Which of these molecules is paramagnetic? \(\mathrm{NO}^{-}, \mathrm{O}_{2}^{2-}, \mathrm{CO}, \text { or } \mathrm{CN}^{-} ?\)
Answer 1:
Only \(\mathrm{NO}^{-}\) is the paramagnet, so \(\text { option A. }\)
Explanation:
This is something you know or you don’t know. You cannot really work it out from the valence electronic structure.
\(N O^{-} \text {has } 5+6+1=12 \text {. valence electrons; }^{-} C \equiv N \text { has }\)
\(4+5+1=10 \cdot \text { valence electrons; }^{-} C \equiv O^{+} \text {has }\)
\(4+6=10 \cdot \text { valence electrons }\)
For \(O=O\) we also have \(12\) valence electrons:
And thus highest occupied molecular orbitals are the degenerate antibonding orbitals, which are \(\pi_{x}^{*}\) and \(\pi_{y}^{*}\).
And, clearly, \(\text { Hund’s rule of maximum multiplicity }\) demands that they equally occupy the degenerate \(\pi^{*}\) orbitals…….and hence a paramagnet.
Note that when we draw an analogous picture for dinitrogen, CONVENTIONALLY, \(\pi_{x}\) and \(\pi_{y}\), are LOWER in energy that \(\sigma_{z}\), But you will have to be in 3rd year inorganic to have to know this.
Answer 3:
It would be \(\mathrm{NO}^{-}\).
Here is a trick to do this question without having to invoke too many MO diagrams, as there is almost no way you could have seen the MO diagram of \(\mathrm{NO}^{-}\). without the impressive ability to draw one from scratch.
(But you’re in luck, because I’ve already drawn it out before if you want to see.)
Basically, know your isoelectronic species: \(\mathrm{O}_{2}^{2+}, \mathrm{CO}, \mathrm{NO}^{+}\), and \(\mathrm{CN}^{-}\) are all diamagnetic with zero \(\pi^{*}\) antibonding electrons.
DIOXIDE DIANION
One of the major breakthroughs of molecular orbital theory was to prove that \(\mathrm{O}_{2}\) was paramagnetic. This MO diagram should be in your textbook, and is also in anor’s answer, where \(\mathrm{O}_{2}\) has two unpaired electrons in its \(\pi^{*}\) antibonding molecular orbitals.
As a result, \(\mathrm{O}_{2}^{2-}\) fills up the \(\pi^{*}\) orbitals with two more electrons, making \(\mathrm{O}_{2}^{2-}\) diamagnetic.
CARBON MONOXIDE
\(CO\) is isoelectronic with \(\mathrm{O}_{2}^{2+}\).
\(CO\) has the second atom as \(C\) instead of \(O\), which gives two less electrons in the structure since \(C\) has two less electrons than \(O\).
Thus, \(CO\) has no unpaired electrons compared to \(\mathrm{O}_{2}\) and is diamagnetic.
CYANIDE
\(\mathrm{CN}^{-}\) is isoelectronic with \(CO\). We know this because \(N\) has one less electron than \(O\), but the
\((−)\) charge on \(\mathrm{CN}^{-}\) adds one electron.
Hence, \(\mathrm{CN}^{-}\) is diamagnetic too.
NITROUS OXIDE ANION
\(\mathrm{NO}^{+}\) is isoelectronic with \(CO\), because \(N\) has one more electron than \(C\), but the \((+)\) charge on \(\mathrm{NO}^{+}\) accounts for one less electron.
Therefore, \(\mathrm{NO}^{-}\) has two unpaired electrons and is paramagnetic.