Problem 25 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{O}_{2}^{2+}\) has a bond order of (a) 1 (b) 2 (c) 3 (d) 4

Step-by-Step Solution

Answer

The bond order of \( \mathrm{O}_2^{2+} \) is 2.

Step 1: Understanding Bond Order

Bond order refers to the number of chemical bonds between a pair of atoms. A higher bond order indicates a stronger bond. Bond order can be calculated using the formula: \[ ext{Bond Order} = rac{ ext{Number of Bonding Electrons} - ext{Number of Antibonding Electrons}}{2} \]

Step 2: Determine Electron Configuration for Dioxygen Ion

The electron configuration of a neutral \( \mathrm{O}_2 \) molecule is \( (\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi^*_{2p_x})^1(\pi^*_{2p_y})^1 \). For \( \mathrm{O}_2^{2+} \), we remove two electrons: one each from the \( \pi^*_{2p} \) orbitals, resulting in \( (\pi^*_{2p_x})^0(\pi^*_{2p_y})^0 \).

Step 3: Count Bonding and Antibonding Electrons

In the modified configuration of \( \mathrm{O}_2^{2+} \), there are 8 bonding electrons (from \( \sigma_{2s} \) and \( \sigma_{2p_z} \), \( \pi_{2p_x} \), \( \pi_{2p_y} \)) and 4 antibonding electrons (from \( \sigma^*_{2s} \)). The remaining antibonding orbitals \( \pi^*_{2p_x} \) and \( \pi^*_{2p_y} \) are now empty.

Step 4: Calculate Bond Order

Using the formula given in Step 1, substitute the values for bonding and antibonding electrons: \[ \text{Bond Order} = \frac{8 - 4}{2} = 2 \] So, the bond order for \( \mathrm{O}_2^{2+} \) is 2.

Key Concepts

Dioxygen ion

The dioxygen ion, often represented as \( \mathrm{O}_2^{2+} \), is an interesting concept in chemistry. This ion is derived from the removal of electrons from the neutral oxygen molecule \( \mathrm{O}_2 \), which results in changes to its bonding and properties. Understanding this ion is key to making sense of its electron configuration and the concept of bond order.

The neutrality of \( \mathrm{O}_2 \) is disrupted in \( \mathrm{O}_2^{2+} \) ; this type of ion is often seen in experimental or high-energy states rather than in nature because of the energy required to remove electrons. In fact, dioxygen ions are crucial in understanding various oxidation processes and play a role in some advanced technologies and chemical reactions.

• This ion is formed by taking away electrons, affecting its bonding.
• It exhibits unique chemical properties compared to its neutral form.
• It helps in studying electron behavior and molecular interactions.

By grasping how dioxygen ions form and behave, you can further extend this understanding to other ionized molecules encountered in chemical studies.

Electron Configuration

The electron configuration of a molecule like dioxygen ion is a key factor in determining its properties. Electron configuration shows how electrons are distributed in molecules and helps in understanding chemical bonding.

For a neutral \( \mathrm{O}_2 \) molecule, its electron configuration can be represented as \( (\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi^*_{2p_x})^1(\pi^*_{2p_y})^1 \). However, once we remove two electrons to form \( \mathrm{O}_2^{2+} \), the new configuration becomes \( (\pi^*_{2p_x})^0(\pi^*_{2p_y})^0 \). This change significantly affects how the ion behaves chemically.

Key Points:

• Neutral \( \mathrm{O}_2 \) has a specific electron arrangement critical for its molecular stability.
• Removing electrons to form \( \mathrm{O}_2^{2+} \) changes the configuration.
• Adjusting electron distribution alters molecular properties and bond strength.

Studying electron configurations aids in predicting reactions and understanding the strength and characteristics of bonds formed in ions like \( \mathrm{O}_2^{2+} \).

Chemical Bonds Calculation

Chemical bond calculation, particularly bond order, is a fundamental concept for understanding molecular stability and reactivity. Bond order indicates the number of bonds between two atoms; it tells us how strong a bond is.

For dioxygen ion \( \mathrm{O}_2^{2+} \), calculating bond order involves counting the bonding and antibonding electrons. The formula used is:\[ \text{Bond Order} = \frac{\text{Number of Bonding Electrons} - \text{Number of Antibonding Electrons}}{2}\]

In the case of \( \mathrm{O}_2^{2+} \):

- **Bonding Electrons:** These are electrons contributing to bond formation. In our dioxygen configuration, there are 8 bonding electrons.- **Antibonding Electrons:** These electrons oppose bond formation. There are 4 antibonding electrons after the adjustments.
Using these, bond order is calculated as:\[ \text{Bond Order} = \frac{8 - 4}{2} = 2\]
Hence, the bond order of 2 signifies a double bond, indicating good stability at this ionic state.

This calculation helps in:

• Predicting molecular stability
• Comparing bond strength between different molecules
• Understanding reaction tendencies

Mastering chemical bonds calculation is crucial for anyone delving into advanced chemistry topics.