To have a color, the transition metal complex
should have an electron in a d-orbital, and an
unoccupied spot in a higher energy d-orbital.
• Practice Problem.
• Predict whether each of the following complexes
could be colored: (a) [Sc(H2O)6]3+ (b)
[Ti(H2O)6]3+ (c) TiO2 (d) CrO2 and, finally, (e)
Cu[(H2O)6]2+.
and
Practice Problem.
• Why is it that the d-electrons may form
weaker or different bonds?
Sc^+3 has no unpaired electrons; therefore, probably is it not colored.
b. Ti(H2O)6^+3. Ti metal is 3d2 4s2. The +3 ion will be 3d1 and probably will be colored.
c. TiO2, has 3d24s2 for the Ti metal. In TiO2 Ti has oxidation state of +4. If we give the 3d2 and the 4s2 to O, it leaves Ti with no unpaired electrons and no color. TiO2, as you know, is a white compound.
Cu metal is 3d9 4s1.
Cy^+2 is 3d8. With H2O, it probably is
6 paired electrons in t2g and 1 unpaired electron in each of the upper level eg. Probably colored since we have electrons at the lower level and two holes at the upper level.
This should give you an idea how to do the others.