Question

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

Answer 1

Only `NO^-` is the paramagnet, so `"option A"`.

Explanation:

This is something you know or you don't know. You cannot really work it out from the valence electronic structure.

`NO^-` has `5+6+1=12*"valence electrons"`; `""^(-)C-=N` has `4+5+1=10*"valence electrons"`; `""^(-)C-=O^+` has `4+6=10*"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, `"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 2

It would be `"NO"^(-)`.

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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 `"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: `"O"_2^(2+)`, `"CO"`, `"NO"^(+)`, and `"CN"^(-)` are all diamagnetic with zero `pi^"*"` antibonding electrons.

DIOXIDE DIANION

One of the major breakthroughs of molecular orbital theory was to prove that `"O"_2` was paramagnetic. This MO diagram should be in your textbook, and is also in anor's answer, where `"O"_2` has two unpaired electrons in its `pi^"*"` antibonding molecular orbitals.

As a result, `"O"_2^(2-)` fills up the `pi^"*"` orbitals with two more electrons, making `"O"_2^(2-)` diamagnetic.

CARBON MONOXIDE

`"CO"` is isoelectronic with `"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 `"O"_2` and is diamagnetic.

CYANIDE

`"CN"^(-)` is isoelectronic with `"CO"`. We know this because `"N"` has one less electron than `"O"`, but the `(-)` charge on `"CN"^(-)` adds one electron.

Hence, `"CN"^(-)` is diamagnetic too.

NITROUS OXIDE ANION

`"NO"^(+)` is isoelectronic with `"CO"`, because `"N"` has one more electron than `"C"`, but the `(+)` charge on `"NO"^(+)` accounts for one less electron.

Therefore, `"NO"^(-)` has two unpaired electrons and is paramagnetic.