Question

How can the colouration (or lack thereof) of the solutions be explained?
How would you expect DELTAO to change as the number of L ligands about the metal increases or decreases, and why?

UJUS ML PULUUPIUIUMLILI There are several types of spectrophotometers. Be sure to ask the instructor if you are unsure how to use the one you are assigned. You will be measuring absorbances at different wavelengths so it is essential to know how to quickly adjust the settings of the assigned spectrophotometer. Preparation of Sample Solutions Eight solutions containing various volumes of the two solutions of Ni2+ and ethylenediamine will be prepared. Use a large test tube to prepare the solutions and be sure to thoroughly mix the contents before measuring the absorbance. Solution Volume Ni2+ (mL) 4.0 #cicici Volume (en) (mL) 0.0 1.2 1.6 2.0 2.0 1.6 0.8 3.2 0.4 3.6 Record the absorbance of the eight solution mixtures at 578 nm. Before recording the absorbance of each solution, rinse the cuvette several times with small portions of the solution to be analysed, wipe clean and then insert into the compartment. After each measurement, discard the solution into a waste container and wash the cuvette with DDW. Make all measurements on the same instrument.

Answer 1

1)Colour of Transition metal ions arises due to d-d transition of the electron in the metal orbitals. When coordination complexes are formed, the d orbital of the metal is forced to split into two sets of eobitsls t2g and eg where t2g has lower energy. Due to partially filled d orbital of the transition metal ions, e excite from lower t2g orbital to higher eg orbital, when it absorbs energy. Upon emission of the absorbed energy, it emits wavelength which lies in the visible region, he ce giving it the complementary colour. The color also depends on the separation of the two orbitals and hence depends on the type of ligand attached. A strong field ligand will cause more separation btw orbitals and thus different intensity of light will be absorbed and emitted.

2) ∆o is the crystal field splitting energy. For various CN , ∆O CHANGES accordingly. For eg. When cn=4 and the symmetry is td, ∆t= 0.44 ∆o. But when cn= 4 and the symmetry is square planar, ∆sp=(1.3≈2)∆O.

When cn=8 and the symmetry is cubic , ∆cu=0.88∆o.. this happens due to change in splitting pattern of the d orbitals in different symmetrys.

Apart from no. Of ligands, ∆o majorly depends on the type of ligand. For a strong field ligand the ∆o will be larger while for a weak field ligand, it will be less.