# Which of these molecules is paramagnetic? "NO"^(-), "O"_2^(2-), "CO", or "CN"^-?

Jul 23, 2017

Only $N {O}^{-}$ 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.

$N {O}^{-}$ has $5 + 6 + 1 = 12 \cdot \text{valence electrons}$; ""^(-)C-=N has $4 + 5 + 1 = 10 \cdot \text{valence electrons}$; ""^(-)C-=O^+ has $4 + 6 = 10 \cdot \text{valence electrons}$.

For $O = O$ we also have $\text{12 valence electrons}$:

And thus highest occupied molecular orbitals are the degenerate antibonding orbitals, which are ${\pi}_{x}^{\text{*}}$ and ${\pi}_{y}^{\text{*}}$.

And, clearly, $\text{Hund's rule of maximum multiplicity}$ demands that they equally occupy the degenerate ${\pi}^{\text{*}}$ 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.

Jul 23, 2017

It would be ${\text{NO}}^{-}$.

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 ${\text{NO}}^{-}$ 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: ${\text{O}}_{2}^{2 +}$, $\text{CO}$, ${\text{NO}}^{+}$, and ${\text{CN}}^{-}$ are all diamagnetic with zero ${\pi}^{\text{*}}$ antibonding electrons.

DIOXIDE DIANION

One of the major breakthroughs of molecular orbital theory was to prove that ${\text{O}}_{2}$ was paramagnetic. This MO diagram should be in your textbook, and is also in anor's answer, where ${\text{O}}_{2}$ has two unpaired electrons in its ${\pi}^{\text{*}}$ antibonding molecular orbitals.

As a result, ${\text{O}}_{2}^{2 -}$ fills up the ${\pi}^{\text{*}}$ orbitals with two more electrons, making ${\text{O}}_{2}^{2 -}$ diamagnetic.

CARBON MONOXIDE

$\text{CO}$ is isoelectronic with ${\text{O}}_{2}^{2 +}$.

$\text{CO}$ has the second atom as $\text{C}$ instead of $\text{O}$, which gives two less electrons in the structure since $\text{C}$ has two less electrons than $\text{O}$.

Thus, $\text{CO}$ has no unpaired electrons compared to ${\text{O}}_{2}$ and is diamagnetic.

CYANIDE

${\text{CN}}^{-}$ is isoelectronic with $\text{CO}$. We know this because $\text{N}$ has one less electron than $\text{O}$, but the $\left(-\right)$ charge on ${\text{CN}}^{-}$ adds one electron.

Hence, ${\text{CN}}^{-}$ is diamagnetic too.

NITROUS OXIDE ANION

${\text{NO}}^{+}$ is isoelectronic with $\text{CO}$, because $\text{N}$ has one more electron than $\text{C}$, but the $\left(+\right)$ charge on ${\text{NO}}^{+}$ accounts for one less electron.

Therefore, ${\text{NO}}^{-}$ has two unpaired electrons and is paramagnetic.