Question

According to molecular orbital theory for $O_2^{+}$:

• A. Bond order is less than $O_2$ and $O_2^{+}$ is paramagnetic
• B. Bond order is more than $O_2$ and $O_2^{+}$ is paramagnetic
• C. Bond order is less than $O_2$ and $O_2^{+}$ is diamagnetic
• D. Bond order is more than $O_2$ and $O_2^{+}$ is diamagnetic

Answer 1

The correct option is B Bond order is more than $O_2$ and $O_2^{+}$ is paramagnetic

For $O_2$
$\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2p_z}{)}^2\right({\pi }_{2p_x}{)}^2=\left({\pi }_{2p_y}{)}^2\right({\pi }_{2p_x}^{\ast }{)}^1=\left({\pi }_{2p_y}^{\ast }{)}^1\right({\sigma }_{2p_z}^{\ast }{)}^0$

For $O_2^{+}$
$\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\sigma }_{2p_z}{)}^2\right({\pi }_{2p_x}{)}^2=\left({\pi }_{2p_y}{)}^2\right({\pi }_{2p_x}^{\ast }{)}^1=\left({\pi }_{2p_y}^{\ast }{)}^0\right({\sigma }_{2p_z}^{\ast }{)}^0$

Bond order for $O_2=2$ and for $O_2^{+}=2.5$
Both are paramagnetic ($O_2$ has 2 unpaired electrons; $O_2^{+}$ has one unpaired electron).