Question

Just really need help on the second one.
Define the terms: a. Ligand. b. Coordination number The word "denticity" has the same root as the word "dentist" and so it's no surprise that it means "toothed". The denticity of a ligand is the number of times a ligand attaches to a metal. For example, a single tridentate ligand attaches to a metal three times. Ligands with a high denticity tend to bind to metals very strongly. Based on your knowledge of thermodynamics, why would a high denticity ligand likely grab onto a metal and not release it? In this lab, half of you will be using potassium dichromate. Look up the SDS for potassium dichromate, determine its safety hazards and appropriate PPE.

Answer 1

high denticity form chelation

chelation is entropy driven

due to chelation entropy increases hence it is favourable condition

Consider two equilibria:

• [Co(H2O)6]^2+ + 6NH3 -----> [Co(NH3)6]^3+ + 6H2O                   

• [Co(H2O)6]^2+ + 3en-----> [Co(en)3]^2+ + 6H2O

In the first case, ∆S= 0(same no: of molecules on either side)

In the second case, ∆S=+ve ( 4 molecules give (6+1)=7 molecules).

Electronically NH3 and en are the same:

Both bind through N atoms.

Lewis base strengths are similar.

Six Co-N bonds formed in each case.

As a result,∆H values of both the reactions are almost the same. But if you look at their equilibrium constants' values, the 2nd equilibrium has a K value which is about 10^5 times than that of the 1st one.

This is because of the entropy change.
( rem: According to the Eyring equation; ∆G°= -RTlnK =∆H- T∆S. So as ∆S increases, K value increases since ∆H valuea are almost the same).

Thus chelate effect is entropy driven. More the no: of binding groups a ligand has; more +ve ∆S is and thus higher the value of Kf will be.