Why would there be differences in UV/Vis spectrum?

$\ce{[NEt4]2[NiBr4]}$ dissolved in $\ce{MeCN}$ and $\ce{[NEt4]2[NiBr4]}$ dissolved in $\ce{H2O}$ have different UV/visible spectra in the visible region. How would they differ, for example would one of them have more peaks? Also how would the molar extinciton coefficients differ and what do these numbers tell you?

• This seems a very strange molecule/complex to me. Assuming bromine has a -1 oxidation state and further assuming $\ce{[NEt_{4}]^{+}}$ it would leave nickel in a -1 oxidation state. I was unable to find this compound, could you add a reference or synthesis? The only compound that I found referenced in Greenwood/Earnshaw with Ni (-I) was $\ce{[Ni2(CO)6]^{2-}}$. – Martin - マーチン Jan 19 '15 at 7:25

$\ce{H2O}$ does not absorb the radiation almost at all, so it will not interfere in the absorbance of the analyte we are analysing, thats why we use water almost always as a solvent in instrumental methods of analysis.Therefore $\ce{MeCN}$ wil absorb some of the radiation so then of course the absorbance of the analyte will also change ( decrease ) and the spectrum will be different ( the peaks of the analyte will be lower ).
I haven't looked for the spectrums for the compound in the different solvents, so I can only theorize what the differences would be. Since the nickel ion has the energy levels for visible light, then the differences would be due to what surrounds the nickel ion. In water, the nickel ion would be surrounded by water, where as in acetonitrile, the molecule $\ce {[NEt4]2[NiBr]}$ would stay together.