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 \(\mathrm{O}_{2}^{1}\) ? (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

Option (b) is correct: paramagnetic and bond order \(>\mathrm{O}_{2}\).

Step 1: Understanding the Molecular Formula

The molecule in question is \(\mathrm{O}_2^{1+}\), which means that one electron has been removed from \(\mathrm{O}_2\), resulting in a positive charge.

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

The electronic configuration for neutral \(\mathrm{O}_2\) based on molecular orbital theory is: \(\sigma_{2s}^2, \sigma^*_{2s}^2, \sigma_{2p_z}^2, \pi_{2p_x}^2, \pi_{2p_y}^2, \pi^*_{2p_x}^1, \pi^*_{2p_y}^1\). This shows \(\mathrm{O}_2\) has 2 unpaired electrons in the antibonding \(\pi^*\) orbitals.

Step 3: Determine the Electronic Configuration for \(\mathrm{O}_2^{+1}\)

By removing one electron from \(\mathrm{O}_2\), the configuration becomes \(\sigma_{2s}^2, \sigma^*_{2s}^2, \sigma_{2p_z}^2, \pi_{2p_x}^2, \pi_{2p_y}^2, \pi^*_{2p_x}^1\). It now has 1 unpaired electron in the \(\pi^*_{2p}\) orbital, making it paramagnetic.

Step 4: Calculate Bond Order of \(\mathrm{O}_2\)

For \(\mathrm{O}_2\), bond order is calculated as \(\text{Bond Order} = \frac{1}{2}(\text{number of bonding electrons} - \text{number of antibonding electrons}) = \frac{1}{2}(10 - 6) = 2\).

Step 5: Calculate Bond Order of \(\mathrm{O}_2^{+1}\)

For \(\mathrm{O}_2^{+1}\), remove one electron from an antibonding orbital. Its bond order becomes \(\text{Bond Order} = \frac{1}{2}(10 - 5) = 2.5\), which is greater than \(\mathrm{O}_2\).

Step 6: Choose the Correct Statement

Comparing \(\mathrm{O}_2^{+1}\) with \(\mathrm{O}_2\), it remains paramagnetic due to an unpaired electron, and its bond order of 2.5 is greater than that of \(\mathrm{O}_2\), which is 2. Therefore, option (b) is correct: paramagnetic and bond order \(>\mathrm{O}_{2}\).

Key Concepts

Paramagnetism

Paramagnetism refers to a property of certain materials that causes them to be attracted to an external magnetic field. This property arises due to the presence of unpaired electrons in the atoms or molecules of the material. In terms of molecular orbital theory, if there are unpaired electrons in the molecular orbitals, the molecule exhibits paramagnetism.

For example,

• In the molecule \(O_2\), the presence of two unpaired electrons in the antibonding \(\pi^*_{2p}\) orbitals renders it paramagnetic.
• Similarly, when \(O_2\) loses an electron to become \(O_2^{1+}\), it still retains one unpaired electron, maintaining its paramagnetic character.

Understanding paramagnetism helps to predict the magnetic behavior of substances in fields like chemistry, material science, and physics. Scientists use this property to probe molecular structure and to gain insights into the behavior of molecules under magnetic influences. Thus, the concept of paramagnetism plays a vital role in understanding the molecular dynamics.

Bond Order

Bond order is an essential concept in chemical bonding that indicates the stability of a molecule and the strength of the bond between atoms. In molecular orbital theory, bond order can be determined by calculating the difference between the number of electrons in bonding and antibonding molecular orbitals.

The formula for bond order is given by: \[ \text{Bond Order} = \frac{1}{2}(\text{number of bonding electrons} - \text{number of antibonding electrons}). \]

• For example, in neutral \(O_2\), with 10 bonding electrons and 6 antibonding electrons, the bond order is calculated as 2. This indicates a double bond.
• However, for \(O_2^{1+}\), after removing one antibonding electron, the bond order increases to 2.5. A higher bond order in \(O_2^{1+}\) indicates a more stable and stronger bond than that in \(O_2\).

Bond order helps chemists understand the bond strength and stability of compounds, influencing the design and synthesis of new materials. By examining bond order, scientists predict the reactivity and properties of molecules, affecting various applications in research and industry.

Electron Configuration

Understanding the electron configuration of molecules is vital in molecular chemistry, as it dictates the molecule's properties and behavior. Electron configuration describes the distribution of electrons in an atom or molecule's orbitals, which ultimately affects its chemical bonding and reactivity.

Using molecular orbital theory, the electron configuration for a molecule like \(O_2\) is specified based on its occupied orbitals: \[ \sigma_{2s}^2, \sigma^*_{2s}^2, \sigma_{2p_z}^2, \pi_{2p_x}^2, \pi_{2p_y}^2, \pi^*_{2p_x}^1, \pi^*_{2p_y}^1. \]

• In molecular form \(O_2^{1+}\), losing an electron results in the change of configuration: \[ \sigma_{2s}^2, \sigma^*_{2s}^2, \sigma_{2p_z}^2, \pi_{2p_x}^2, \pi_{2p_y}^2, \pi^*_{2p_x}^1. \]
• This reduced number of electrons in the antibonding \(\pi^*_{2p}\) orbitals accounts for its increased bond order and paramagnetic property.

Delving into electron configuration helps elucidate the underlying reason for a molecule's magnetic and chemical characteristics, providing critical insights into structural, physical, and reactive aspects. Such understanding is fundamental in predicting reactions, designing experiments, and advancing technological applications based on molecular systems.