Question

In which of the following ionization processes the bond energy increases and the magnetic behaviour changes from paramagnetic to diamagnetic :

• A. $O_2 \to O_2^+$
• B. $C_2 \to C_2^+$
• C. $NO \to NO^+$
• D. $N_2 \to N_2^+$

Answer 1

The Correct Option is C

To determine in which ionization process the bond energy increases and the magnetic behavior changes from paramagnetic to diamagnetic, let's analyze each given option based on molecular orbital (MO) theory and electronic configuration changes.

Concepts Required:

• Paramagnetism vs. Diamagnetism: A molecule is paramagnetic if it has unpaired electrons and diamagnetic if it has all paired electrons.
• Bond Energy and Bond Order: Bond energy generally increases with bond order, which is the number of covalent bonds between two atoms. Bond order is related to the stability and strength of a bond.

Solution Analysis:

1. \( O_2 \to O_2^+ \):
   • Oxygen (\(O_2\)) has a molecular electronic configuration of: \( \sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^2 = \pi(2p_y)^2 \pi^*(2p_x)^1 \pi^*(2p_y)^1 \).
   • It has unpaired electrons in the \(\pi^*\) orbitals, making it paramagnetic. Bond order = 2.
   • Upon ionization to \(O_2^+\), it loses one electron from a \(\pi^*\) orbital. The bond order becomes 2.5 (more stable), but it remains paramagnetic as there is still an unpaired electron.
2. \( C_2 \to C_2^+ \):
   • Carbon (\(C_2\)) has a molecular electronic configuration: \( \sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x)^2 \pi(2p_y)^2 \sigma(2p_z)^0 \).
   • All electrons are paired, making it diamagnetic. Bond order = 2.
   • After removing an electron (ionization to \(C_2^+\)), the bond order decreases, which typically corresponds to reduced bond energy. Magnetic property does not show a change from paramagnetic to diamagnetic.
3. \( NO \to NO^+ \):
   • Nitric oxide (\(NO\)) has a configuration: \(\sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^2 = \pi(2p_y)^2 \pi^*(2p_x)^1 \).
   • It has an unpaired electron in a \(\pi^*\) orbital, making it paramagnetic.
   • When ionized to \(NO^+\), it loses the unpaired electron, resulting in a fully paired electronic configuration. This changes its behavior to diamagnetic. Simultaneously, the bond order increases from 2.5 to 3, indicating increased bond energy.
4. \( N_2 \to N_2^+ \):
   • Nitrogen (\(N_2\)) has the configuration: \( \sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x)^2 \pi(2p_y)^2 \sigma(2p_z)^2 \).
   • All electrons are paired, making it diamagnetic. Bond order = 3.
   • Upon ionization to \(N_2^+\), the bond order decreases to 2.5 leading to a decrease in bond energy. The magnetic behavior remains unchanged as diamagnetic.

Conclusion:

The correct answer is \(NO \to NO^+\) because this ionization process results in an increase of bond energy and a change from paramagnetic to diamagnetic behavior.