Which of the following complexes exhibit the highest paragmagnetic behavior? a. [Co(ox)2(OH)2]^- b. [Ti(NH3)6]^3+ c. [V(gly)2(OH)2(NH3)2]^+ d. [Fe(en)(bpy)(NH3)2]^2+

1 Answer
Mar 11, 2018

I found #["Co"("ox")_2("OH")_2]^(-)# to have four unpaired electrons. That would make it the most paramagnetic.


I would first reference your text to check the charge of the ligands, but you would have to memorize these in an exam. I have a naming guide here.

  • #"ox"# is oxalato, a #-2# bidentate ligand.
  • #"gly"# is glycine, a neutral ligand at #"pH"# #7#. Since it has a carboxyl group, it is possible that it binds #eta^2#. It is also possible that it binds bidentate at basic #"pH"#.
  • #"OH"# is hydroxide, a #-1# monodentate ligand.
  • #"en"# and #"bipy"# are both neutral bidentate ligands.

From that, we can then determine the oxidation states on each metal and thus their number of unpaired electrons.

  • #["Co"("ox")_2("OH")_2]^(-)# has two #-2# bidentate and two #-1# monodentate ligands, giving a #+3# oxidation state on cobalt, making it a #d^6# metal in the octahedral ligand field.

Since #"ox"^(2-)# and #"OH"^(-)# are both fairly weak-field ligands (weaker sigma donors than water), this is likely a high spin #d^6# complex, with four unpaired electrons.

https://chem.libretexts.org/

  • #["Ti"("NH"_3)_6]^(3+)# has three neutral ammine ligands, each strong-field, making this a low-spin complex involving a #d^1# metal. Only one unpaired electron.

  • #["V"("gly")_2("OH")_2("NH"_3)_2]^(+)# has two #-1# hydroxo ligands, two neutral ammine ligands, and two neutral monodentate #eta^2# glycine ligands (at #"pH"# #7#), making this #"V"^(+3)#, a #d^2# metal with two unpaired electrons in the octahedral ligand field splitting diagram.

https://chem.libretexts.org/

I did not need to think about strong- or weak-field for either of the previous two complexes. Why is that?

  • #["Fe"("en")("bpy")("NH"_3)_2]^(2+)# has two neutral bidentate ligands (ethylenediamine and 2,2'-bipyridine), and two neutral monodentate ammine ligands, making iron a #+2# oxidation state. We therefore have a #d^6# metal in an octahedral ligand field consisting of three species of strong-field ligands, giving a low-spin #d^6# configuration, having diamagnetism.

https://chem.libretexts.org/