[Fal]

How exactly do UV-Vis spectra and magnetic measurements reveal information regarding co-ordination geometry? Is there a direct link or more of a long process to understanding this?

[AWK]

http://www.chemicalforums.com/index.php?topic=25016.0

[cheese (MSW)]

Let me answer the original 2008 question.
Immediately recognize that the formula [Co2(MImT)2(NO3)2] is incorrect.  If there were two Co atoms you would get antiferromagnetic coupling between the unpaired e⁻s.  This would completely or partially quench the magnetic moment of the cmplx, and make the interpretation of the UV-vis spectrum extremely difficult.  Conclude formula is [Co(MImT)2(NO3)2].  This is a Co(II) cmplx that is [Ar] 3d^7.  We now have to determine the nature of the MImT ligand: is it monodentate or bidentate and what is the donor atom?  A survey of the literature reveals it to be a mondentate ligand that binds to Co^2+ via a S atom.  S is a soft base and Co^2+ is borderline H/S so consistent with HSAB theory (Wikipedia).  The [NO3]^- ligands presumably coordinate to the Co^2+ via a hard O donor atom (the cmplx is made in a nonaqueous solv).
What the problem is asking is [Co(MImT)4(NO3)2] octahedral (six coordinate) or tetrahedral: [Co(MImT)4]^2+ (NO3)2?
In the ligand environment (the crystal field) the Co^2+ 3d AOs do not have the same energies (they are not degenerate).  We will only consider the tetrahedral and octahedral cases (square planar is also a common possibility).
In the tetrahedral ligand field we have a d AO splitting diagram of e(↑↓) (↑↓)→Δtet→ (↑)(↑)(↑)t ; for the octahedral case it is 
t2g (↑↓) (↑↓)(↑)→Δoct→(↑)(↑)eg.   In both cases Δ is less than the pairing energy and so the cmplxs are high spin (true for all classical tet cmplxs and most M^2+ cmplxs; M = 1st Row TM).  As you can see both cases have 3 unp e⁻s and cannot at an undergrad level be distinguished by their magnetic moments (actually they can but let us not muddy the waters).  It is however found experimentally (and predicted by CFT) that Δtet ≈ Δoct/2 (4/9 in CFT).
We would therefore predict that if [Co(MImT)4(NO3)2] were octahedral the d-d transition would occur at significantly higher energies than the tet cmplx, that is, shorter wavelengths (the molar absorptivity also decreases).  This is not observed: the d-d transition is actually observed at slightly longer wavelengths that is Δtet for [Co(MImT)4]^2+ is less than that for Co(MImT)2(NO3)2.    Ligands can be arranged in order of their ligand field strengths; the order is called the spectrochemical series (Wikipedia) and S donor ligands are indeed weaker ligands than O donor ligands.  The UV-vis spectrum clearly indicates that the  [Co(MImT)4(NO3)2] is terahedral:  [Co(MImT)4]^2+ (NO3)2
Note that 525 nm corresponds to the yellow region of the visble spectrum and hence the complementary color is observed: these cmplxs are a deep blue.
Coordination compounds of 1-methylimidazoline-2-(3H)-thione and metal nitrates [M(II) = Co, Zn and Cd].  A spectroscopic, thermal analysis and x-ray diffraction study.  Raper, E. S.; Nowell, I. W. Inorganica Chimica Acta (1980), 43(2), 165-72.
...A single-crystal x-ray study confirmed the presence of an uncoordinated H2O mol. in Co(MImT)4(NO3)2.H2O.  The compd. also contains ionic nitrate groups and a slightly distorted tetrahedral cation with Co-S bonds..   
This is nicely laid out for the equilibrium
[Co(H2O)6]^2+  +  4Cl^-  ⇋  [CoCl4]^2-  + 6H2O
The aqua ion is pale pink (absorption in higher energy blue-green); [CoCl4]2- deeper blue (abs in the red)   
F. A. Cotton, G. Wilkinson, C. A. Murillo, M. Bochmann,  Advanced Inorganic Chemistry 6th ed (1999). P 820-821