# The ways to determine if an ionic compound is paramagnetic or not

Find which of $$\ce{Na2O2}$$ and $$\ce{CsO2}$$ is paramagnetic.

The given solution involves removing the cations and taking the anionic part, i.e. $$\ce{O2^2-}$$ and $$\ce{O2-}$$, and then calculating the MO diagrams of the anions.

However, why is it that we don't have to consider the cationic part for magnetic behaviour? The calculations I had in my textbook were more direct because I'd have binary compounds of form $$\ce{X-Y}$$, then I'd take MO of the whole compound and see if it has odd or even electron, and see if the total compound has a bond order $$\gt 0$$.

Why do we only have to look at MO diagram of anion to determine if an ionic compound is paramagnetic or not in this case?

• The cationic part has the electron configuration of an inert gas, hence it does not matter. These are alkaline metals, after all. Aug 2, 2021 at 13:01
• So, the method described is not a general procedure? @IvanNeretin Aug 2, 2021 at 13:02
• With transition metals, you might want to consider the MO of the whole compound, like you said. Aug 2, 2021 at 13:03
• To put it another way you always have to look at the whole system, but the ionic model factors it into two (or more) parts which can be considered independently. Na+ and Cs+ are "clearly" diamagnetic being isoelectronic with noble gases, so in such cases you only need consider the anion. Aug 2, 2021 at 15:40
• I just can not understand how this question was marked as needs more focus... Aug 3, 2021 at 3:04

As mentioned by @IvanNeretin and @IanBush in the comments, the cation in this case can be neglected as they have noble gas configuration, and therefore are diamagnetic.

In general, for an ionic solid, it is possible to consider the magnetic effect of the ions separately. As there is no covalent bond, it isn't particularly necessary to look at the molecular orbitals (MO) of the system. This is true, even for many transition metal compounds, such as magnetite ($$\ce{Fe3O4}$$), which has $$\ce{Fe(II)}$$ and $$\ce{Fe(III)}$$ centres. The magnetic behaviour of the solid can be reasonably explained by considering each of the iron ions as a separate magnetic dipole. The oxygen stays as oxide ($$\ce{O^2-}$$) which has no unpaired electrons, and is diamagnetic.

There are some interesting cases, where covalent molecules show magnetic behaviour due to unpaired electrons. In those cases, you have to consider the MO of the whole molecule, there is no way to consider the individual atoms as magnetic.

For example, this compound:

This is a spontaneous charge transfer salt that is produced when you react decamethyl ferrocene with tetracyanoethylene under the right conditions. The ferrocene moiety gives up an electron to the $$\pi^*$$ orbitals of tetracyanoethylene. The tetracyanoethylene radical anion formed is magnetic due to the unpaired electron. The iron is in $$\ce{Fe(III)}$$ state and is also magnetic.

In solid state, the ions are close enough for the spins to couple, and the solid becomes a ferromagnet below $$\pu{4.8 K}$$.

Incidentally, this is the first compound isolated where part of the magnetism is coming from unpaired electrons that are in p-orbitals (i.e. MOs formed from p-orbitals). All previously known magnetic materials had d-block elements.

• If one ion is diamagnetic and another is para, then what is the magnetic prop of compound in total? Aug 11, 2021 at 5:21
• @Buraian Paramgnetism is almost always stronger than diamagnetism, and overrides that effect. I use the word "almost" because I am not completely sure, but I have never come across any instance where diamagnetism was stronger than para. Think about it—paramagnetic materials also have paired electrons, if their effect was stronger than the one or two unpaired electrons, there wouldn't be any paramagnetic material!! Aug 11, 2021 at 8:10
• KMnO4 is an example where the paramagnetic and diamagnetic terms are comparable - but this is not the usual unpaired electron paramagnetism, which as you say (always?) trumps the diamagnetic term - tandfonline.com/doi/abs/10.1080/00268976000100311 en.wikipedia.org/wiki/Van_Vleck_paramagnetism Aug 12, 2021 at 8:22