Problem 143 - A Complete Resource Book in Chemistry for JEE Main

Question

According to molecular orbital theory which of the following statement about the magnetic character and bond order is correct regarding \(\mathrm{O}_{2}^{+} ?\) (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 \(>\mathrm{O}_{2}\)

Step-by-Step Solution

Answer

The correct answer is (b) paramagnetic and bond order \( > \mathrm{O}_2 \).

Step 1: Understanding Molecular Orbital Theory

Start by understanding that molecular orbital theory explains the bonding by considering molecular orbitals that spread over the atoms in a molecule. The molecular orbital diagram for oxygen molecules will help us analyze the magnetic properties and bond order.

Step 2: Determine the Electron Configuration for \( \mathrm{O}_2 \)

Oxygen, \( \mathrm{O}_2 \), has 16 electrons in total. Its electronic configuration in molecular orbital theory is \((\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^1(\pi_{2p_y}^*)^1\). This configuration gives \( \mathrm{O}_2 \) a bond order of 2 and makes it paramagnetic due to the presence of two unpaired electrons.

Step 3: Adjust Electron Count for \( \mathrm{O}_2^+ \)

\( \mathrm{O}_2^+ \) is formed by removing one electron from \( \mathrm{O}_2 \). Remove an electron from one of the antibonding \( \pi_{2p}^* \) orbitals, giving \((\pi_{2p_x}^*)^1\) or \((\pi_{2p_y}^*)^1\). Now, the electronic configuration is \((\pi_{2p_x}^*)^1\) or \((\pi_{2p_y}^*)^1\), with only one unpaired electron.

Step 4: Calculate Bond Order for \( \mathrm{O}_2^+ \)

The bond order formula is \( \text{Bond Order} = \frac{1}{2} (\text{Electrons in Bonding Orbitals} - \text{Electrons in Antibonding Orbitals}) \). For \( \mathrm{O}_2^+ \), there are 10 electrons in bonding orbitals and 5 in antibonding orbitals. Thus, \( \text{Bond Order} = \frac{1}{2} (10 - 5) = 2.5 \).

Step 5: Compare with \( \mathrm{O}_2 \)

\( \mathrm{O}_2^+ \) has a bond order of 2.5, which is greater than the bond order of \( \mathrm{O}_2 \) (which is 2). \( \mathrm{O}_2^+ \) is paramagnetic because it has one unpaired electron compared to \( \mathrm{O}_2 \), which has two unpaired electrons.

Key Concepts

Magnetic Properties

Magnetic properties of molecules can be intriguing, as they describe how a molecule interacts with an external magnetic field. In the context of molecular orbital theory, this interaction is determined by the presence or absence of unpaired electrons in the molecular orbitals.

When electrons are paired in all occupied orbitals, the molecule is considered diamagnetic and does not exhibit magnetic behavior. Conversely, if there are unpaired electrons, the molecule is paramagnetic and is attracted to a magnetic field.

This difference in behavior is due to the magnetic moments of unpaired electrons, which align with the external magnetic field, causing paramagnetic attraction. In molecules like the oxygen ion,

• Unpaired electrons lead to paramagnetism.
• Paired electrons cause diamagnetism.

Bond Order

Bond order is a concept that helps us understand the strength and stability of a chemical bond within molecular orbital theory. It's calculated using the formula:

\[ \text{Bond Order} = \frac{1}{2} (\text{Number of bonding electrons} - \text{Number of antibonding electrons}) \]

A higher bond order indicates stronger and more stable bonds. For instance, a molecule with a bond order of 3 is triple-bonded and thus very stable. A bond order of zero suggests no bond formation.

In comparing oxygen ( \(\mathrm{O}_2\) ) and its ion counterpart ( \(\mathrm{O}_2^+\) ), the bond order changes due to variations in electron configurations.

• \(\mathrm{O}_2\) has a bond order of 2.
• \(\mathrm{O}_2^+\) has a bond order of 2.5, reflecting a stronger bond due to fewer antibonding electrons.

Oxygen Molecule Chemistry

The chemistry of oxygen molecules is essential in understanding their role in natural processes and reactions. Molecular oxygen, \(\mathrm{O}_2\), is vital for life on Earth, participating in respiration and combustion.

According to molecular orbital theory, the electronic configuration of \(\mathrm{O}_2\) reveals why it is paramagnetic and has a bond order, indicative of its double-bonded nature. In its ground state, oxygen has 16 electrons allocated in a specific order, filling both bonding and antibonding orbitals.

When oxygen forms a positive ion \(\mathrm{O}_2^+\), it loses one electron, commonly from the antibonding orbital, affecting both its magnetic properties and bond order. This ionization typically results in:

• A single unpaired electron in \(\mathrm{O}_2^+\).
• An increased bond order compared to the neutral \(\mathrm{O}_2\).

Paramagnetism

Paramagnetism emerges from the presence of unpaired electrons, which create magnetic moments that align with an external magnetic field, leading to attraction. This is a key feature of certain molecules, including \(\mathrm{O}_2\) and \(\mathrm{O}_2^+\).

For a molecule to be paramagnetic, not all of its electron spins should be paired. In this context, the presence of unpaired electrons in the \(\pi_{2p}^*\) antibonding orbitals in oxygen and its ions is what results in paramagnetic behavior.

• In \(\mathrm{O}_2\), two unpaired electrons contribute to its paramagnetism.
• For \(\mathrm{O}_2^+\), the paramagnetism is due to a single unpaired electron, as removing one electron decreases the number of unpaired electrons.

Understanding the electron configuration through molecular orbital theory allows us to predict and explain these magnetic properties effectively.