Question

4. (15) Predict the following equilibria and briefly explain your rational in the allocated space H2 (a) [K(OH2)c)* + 25 Luminor, + 2 H2O Your answer: oth Hz (b) Ca(SR)2 + 2 NOOR Ca(OR)2 + 2 NASR Your answer: NH ΝΗ, + Feci „NH ci NH- Your answer: Met Met Protein (d) + HER2 HER2 Met Met Your answer: (e) TICLA Your answer:

Answer 1

Solution: OH2 он, a) [K(OH2)c]* +2 S S H2O OH2 + 2 H2O Answer: The equilibria for the above reaction will be shifted towards the left i.e. to the reactant side as Kt is a hard center and sis soft center. Therefore, S will not prefer to bind with K+. Besides, the total entropy change for the reaction is zero and doesn't drive the equilibria to the right. b) Cd(SR)2 + 2 NaOR Cd(OR)2 +2 NaSR Answer: The equilibria for the above reaction will be shifted towards the left i.e. to the reactant side as Cd+2 is a soft center, S is soft center as well unlike the case for O atom which is considered to be hard and attached with another hard center, Sodium (Na). Therefore, Cd2+ will not prefer to bind with O as it is already stabilized with S-atom. The total entropy change doesn't play a role for this reaction (AS = 0).

c) + FeCl3 NH + FeCl3 NH - NH- Answer: The equilibria for the above reaction will be shifted towards the left i.e. to the reactant side. Because, even after forming a co-ordination complex product, the oxidation state of Fe doesn't change (+3 in both cases). Both N and Fe center here are neither extremely hard nor soft. Therefore, the equilibria will be determined by entropy which is not favourable to the reactant side (AS = -ve). + HgR2 > Net Net HgR2 Met Met Answer: The equilibria for the above reaction will be shifted towards the left i.e. to the reactant side. Both "Met" and Hg2+ are soft center, therefore, bond formation is feasible. However, the total entropy change for the reaction is negative which is not favourable. Besides, addition of bulky HgR2 via co-ordinate bond can create some strain in the cyclic protein structure. So the equilibria will be shifted towards left.

e) Ticla + 2 0 H cí Answer: The equilibria for the above reaction will be shifted towards the left i.e. to the reactant side. Although, both Ti4+ and O are hard center and formation of the co-ordinated compound is feasible but it is entropically forbidden. Hence, the reaction equilibria will mostly lie to the left.