Question

In the formation of $O_2^{+}$ from $O_2$, the electron is removed from:

• A. $\sigma -$ orbital
• B. $\pi -$ orbital
• C. ${\sigma }^{\ast }-$ orbital
• D. ${\pi }^{\ast }-$ orbital

Answer 1

The correct option is D ${\pi }^{\ast }-$ orbital

The electronic configuration of $O_2$ is: $\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2pz}{)}^2\right({\pi }_{2px}{)}^2\left({\pi }_{2py}{)}^2\right({\pi }_{2px}^{\ast }{)}^1({\pi }_{2py}^{\ast }{)}^1$

So, in the formation of $O_2^{+}$ from $O_2$, the electron will be removed from ${\pi }^{\ast }$ orbital because it is the highest occupied molecular orbital.