Question

The pairs of species of oxygen and their magnetic behaviours are noted below. Which of the following presents the correct description?

• A. $O^{+},O_2^{2-}$ - Both paramagnetic
• B. $O,O_2^{2-}$ - Both paramagnetic
• C. $O_2^{-},O_2^{2-}$ - Both diamagnetic
• D. $O_2^{+},O_2$ - Both paramagnetic

Answer 1

Answer: (D)

To determine the magnetic behavior of the molecules, you will first have to draw the molecular orbital diagram depending on the no. of paired or unpaired no. of electrons the molecule orbital, we can conclude that the molecule is diamagnetic or paramagnetic respectively.

Whenever two electrons are paired together in an orbital, or their total spin is $0$,they are diamagnetic.

A paramagnetic electron is an unpaired electron. So the orbital in this case will have a net spin

A. $O$ and $O_2^{2-}$

Let us see the electronic configuration of Oxygen$=1s^{2}2s^{2}2p^2$ (Image 1)

The oxygen atom has $2$ unpaired electrons in the $2p$ orbital and based on the above discussion, we can say that it is paramagnetic.

Now the $O_2^{2-}$ molecular ions magnetism can be checked by drawing the molecular orbital diagram.

$O_2^{2-}$ (image 2)

As you see the $\pi 2p_z$ and $\pi 2p_x$ have paired no. of electrons. Hence $O_2^{2-}$ is diamagnetic

B. $O_2^{-}$ and $O_2^{2-}$

$O_2^{2-}$ is diamagnetic as we have already seen above. Now let us see if $O_2^{-}$ is diamagnetic too.

$O_2^{-}$ (image 3)

$\pi 2p_y$ has one unpaired electron, hence $O_2^{-}$ is paramagnetic

C.$O^{+}$ and $O_2^{2-}$

$O_2^{2-}$ is diamagnetic which has already been proved earlier and $O^{+}$ atom will have an electronic configuration as $=1s^{2}2s^{2}2p^1$ (Image 4)

So as you can see, $O^{+}$ is paramagnetic, but since $O_2^{2-}$ is diamagnetic, this option is also not correct.

A. $O_2^{+}$ and $O_2$

(Image 5)

Both $O_2^{+}$ and $O_2$ has one and two unpaired electrons in their molecular orbitals, which means both the molecules have a net spin and both are paramagnetic.

Ans: The correct description is given in option D only.