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Chelating ligands often form more stable complex ions than the corresponding monodentate ligands with the same donor atoms. For example, Ni 2 + ( a q ) + 6 NH 3 ( a q ) ⇌ Ni ( NH 3 ) 6 2 + ( a q ) K = 3.2 × 10 8 Ni 2 + ( a q ) + 3 en ( a q ) ⇌ Ni ( e n ) 3 2 + ( a q ) K = 1.6 × 10 18 Ni 2 + ( a q ) + penten ( a q ) ⇌ Ni ( penten ) 2 + ( a q ) K = 2.0 × 10 19 where en is ethylenediamine and penten is This increased stability is called the chelate effect. Based on bond energies, would you expect the enthalpy changes for the above reactions to be very different? What is the order (from least favorable to most favorable) of the entropy changes for the above reactions? How do the values of the formation constants correlate with ∆S°? How can this be used to explain the chelate effect?

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 20, Problem 93CP
Textbook Problem
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Chelating ligands often form more stable complex ions than the corresponding monodentate ligands with the same donor atoms. For example,

Ni 2 + ( a q ) + 6 NH 3 ( a q ) Ni ( NH 3 ) 6 2 + ( a q ) K = 3.2 × 10 8 Ni 2 + ( a q ) + 3 en ( a q ) Ni ( e n ) 3 2 + ( a q ) K = 1.6 × 10 18 Ni 2 + ( a q ) + penten ( a q ) Ni ( penten ) 2 + ( a q ) K = 2.0 × 10 19

where en is ethylenediamine and penten is

Chapter 20, Problem 93CP, Chelating ligands often form more stable complex ions than the corresponding monodentate ligands

This increased stability is called the chelate effect. Based on bond energies, would you expect the enthalpy changes for the above reactions to be very different? What is the order (from least favorable to most favorable) of the entropy changes for the above reactions? How do the values of the formation constants correlate with ∆S°? How can this be used to explain the chelate effect?

Interpretation Introduction

Interpretation: Reactions corresponding to the formation of chelating ligands are given.

Based on bond energies, weather the given reactions will have different enthalpy changes or not. Order of the entropy changes for the given reactions is to be stated. The relationship between ΔSο and values of formation constants and its usage to explain chelate effect is to be stated.

Concept introduction: The electrons in the d orbital of a transition metal split into high and low energy orbitals when ligands are attached to it. The energy difference between these two levels depends upon the properties of both metal and the ligands. If the ligand is strong, then splitting will be high and the complex will be low spin. If the ligand is weak, then splitting will be less and the complex will be high spin.

To determine: If the given reactions will have different enthalpy changes or not.

Explanation of Solution

Explanation

The change in enthalphy depends upon the nature of ligand, nature of metal and size of ligand. There is difference in the size of ligands amine, ethylenediammine and penten, therefore there enthalphy changes of the given reactions are different.

The order of the entropy changes for the given reactions is Ni(NH3)62+(aq)<Ni(en)32+(aq)<Ni(penten)2+(aq) .

The change in entropy is directly proportional to the change in the number of molecules in the reactants and in the product.

For the reaction,

Ni(H2O)6+2(aq)+6NH3(aq)Ni(NH3)62+(aq)+6H2O

The number of molecules in the reactant is 1+6=7 . The number of molecules in products is 1+6=7 . There is no difference in number of molecules.

For the reaction,

Ni(H2O)6+2(aq)+3enNi(en)32+(aq)+6H2O

The number of molecules in the reactant is 1+3=4

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Chemistry: An Atoms First Approach
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