Question

3)      a) When [Ni(NH₃)₄]²⁺ is treated with HCl, two compounds having the formula [Ni(NH₃)₂Cl₂] (designed X and Y) are formed. X can be converted into Y by diluting in dilute HCl. A solution of X reacts with oxalic acid to form Z complex [Ni(NH₃)₂C₂O₄], Y does not react with oxalic acid. Identify X, Y and Z complexes and draw the geometrical structure of them.

Answer 1

Нет a) [Ni(NH3)4]2+ → [Ni(NH3)242] This compound has two isomers (i) cis-[Ni(NH3), Ul.] and (ii) trans - [ Ni (NH3), eld] A solution of x reacts with oxalic acid to form z. oxalic acid is a bidentate ligand. highly probable to remove the by bidentate ligand. Hence cis - [Ni(NH3)2 Cl2]. The eis isomers provide to accomodate a x is It is usually eis ligands proper bite angles bidentate ligand NH3 the a NH₃ HQ C2 04 Ni Ni NH3 СА NH3 Х [Ni (NH3)2 (204) cu can be Hence Y is the trans is omer. U NH₃ since two ce groups are in Ni trans orientation it is not NH possible for C2042- to approach- y and bind with Ni: cis is omer easily converted to trans isomer since in case of cis is omer two u groups suffer Steric hinderence. convert to more stable There fore it will trans isomer by diluting in dilute Hel. He X cis and trans isomers of complex is as follows: OH2 H₂O b) The 0 K+ kt cm С» 0 HO OH2 trans cia

diagram isomer since the bottes are not rebelled we will predict the isomers by spectroscopie techniques. UV-vis spectroscopy will provide the information. 1) energy splitting of cis and trans isomer is reverse 2) Extent of splitting of cis is half to that of the trans is omer Number of peaks in trans is omer is double to the number of cis isomer. There fore shall first conduct the UV-vis spectroscopic study. The compound having higher number trans is omer of the compound and the other cis is omer. The ex periment should carried out using equal molar concentration. we of peaks i one be