Question:

Total number of electrons present in \( (\pi^*) \) molecular orbitals of \( \text{O}_2 \), \( \text{O}_2^+ \), and \( \text{O}_2^- \) is __________ .

Updated On: Nov 3, 2025
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Correct Answer: 6

Approach Solution - 1

To calculate the total number of electrons in the \( (\pi^*) \) molecular orbitals of \( \text{O}_2 \), \( \text{O}_2^+ \), and \( \text{O}_2^- \), we need to first understand the electronic configuration of molecular oxygen and its ions: \( \text{O}_2 \): The electronic configuration is \((\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi^*_{2p_x})^1(\pi^*_{2p_y})^1\). Thus, there are 2 electrons in the \( \pi^* \) orbitals. \( \text{O}_2^+ \): Removing an electron gives \((\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi^*_{2p_x})^1(\pi^*_{2p_y})^0\), with 1 electron in the \( \pi^* \) orbitals. \( \text{O}_2^- \): Adding an electron gives \((\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi^*_{2p_x})^2(\pi^*_{2p_y})^1\), with 3 electrons in the \( \pi^* \) orbitals. Adding the \( \pi^* \) electrons across these species: \(2+1+3=6\). The total number of \( \pi^* \) electrons is 6, which is within the provided range.

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Approach Solution -2

The electronic configuration of O$_2$ and its ions in terms of molecular orbitals is as follows: 
$\sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^2 \pi(2p_y)^2 = \pi(2p_y)^2 \pi^*(2p_x)^1 \pi^*(2p_y)^1$ 
For O$_2$: There are 2 electrons in $\pi^*$ orbitals. 
For O$_2^+$: 1 electron is removed from the $\pi^*$ orbitals, leaving 1 electron. 
For O$_2^-$: 1 electron is added to the $\pi^*$ orbitals, making it 3 electrons.
Total electrons in ($\pi^*$) molecular orbitals $= 2 + 1 + 3 = 6$
Final Answer: (6)

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