself.

Short Answer

Expert verified
(a) The peroxide ion (O2^2-) has a longer bond length than the superoxide ion (O2^-) because it has two additional electrons in the antibonding π2p* orbital, which weakens the O-O bond more than the one additional electron in superoxide ion. (b) The magnetic properties of B2 are consistent with the π2p MOs being lower in energy than the σ2p MO because there are two unpaired electrons in the π2p orbitals, giving rise to the magnetic properties. (c) The O2^2+ ion has a stronger O-O bond than O2 because it has two electrons removed from the antibonding π2p orbitals, reducing the antibonding effect and strengthening the O-O bond.

Step by step solution

01

(a) Peroxide ion (O2^2-) and superoxide ion (O2^-) bond lengths

Both ions are formed from the same parent molecule, O2. In O2^2-, two electrons are added, while in O2^-, only one electron is added. In molecular orbital theory, we can represent the electron configurations for these molecules as: - O2: σ1s2 σ1s*2 σ2s2 σ2s*2 σ2p2 π2p4 - O2^2-: σ1s2 σ1s*2 σ2s2 σ2s*2 σ2p2 π2p6 - O2^-: σ1s2 σ1s*2 σ2s2 σ2s*2 σ2p2 π2p5 In peroxide ion, O2^2-, the additional two electrons occupy the antibonding π2p* orbital. This weakens the O-O bond, resulting in a longer bond length compared to O2. In contrast, in superoxide ion, O2^-, only one additional electron occupies the antibonding π2p* orbital, which creates a weaker bond length increase compared to O2^2-.
02

(b) Magnetic properties of B2

The molecular orbital configuration for a B2 molecule is: - B2: σ1s2 σ1s*2 σ2s2 σ2s*2 π2p2 Considering the observed magnetic properties of B2, they are consistent with the π2p MOs being lower in energy than the σ2p MO. This is because two unpaired electrons are present in π2p orbitals, which gives rise to the magnetic properties of B2.
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(c) Comparison of O2^2+ ion and O2 bond strength

The electron configurations for the O2^2+ ion and O2 molecule are: - O2: σ1s2 σ1s*2 σ2s2 σ2s*2 σ2p2 π2p4 - O2^2+: σ1s2 σ1s*2 σ2s2 σ2s*2 σ2p2 π2p2 When O2 loses two electrons to form O2^2+, the two electrons are removed from the π2p antibonding orbitals. This reduces the antibonding effect, leading to a stronger O-O bond in O2^2+ than in the parent O2 molecule.

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Key Concepts

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

Bond Length
In the realm of molecular orbital theory, understanding bond length is crucial as it gives us insight into the strength and stability of chemical bonds. Molecules are not just static entities; their bond lengths can tell us a lot about their electronic structure. For example, the peroxide ion, \( \mathrm{O}_{2}^{2-} \), has a longer bond length compared to the superoxide ion, \( \mathrm{O}_{2}^{-} \). This difference can be explained by examining the electrons and orbitals involved.
  • The peroxide ion, \( \mathrm{O}_{2}^{2-} \), contains two extra electrons in the antibonding \( \pi_{2p}^{*} \) orbital. This orbital, when filled with electrons, weakens the bond between the oxygen atoms.
  • In contrast, the superoxide ion, \( \mathrm{O}_{2}^{-} \), has only one additional electron in the antibonding \( \pi_{2p}^{*} \) orbital, making the bond slightly stronger than in the peroxide ion but weaker than in a neutral \( \mathrm{O}_{2} \) molecule.
Thus, more electrons in antibonding orbitals lead to weaker bonds, increasing bond length.
Magnetic Properties
Magnetic properties of molecules can reveal much about their electronic configurations. Take the \( \mathrm{B}_{2} \) molecule, for example. Unlike what you might expect from a typical molecule, \( \mathrm{B}_{2} \) exhibits paramagnetic behavior. This occurs because of the arrangement of its molecular orbitals. Within \( \mathrm{B}_{2} \), the \( \pi_{2p} \) molecular orbitals are filled with two unpaired electrons. These unpaired electrons contribute to its magnetic nature. This is in contrast to situations where all electrons are paired, resulting in diamagnetism.
  • The order of molecular orbitals is a crucial factor. For \( \mathrm{B}_{2} \), the \( \pi_{2p} \) orbitals are lower in energy than the \( \sigma_{2p} \) orbitals, preserving unpaired electrons and enhancing its magnetic properties.
  • This distinct ordering is specific to certain molecules like \( \mathrm{B}_{2} \), displaying the fascinating variation in molecular orbital theory.
Understanding these properties aids in the exploration of molecular behavior and potential applications in magnetic materials.
O-O Bond Strength
Examining the strength of an \( \mathrm{O-O} \) bond highlights the impact of electronic configuration on molecular stability. Comparing the \( \mathrm{O}_{2}^{2+} \) ion with the \( \mathrm{O}_{2} \) molecule provides an insightful example.The \( \mathrm{O}_{2}^{2+} \) ion is formed by the removal of two electrons from the neutral \( \mathrm{O}_{2} \) molecule. These electrons specifically come out of the \( \pi_{2p}^{*} \) antibonding orbitals:
  • The removal of electrons from antibonding orbitals decreases their destabilizing effect, leading to a much stronger bond.
  • As a result, the \( \mathrm{O-O} \) bond in \( \mathrm{O}_{2}^{2+} \) is significantly stronger than in neutral \( \mathrm{O}_{2} \), where these orbitals are occupied.
Consequently, understanding electron occupation in specific orbitals is essential in predicting bond strength and overall molecular properties.

