Question

Among the following molecules/ions, $$C_2^{2-},N_2^{2-},O_2^{2-},O_2$$ which one is diamagnetic and has the shortest bond length?
A) $$C_2^{2-}$$
 B)$$N_2^{2-}$$
 C)$$O_2$$
 D)$$O_2^{2-}$$

Answer 1

Hint: If a molecule/ion has an unpaired electron then we can say it is paramagnetic, and if it has all paired electrons, then it will be diamagnetic. The bond length of the molecule depends on the bond order of the molecule with an inverse proportional relationship.

Complete step-by-step answer:
We can easily differentiate paramagnetic and diamagnetic molecules or ion on the basis of their electronic configuration. Paramagnetic molecules are attracted to magnetic field because they have unpaired electrons, whereas diamagnetic molecules are repelled by the magnetic field and all the electrons have paired spins.
We can also say that an odd number of electrons can produce paramagnetic ion or molecule and sometimes molecules with even number of electrons may also be paramagnetic such as oxygen molecule. To find out which molecule is diamagnetic or paramagnetic, let us first look at their electronic configuration in accordance with Molecular Orbital Theory.
Electron spin is very important in order to determine the magnetic properties of an atom. In diamagnetic case, all of the electrons are paired up and share their orbital with another electron, due to this their total spin becomes zero as they are aligned in anti-parallel fashion, so the atom becomes diamagnetic.
 $$C_2^{2-}$$ , the total number of electrons are 14 i.e., 6 from each carbon atom and two from negative charges. The electronic configuration is $$(\sigma 1s{)}^2({\sigma }^{\ast }1s{)}^2(\sigma 2s{)}^2({\sigma }^{\ast }2s{)}^2(\sigma 2p_z{)}^2(\pi 2{p_x}^2)(\pi 2p_y{)}^2$$ . We can see that there are no unpaired electrons, so it a diamagnetic ion.
 $$N_2^{2-}$$ , the total number of electrons are 16 i.e., 7 from each nitrogen atom and two from negative charges. The electronic configuration is $$(\sigma 1s{)}^2({\sigma }^{\ast }1s{)}^2(\sigma 2s{)}^2({\sigma }^{\ast }2s{)}^2(\sigma 2p_z{)}^2(\pi 2{p_x}^2=\pi 2p_y{)}^2(\pi 2{p_x}^1=\pi 2{p_y}^1)$$ . We can see that the last two electrons enter in different degenerate orbitals and are unpaired, so, it is a paramagnetic molecule.
 $$O_2$$ , the total number of electrons are 16 i.e., 8 from each oxygen atom. The electronic configuration is $$(\sigma 1s{)}^2({\sigma }^{\ast }1s{)}^2(\sigma 2s{)}^2({\sigma }^{\ast }2s{)}^2(\sigma 2p_z{)}^2(\pi 2{p_x}^2=\pi 2p_y{)}^2(\pi 2{p_x}^1=\pi 2{p_y}^1)$$ . We can see that the last two electrons enter in different degenerate orbitals and are unpaired, so, it is a paramagnetic molecule.
 $$O_2^{2-}$$ , the total number of electrons are 18 i.e., 8 from each oxygen atom and two from negative charges. The electronic configuration is $$(\sigma 1s{)}^2({\sigma }^{\ast }1s{)}^2(\sigma 2s{)}^2({\sigma }^{\ast }2s{)}^2(\sigma 2p_z{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2({\pi }^{\ast }2p_x{)}^2({\pi }^{\ast }2p_y{)}^2$$ . We can see there are no unpaired electrons, so it is a diamagnetic ion.
Now we know that two of them are diamagnetic and the rest two are paramagnetic, let us look at their bond order from their electronic configurations.
 $$bondorder={\displaystyle \frac{(bondingelectrons-antibondingelectrons)}{2}}$$
Bond order is inversely proportional to the bond length. So, the one which will have highest bond order will tend to have shortest bond length. From the below table, we can see that $$C_2^{2-}$$ has highest bond order of 3, so it will have shortest bond length.


We know now that $$C_2^{2-}$$ is diamagnetic as well as it has shortest bond length. Hence, the correct option is (A).

Note: In case we don’t remember the Molecular Orbital Theory, we can determine experimentally if a substance is diamagnetic or paramagnetic by using Gouy balance. It measures the apparent change in the mass of the substance as it is attracted or repelled by the region of high magnetic field between the poles.