Problem 29
Question
According to molecular orbital theory which of the following statement about the magnetic character and bond order is correct regarding \(\quad\) [2004S] \(\mathrm{O}_{2}^{+}\) (a) Paramagnetic and Bond order \(<\mathrm{O}_{2}\) (b) Paramagnetic and Bond order \(>\mathrm{O}_{2}\) (c) Diamagnetic and Bond order \(<\mathrm{O}_{2}\) (d) Diamagnetic and Bond order \(>\mathrm{O}\),
Step-by-Step Solution
Verified Answer
(b) Paramagnetic and Bond order \( > \mathrm{O}_2 \).
1Step 1: Identify the Electron Configuration of O2
Determine the electron configuration of an \( \mathrm{O}_2 \ \) molecule. Oxygen molecule, \( \mathrm{O}_2 \ \), has 16 electrons (8 from each oxygen atom). Its molecular orbital configuration is \( (\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi_{2p}^*)^2 \ \).
2Step 2: Determine Electron Configuration of O2^+
For \( \mathrm{O}_2^+ \ \), remove one electron from \( \mathrm{O}_2 \ \) since it has a positive charge. The electron is removed from the highest energy occupied molecular orbital, which is \( \pi_{2p}^* \ \). The configuration becomes \( (\pi_{2p}^*)^1 \ \).
3Step 3: Calculate Bond Order
Bond order is defined as \( \frac{1}{2}(\text{number of bonding electrons} - \text{number of antibonding electrons}) \ \). For \( \mathrm{O}_2 \ \), bond order is \( \frac{1}{2} (10 - 6) = 2 \ \). For \( \mathrm{O}_2^+ \ \), bond order is \( \frac{1}{2} (10 - 5) = 2.5 \ \). Hence, bond order of \( \mathrm{O}_2^+ \ \) is greater than \( \mathrm{O}_2 \ \).
4Step 4: Determine Magnetic Character
The presence of unpaired electrons in the configuration indicates paramagnetism. \( \mathrm{O}_2^+ \ \) has one unpaired electron in \( \pi_{2p}^* \ \), making it paramagnetic.
5Step 5: Compare Findings with Given Options
Paramagnetic and bond order \( > \mathrm{O}_2 \ \) matches with option (b). None of the other options correctly describe the magnetic character and bond order.
Key Concepts
ParamagnetismBond OrderElectron Configuration
Paramagnetism
Paramagnetism is a property of substances with unpaired electrons in their molecular orbitals. These unpaired electrons cause the substance to be attracted to external magnetic fields. Unlike diamagnetic materials that are repelled by magnetic fields, paramagnetic materials show an attractive response due to these unpaired electrons.
In the context of molecular orbital theory, if the molecular orbital (MO) diagram of a molecule has one or more unpaired electrons, it is characterized as paramagnetic. In our example, \( \mathrm{O}_2^+ \) shows paramagnetism because of an unpaired electron residing in the \(\pi_{2p}^*\) orbital.
Understanding paramagnetism is crucial for predicting how molecules will behave in the presence of a magnetic field, which can also influence the structure and reactivity of the molecule.
In the context of molecular orbital theory, if the molecular orbital (MO) diagram of a molecule has one or more unpaired electrons, it is characterized as paramagnetic. In our example, \( \mathrm{O}_2^+ \) shows paramagnetism because of an unpaired electron residing in the \(\pi_{2p}^*\) orbital.
Understanding paramagnetism is crucial for predicting how molecules will behave in the presence of a magnetic field, which can also influence the structure and reactivity of the molecule.
Bond Order
Bond order is a concept used in molecular orbital theory to predict the stability of a molecule. It gives insight into the strength of the bond between atoms by considering both bonding and antibonding electrons.
Bond order is calculated with the formula:\[\text{Bond Order} = \frac{1}{2} (\text{number of bonding electrons} - \text{number of antibonding electrons})\]
A higher bond order implies a stronger, more stable bond. In the case of \( \mathrm{O}_2 \), the bond order is 2, but when an electron is removed to form \( \mathrm{O}_2^+ \), the bond order increases to 2.5. This implies that \( \mathrm{O}_2^+ \) has a stronger bond compared to \( \mathrm{O}_2 \).
Understanding bond order helps in determining not only the bond strength but also the molecular vibration frequencies, and reactivity in chemical reactions.
Bond order is calculated with the formula:\[\text{Bond Order} = \frac{1}{2} (\text{number of bonding electrons} - \text{number of antibonding electrons})\]
A higher bond order implies a stronger, more stable bond. In the case of \( \mathrm{O}_2 \), the bond order is 2, but when an electron is removed to form \( \mathrm{O}_2^+ \), the bond order increases to 2.5. This implies that \( \mathrm{O}_2^+ \) has a stronger bond compared to \( \mathrm{O}_2 \).
Understanding bond order helps in determining not only the bond strength but also the molecular vibration frequencies, and reactivity in chemical reactions.
Electron Configuration
Electron configuration in molecular orbital theory describes how electrons are distributed among molecular orbitals. This information is crucial for predicting molecular properties like magnetism and bond strength.
Consider the oxygen molecule \( \mathrm{O}_2 \), which has a total of 16 electrons. Its electron configuration in the molecular orbital terms is \((\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi_{2p}^*)^2\). Now, when electron is removed to form \( \mathrm{O}_2^+ \), this configuration changes to \((\pi_{2p}^*)^1\), indicating fewer electrons in the antibonding molecular orbital.
The changes in electron configuration upon ionization or other reactions significantly affect a molecule's magnetic and chemical properties. In \( \mathrm{O}_2^+ \), the removal of one electron increases bond order, illustrating a subtle but important shift in molecular structure.
Consider the oxygen molecule \( \mathrm{O}_2 \), which has a total of 16 electrons. Its electron configuration in the molecular orbital terms is \((\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi_{2p}^*)^2\). Now, when electron is removed to form \( \mathrm{O}_2^+ \), this configuration changes to \((\pi_{2p}^*)^1\), indicating fewer electrons in the antibonding molecular orbital.
The changes in electron configuration upon ionization or other reactions significantly affect a molecule's magnetic and chemical properties. In \( \mathrm{O}_2^+ \), the removal of one electron increases bond order, illustrating a subtle but important shift in molecular structure.
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