E FOLLOWING QUESTIONS MAY BE COMPLETED PRIOR TO LAB OR DURING LAB YOUMAY SEEK AS
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E FOLLOWING QUESTIONS MAY BE COMPLETED PRIOR TO LAB OR DURING LAB YOUMAY SEEK ASSISTANCE FROM AN INSTRUCTOR IF NEEDED. an solution of aickel nitrate will be available in the laboratory. Observe this solution and then the following questions. Color of nickel nitrate solution tiewe ihely bhoien (Refer to Table 1, page 1.) Predict the color that nickel nitrate likely absorbs in terms of energy, explain what difference must exist between ions that produce different colored solutions. Once a spectrophotometer has been "blanked" at the desired wavelength and is ready to measure the absorbance of your sample, explain how the spectrophotometer takes this measurement once your sample has been inserted into the instrument. Lab 3-Visible Spectroscopy-An Introduction to Beer's Law Pa lisbury University Department of ChemistryExplanation / Answer
Color of nickel nitrate solution is green
Wavelength range, where Nickel nitrate mostly absorb is in between 350 and 450 nm with a maximum absorbance at 393 nm [ Here table 1 is not provided]
Nickel nitrate appears green because it absorbs red and blue color of the visible spectrum.
For ions to exhibit color, the energy required to transfer an electron from one orbital to another must be in the visible range. This energy difference will give characteristic color to the particular ion. This is why zinc sulphate is white, here the d orbitals are completely filled and it is not possible for the d electron transition. On the other hand, nickel nitrate is green colored, here electronic transition in the d orbital is possible and the corresponding energy difference will result a green colored compound.
The spectrophotometer first measures the intensity of light passing through the blank solution and then the intensity of light passing a sample. Percentage of transmittance is obtained as the percentage of light of a certain wavelength that passed through the solution relative to the blank (Solvent). Absorbance is obtained as the negative logarithm of percent transmittance divided by 100.
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