Is #"V"("CN")_6^(3-)# diamagnetic or paramagnetic?

1 Answer
Truong-Son N.
Oct 31, 2017

To know that, determine the #d#-electron count. Consider the ligands.

  1. #"CN"^(-)# is a #sigma# donor and #pi# acceptor, donating #2# electrons into the metal.
  2. There are six #"CN"^(-)#, so they donate a total of #12# electrons into the molecular orbitals belonging to the ligands.
  3. The six #"CN"^(-)# contribute a total charge of #-6#.

Therefore, since the total complex charge is #-3#, #"V"# must be #+3#, i.e. a #d^2# metal. That makes this a #bb14#-electron complex.

Therefore, this is paramagnetic, because the #d# orbital splitting diagram only has two electrons, which stay unpaired whether this is high spin or low spin (it is low spin; the #Delta_o# is large from the #pi# acceptor and #sigma# donor nature of the cyanide ligand).

#" "" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "e_g#

#ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(color(white)(uarr darr))" "t_(2g)#

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