Before You Begin

Question #44056

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following ligand exchange reaction for a \ce{Ni^{2+}} ( $d^8$ ) complex in aqueous solution at $300 \text{ K}$ :

$\ce{[Ni(A)6]^{2+}(aq) + 3B(aq) <=> [Ni(B)3]^{2+}(aq) + 6A(aq)}$

where A is a neutral monodentate ligand and B is a neutral bidentate ligand.

The crystal field splitting energy ( $\Delta_o$ ) for \ce{[Ni(A)6]^{2+}} is $150 \text{ kJ mol}^{-1}$ and for \ce{[Ni(B)3]^{2+}} is $120 \text{ kJ mol}^{-1}$ . Assume that the standard enthalpy change ( $\Delta H^\circ$ ) of the exchange reaction is determined entirely by the difference in Crystal Field Stabilization Energy (CFSE) between the two complexes.

If the standard entropy change ( $\Delta S^\circ$ ) for the exchange reaction is $+140 \text{ J K}^{-1} \text{ mol}^{-1}$ , calculate the standard Gibbs free energy change ( $\Delta G^\circ$ ) and identify the thermodynamically more stable complex at $300 \text{ K}$ .

Options

A.
$-6 \text{ kJ mol}^{-1}$ , \ce{[Ni(B)3]^{2+}}
B.
$+36 \text{ kJ mol}^{-1}$ , \ce{[Ni(A)6]^{2+}}
C.
$-72 \text{ kJ mol}^{-1}$ , \ce{[Ni(B)3]^{2+}}
D.
$+78 \text{ kJ mol}^{-1}$ , \ce{[Ni(A)6]^{2+}}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Standard State Conditions|Crystal Field Splitting (Octahedral)|Werner’s Theory Postulates

Why This Question Is Tricky

Common Mistakes

The problem tests the critical distinction between electronic stabilization (CFSE) and overall thermodynamic stability ( $\Delta G^\circ$ ).

Students often fall into the trap of assuming that a stronger ligand field (higher $\Delta_o$ and more negative CFSE) strictly dictates the most stable complex.

Here, the monodentate ligand A creates a stronger field, making \ce{[Ni(A)6]^{2+}} enthalpically favored.

However, the hidden variable is the 'Chelate Effect' introduced by the bidentate ligand B, which provides a massive positive entropy change ( $\Delta S^\circ$ ) due to the increase in the number of free particles.

By forcing the calculation of $\Delta G^\circ$ , the problem demonstrates that entropy can override a weaker crystal field, making the chelate complex thermodynamically dominant.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Thermodynamics - Class 11 Part 1
Concept	Standard State Conditions
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free