Download Activity 2 - Why Do the D-orbitals Give us Colour

Activity 2 - Why do the d-orbitals give us colour?
So now we know how we fill the electrons into the different orbitals we have to think about
the actual electronic configuration of some transition metal complexes. Let’s go back to
copper, and we’re going to think about the nice simple form, hexaqua copper, [Cu(H2O)6]2+
(here we have a Cu2+ ion and 6 water molecules connected to it).
When we have a positive ion, for example the Cu2+ above, the positive charge means we
have lost two electrons. When we lose electrons from most ions we simply reverse the order
we put them in, however in the transition metals we lose any electrons from the 4s shell
before we lose the 3d electrons. So for Cu2+ we lose 1 4s and 1 3d electron, which gives us
a 3d9 configuration.
Question: Draw or write out the d-electron configuration for 1) Mn5+ 2) V2+ 3) Cr3+
Going back to our hexaqua copper (II) complex, the copper in the centre is surrounded by
the six water molecules which are spread out as far as possible from each other, giving an
octahedral arrangement with the ligands all 90˚ from each other, the most sterically
favourable position, i.e. the position where the ligands are furthest from each other and
interact with each other least (see Figure 1).
Figure 1. The shape of the hydrated copper ion: the dashed lines are pointing backwards
into the page whereas the solid triangles are pointing out of the paper.
Question: What would be the most stable shape arrangement of ligands around a metal for 3
or 4 ligands? See if you can work out the angles the ligands would be from each other, you
might need to look up the names of the arrangements.
In the octahedral arrangement it is found that the different d-orbitals (remember Figure 3)
interact differently with the electrons in the ligands (remember the electrons in the ligands
will repel the electrons in the metal centre, hence raising the energy of these metal
electrons). To work out how the orbitals differ we have to consider that orbitals which point
directly towards the ligands, which lie on the axis, will feel more repulsion than ligands which
do not. From Figure 2 it can be seen that three of the orbitals, dxz, dxy, dyz, point between
the axis, i.e. away from the ligands, whilst two, dx2-y2 and dz2, point along the axis, i.e.
directly towards the ligands. The two orbitals which point towards the ligands are higher in
energy whilst the three which point between the ligands are lower in energy.
Figure 2. Showing the d-orbitals and how they interact with the ligands (small circles) which
lie on the axis.
This difference between the energy of the orbitals is called Δ, delta, and corresponds to the
energy of light in the visible part of the electromagnetic spectrum (see Figure 3). This means
that for an electron to be promoted (moved from the lower to the higher energy level) it will
absorb light from within the visible spectrum. As part of the spectrum has been absorbed
what is seen is the combination of the remaining parts of the spectrum. For example, if Δ
equates to the frequency of orange light then the solution will appear blue, the opposite
colour on the colour wheel (for a colour wheel see Figure 4).
Figure 3. Showing the energy gap, Δ, which corresponds to the energy needed for promotion
of an electron.
Figure 4. Simple colour wheel showing the opposite colours. For example if green is
absorbed the solution will appear red.
Question: Draw the electron configuration both before and after promotion of an electron for
V3+. Remember to draw the energy of the d-orbitals as has been done in Figure 3