The question asks about the origin of the electron lost in the process \(\text{O}_2^+ \rightarrow \text{O}_2^{2+} + e^-\). To solve this question, we should analyze the electronic configuration of the \(\text{O}_2^+\) molecule.
To clarify which orbital loses the electron, let's follow these steps:
- The molecular orbital (MO) configuration for an oxygen molecule, \(\text{O}_2\), is: \(\sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^2 = \pi(2p_y)^2 \pi^*(2p_x)^1 = \pi^*(2p_y)^1\).
- For \(\text{O}_2^+\), one electron is removed, so the configuration becomes: \(\sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^2 = \pi(2p_y)^2 \pi^*(2p_x)^0 = \pi^*(2p_y)^1\).
- In the process \(\text{O}_2^+ \rightarrow \text{O}_2^{2+} + e^-\), another electron is removed.
- The electron removed from \(\text{O}_2^+\) to become \(\text{O}_2^{2+}\) is one from the bonding \(\pi\)-orbital (either \(\pi(2p_x)\) or \(\pi(2p_y)\)), leading to a change in the bonding characteristics of the molecule.
Hence, the electron that is removed is from the bonding \(\pi\)-orbital, making the correct answer:
Correct Answer: bonding \(\pi\)-orbital