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The ways to determine if an ionic compound is paramagnetic or not
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Kevin Fisher
The ways to determine if an ionic compound is paramagnetic or not
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.
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.
@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 wouldnt be any paramagnetic material!!More
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 - More
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.
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.
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