1 answer

Which of these complexes has the lowest-energy #d-d# transition band in a UV-Vis spectrum?

Question:

#a)# #["CoBr"("NH"_3)_5]^(3+)# #b)# #["Co"("H"_2"O")("NH"_3)_5]^(3+)# #c)# #["Co"("CN"-kappa""C)("NH"_3)_5]^(2+)# #d)# #["Co"("NH"_3)_6]^(3+)#


Answers

The lowest-energy #d-d# transition band would mean that the ligand field splitting energy for the octahedral complex, #Delta_o#, is small.

That is when the complex has a high-spin configuration, and the complex has few strong-field ligands or many weak-field ligands.

In the spectrochemical series, the ligand field strength of each ligand is:

#overbrace("Br"^(-))^("pi donor")# #"<<"# #overbrace("H"_2"O")^("mostly sigma donor") < overbrace("NH"_3)^"sigma donor"# #"<<"# #overbrace("CN"^(-))^"pi acceptor AND sigma donor"#

And so we have...

#barul(|stackrel(" ")(" "Delta_o(A) < Delta_o(B) < Delta_o(D) < Delta_o(C)" ")|)#


First off, let's define #pi# acceptors, #sigma# donors, and #pi# donors... You may also want to read the this answer for a brief review of crystal field theory vs. ligand field theory.

Inorganic Chemistry, Miessler et al., pg. 371

  • #bbpi# acceptors interact in a backbonding interaction, where the metal #t_(2g)# orbitals donate electron density back into the ligand's antibonding orbitals.

    As a result, the #t_(2g)# orbitals are lowered in energy (because repulsions are lessened), increasing #Delta_o#.

  • #bbpi# donors donate electron density into the #t_(2g)# orbitals (which are the triply-degenerate, #pi#-compatible orbitals), destabilizing those orbitals and slightly decreasing #Delta_o#.
  • #bbsigma# donors donate electron density into the metal #sigma#-compatible orbitals, raising the energy of the metal antibonding orbitals, the #e_g^"*"# (in crystal field theory they may be labeled simply #e_g#), thus increasing #Delta_o#.

In short... having a lot of #pi# acceptors promotes low spin, having a lot of #pi# donors promotes high spin, and having a lot of #sigma# donors promotes low spin.

You can look at the spectrochemical series to check which ligands are strong-field and which are weak-field.
https://en.wikipedia.org/wiki/Spectrochemical_series

The one different ligand is highlighted in red and shall be considered.

#A)#

#["Co"color(red)("Br")("NH"_3)_5]^(2+)#, or pentamminebromocobalt(III), contains:

  • #"Br"^(-)# is a weak-field ligand, because it is a #bbpi# donor.
  • #"NH"_3# is a strong-field ligand, because it is a #bbsigma# donor.

Since #A# contains a weak-field ligand, it is expected to have a #Delta_o# that is the smallest among #A-D#.

#B)#

#["Co"color(red)("H"_2"O")("NH"_3)_5]^(3+)#, or pentammineaquacobalt(III), likewise has many strong-field ammine ligands, so this is going to be low spin, i.e. #Delta_o# is large.

In fact, water is a #sigma# donor (primarily), so #Delta_o# for this complex is larger than for #A#.

#C)#

#["Co"color(red)("CN")("NH"_3)_5]^(2+)#, or pentamminecyanocobalt(III), has five #"NH"_3# ligands, so you again know that this is going to have a large #Delta_o#. However, #"CN"^(-)#, cyanide, is both a #pi# acceptor and #sigma# donor, making it VERY strong-field.

Hence, this has the largest #Delta_o# among #A-C#.

#D)#

#["Co"color(red)("NH"_3)("NH"_3)_5]^(3+)#, or hexamminecobalt(III) (I wrote the formula like that on purpose), has all six #sigma# donors, and so it has a large #Delta_o#, being a low-spin complex.

Comparing, this has a larger #Delta_o# than #A# and #B#, but smaller than #C#.

Overall:

#color(blue)barul(|stackrel(" ")(" "Delta_o(A) < Delta_o(B) < Delta_o(D) < Delta_o(C)" ")|)#

.

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