Problem 138 - A Complete Resource Book in Chemistry for JEE Main

Question

Match the following:. List I (Species) 1\. \(\mathrm{O}_{2}^{2+}\) 2\. \(\mathrm{O}_{2}\) 3\. \(\mathrm{F}_{2}\) 4\. \(\mathrm{O}_{2}^{+}\) List II (Bond order) (i) \(1.0\) (ii) \(2.0\) (iii) \(2.5\) (iv) \(3.0\)

Step-by-Step Solution

Answer

1 - (ii), 2 - (ii), 3 - (i), 4 - (iii)

Step 1: Understanding Bond Order

Bond order is defined as the number of chemical bonds between a pair of atoms. It indicates the stability of a bond in a molecule, calculated as \(( ext{number of bonding electrons} - ext{number of antibonding electrons})/2\). Higher bond order means a more stable bond.

Step 2: Determine Bond Order for \(\mathrm{O}_{2}^{2+}\)

For \(\mathrm{O}_{2}^{2+}\), remove 2 electrons from \(\mathrm{O}_{2}\) which leaves 10 total electrons. The molecular orbital configuration is \((\sigma_{1s})^2(\sigma^{*}_{1s})^2(\sigma_{2s})^2(\sigma^{*}_{2s})^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^0\). Calculating bond order: \((8 - 4)/2 = 2.0\). Therefore, \(\mathrm{O}_{2}^{2+}\) corresponds to bond order (ii) 2.0.

Step 3: Determine Bond Order for \(\mathrm{O}_{2}\)

\(\mathrm{O}_{2}\) has 12 electrons. The molecular orbital configuration is \((\sigma_{1s})^2(\sigma^{*}_{1s})^2(\sigma_{2s})^2(\sigma^{*}_{2s})^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2\). Calculating bond order: \((10 - 6)/2 = 2.0\). Thus, \(\mathrm{O}_{2}\) corresponds to bond order (ii) 2.0.

Step 4: Determine Bond Order for \(\mathrm{F}_{2}\)

\(\mathrm{F}_{2}\) has 14 electrons. The molecular orbital configuration is \((\sigma_{1s})^2(\sigma^{*}_{1s})^2(\sigma_{2s})^2(\sigma^{*}_{2s})^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi^{*}_{2p_x})^2(\pi^{*}_{2p_y})^2\). Calculating bond order: \((8 - 6)/2 = 1.0\). Hence, \(\mathrm{F}_{2}\) corresponds to bond order (i) 1.0.

Step 5: Determine Bond Order for \(\mathrm{O}_{2}^{+}\)

For \(\mathrm{O}_{2}^{+}\), remove 1 electron from \(\mathrm{O}_{2}\) which leaves 11 total electrons. The molecular orbital configuration is \((\sigma_{1s})^2(\sigma^{*}_{1s})^2(\sigma_{2s})^2(\sigma^{*}_{2s})^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^1\). Calculating bond order: \((10 - 5)/2 = 2.5\). Therefore, \(\mathrm{O}_{2}^{+}\) corresponds to bond order (iii) 2.5.

Key Concepts

Molecular Orbital Theory

The Molecular Orbital (MO) Theory is a fundamental concept in understanding the structure and behavior of molecules. Unlike the localized bonding theories, like the Valence Bond (VB) Theory, MO Theory delocalizes the electrons over the entire molecule. This means electrons are not assigned to individual bonds but rather spread out over various molecular orbitals.

• Molecular orbitals are formed by the linear combination of atomic orbitals (LCAO).
• These orbitals are spread across the entire molecule, which allows for a more accurate representation of electron behavior in molecules.
• They can be classified into bonding, antibonding, and non-bonding orbitals.

In MO Theory, bonding molecular orbitals result in lower energy and greater stability, as they increase the electron density between the nuclei. Antibonding orbitals, marked with an asterisk (*), have higher energy and decrease stability because they decrease electron density between the nuclei.
The energy levels of these molecular orbitals determine the molecular bond order, which is crucial in predicting the stability and bond strength within a molecule.

Stability of Chemical Bonds

Understanding the stability of chemical bonds is essential for predicting the behavior of molecules in reactions. Bond order is a key factor that influences this stability. The bond order is half the difference between the number of electrons in bonding orbitals and antibonding orbitals:
\[\text{Bond Order} = \frac{\text{Number of Bonding Electrons} - \text{Number of Antibonding Electrons}}{2}\]

• A higher bond order indicates a more stable and stronger bond.
• Stable molecules tend to have higher bond orders, while less stable molecules have lower bond orders.
• Molecules with a bond order of zero do not exist under normal conditions because they lack the stability provided by adequate bonding electrons.

For example,

• \(\mathrm{O}_2\) has a bond order of 2, indicating a double bond and stability.
• \(\mathrm{F}_2\) has a bond order of 1, denoting a single bond and relatively lower stability compared to \(\mathrm{O}_2\).

Recognizing the bond order helps in understanding reactivity and the type of bond (single, double, or triple) in a species.

Species and Electron Configuration

Electronic configurations are the backbone for determining how electrons are distributed in molecular orbitals. Each species, whether it is neutral or charged, has a unique electron configuration that influences its molecular geometry, reactivity, and overall properties.

• The electron configuration provides a map of where each electron in a molecule is likely to be found.
• This distribution helps us determine the bond length, bond angle, and potential energy of the molecule.
• Changes in the electron configuration, such as losing or gaining electrons, lead to different species with altered properties.

For instance, removing an electron from

• \(\mathrm{O}_2\) results in \(\mathrm{O}_2^+\), thereby increasing its bond order to 2.5. This change reflects an increase in bonding strength due to reduction in electron repulsion.
• Conversely, adding electrons leads to more antibonding interactions, decreasing stability, as seen in \(\mathrm{O}_2^{2+}\) with a bond order of 2.

Thus, by recognizing the electron configuration of different species, we can predict their stability and chemical behavior in various reactions.