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Most popular questions from this chapter

Butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6},\) is a planar molecule that has the following carbon-carbon bond lengths: $$ \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{134 \mathrm{pm}} \mathrm{CH}=\mathrm{CH}_{2} $$ (a) Predict the bond angles around each of the carbon atoms and sketch the molecule. (b) From left to right, what is the hybridization of each carbon atom in butadiene? (c) The middle \(\mathrm{C}-\mathrm{C}\) bond length in butadiene \((148 \mathrm{pm})\) is a little shorter than the average \(\mathrm{C}-\mathrm{C}\) single bond length (154 pm). Does this imply that the middle \(\mathrm{C}-\mathrm{C}\) bond in butadiene is weaker or stronger than the average \(\mathrm{C}-\mathrm{C}\) single bond? (d) Based on your answer for part (c), discuss what additional aspects of bonding in butadiene might support the shorter middle \(\mathrm{C}-\mathrm{C}\) bond.

Draw sketches illustrating the overlap between the following orbitals on two atoms: (a) the \(2 s\) orbital on each atom, (b) the \(2 p_{z}\) orbital on each atom (assume both atoms are on the \(z\) -axis), \((\mathbf{c})\) the 2 s orbital on one atom and the \(2 p_{2}\) orbital on the other atom.

An \(\mathrm{AB}_{2}\) molecule is described as having a tetrahedral geometry. (a) How many nonbonding domains are on atom A? (b) Based on the information given, which of the following is the molecular geometry of the molecule: (i) linear, (ii) bent, (iii) trigonal planar, or (iv) tetrahedral?

For each statement, indicate whether it is true or false. (a) The greater the orbital overlap in a bond, the weaker the bond. (b) The greater the orbital overlap in a bond, the shorter the bond. \((\mathbf{c})\) To create a hybrid orbital, you could use the \(s\) orbital For each statement, indicate whether it is true or false. (a) The greater the orbital overlap in a bond, the weaker the bond. (b) The greater the orbital overlap in a bond, the shorter the bond. \((\mathbf{c})\) To create a hybrid orbital, you could use the \(s\) orbital

For each statement, indicate whether it is true or false. (a) \(\ln\) order to make a covalent bond, the orbitals on each atom in the bond must overlap. (b) A \(p\) orbital on one atom cannot make a bond to an \(s\) orbital on another atom. \((\mathbf{c})\) Lone pairs of electrons on an atom in a molecule influence the shape of a molecule. (d) The 1 s orbital has a nodal plane. \((\mathbf{e})\) The \(2 p\) orbital has a nodal plane.

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