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

2 Answers

Answer:

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"#:

http://en.wikipedia.org

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.

Jul 23, 2017

It would be #"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 #"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.