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)
The total number of molecular orbitals formed from 2s and 2p atomic orbitals of a diatomic molecule is _________.