[lrw1793]

Just working through some past papers and just not sure if I've got the right idea here, thanks if you can *delete me*

1. Explain why solutions of ferrocyanide ion, [Fe(CN)6]2–, are colourless, but solutions of ferricyanide, [Fe(CN)6]3–, appear strongly red coloured.

[Fe(CN)6]2– CN weak field - high spin. Fe d4
[Fe(CN)6]3– d5 high spin
Surely [Fe(CN)6]3– should be colourless as there are no spin-allowed d-d transitions?

2. Which has the lowest energy d-d transition?
[Co(CN)6]3– or [Ir(CN)6] 3–
both high spin d6. Ir has a larger octahedral splitting as it has a higher principal quantum number - why is this? [Co(CN)6]3– will have lower energy d-d transition

[Co(NH3)5Cl]2+ or [Co(NH3)5(NO2)]2+
[Co(NH3)5Cl]2+ NH3 is strong field, Cl is weak field, so this will have the lowest energy transition
[Co(NH3)5(NO2)]2+ NH3 strong field, NO2 stronger field. This will therefore have greater splitting

3. In each of the following complexes, state the type of the transitions you expect to observe in their UV-vis absorption spectra and comment on the strength of the transitions.

[Mn(H2O)6]2+ d5 high spin - only weak spin forbidden and Laporte forbidden transitions
[MnO4]– d0 no d-d transitions
[Ru(NH3)6]2+ d6 low spin - d-d transitions observed
[NiBr4]2- d8 low spin tetrahedral - d-d transitions observed
How do I know how strong d-d transitions are? Is charge transfer observed for every TM complex?

4. Explain why solutions of [Ni(NH3)6]2+ and [Ni(H2O)6]2+ appear violet and green,
respectively, while solutions of [NiCl4]2– are blue.

[Ni(NH3)6]2+ strong field ligand. d8 low spin. Increased octahedral splitting. Absorbs yellow light
[Ni(H2O)6]2+ weak field ligand. d8 low spin. Decreased octahedral splitting. Absorbs lower energy red light
[NiCl4]2– tetrahedral. d8 low spin. Tetrahedral splitting much smaller than octahedral therefore absorbs lowest energy light. However to appear blue it must absorb orange which is higher energy?

5. In the following pairs of complexes, determine the d orbital configurations of
the metals and identify which one has the higher Ligand Field Stabilisation
Energy (LFSE). Explain your answer.
[TiF6]3– and [Ti(H2O)6]3+
both d1, F is weaker field than H2O, [Ti(H2O)6]3+ has largest splitting therefore higher LFSE
[Mn(H2O)6]3+ and [Mn(CN)6]3-
both d3 CN is strong field so has larger splitting and higher LFSE.
[Fe(CN)6]2– and [Ru(CN)6]2–
Both d6, Ru has larger n therefore larger splitting and higher LFSE
[Co(NH3)6]2+ and [CoCl4] 2–
both d7, octahedral splitting is larger so [Co(NH3)6]2+

[Ni(CN)4]2– and [NiCl4] 2–
[Ni(CN)4]2–  is square planar and [NiCl4] 2– is tetrahedral – not sure on this one

5. Which of the following transitions would you expect to be the strongest in the
absorption spectra of the given complexes? Explain your answer.
d-d or charge transfer in [NiCl4] 2–
charge transfer is always stronger as it is both Laporte and spin allowed, d-d transitions are only spin allowed
d-d in [NiI4]2– or [Ni(NH3)6]2+
[NiI4]2– d8 tetrahedral
[Ni(NH3)6]2+ d8 octahedral – this will be strongest as octahedral splitting>tetrahedral splitting

6. V3+ ions form octahedral complexes with ligands X, Y. The complex with X,
[VX6]3+, is yellow and with Y, [VY6]3+, is blue green. Which of the two
ligands leads to higher LFSE? Rationalise your answer.
V3+ d2
[VX6]3+ appears yellow, absorbs violet
[VY6]3+ appears blue green, absorbs yellow/orange
Violet is a higher energy than yellow orange, therefore [VX6]3+ has larger octahedral splitting and the LFSE will be higher.

[kelvinLTR]

Given the ability of backbonding, shouldn't Cyanido ligand be a strong ligand?

[kelvinLTR]

2. Which has the lowest energy d-d transition?
[Co(CN)6]3– or [Ir(CN)6] 3–
both high spin d6. Ir has a larger octahedral splitting as it has a higher principal quantum number - why is this? [Co(CN)6]3– will have lower energy d-d transition

Ir, being the larger metal ion (higher principle quantum number) has a higher surface area for the ligands to be attached to. Therefore the ligands, without having to face ligand ligand repulsions, will move closer to metal empty d orbitals, thus increasing the repulsions between metal d orbitals and ligand orbitals. This will mean the splitting would be higher for orbitals along the axes, and thus a higher crystal filed splitting energy.

[kriggy]

Ferrocyanide is 4- not 2-. And the other thing, CN makes the strongest ligan fields so where did you get the information that 
Fe(CN)6]2– has weak field?

[lrw1793]

Ferrocyanide is 4- not 2-. And the other thing, CN makes the strongest ligan fields so where did you get the information that 
Fe(CN)6]2– has weak field?

That's the question as it is in the exam paper  Oh yeah, my bad don't know why I put that!