Question

Arrange the following Cobalt complexes in the order of increasing Crystal Field Stabilization Energy (CFSE) value. 

 

• A. A < B < C < D
• B. B < C < D < A
• C. C < D < B < A
• D. B < A < C < D

Hint:

To compare CFSE, first check the oxidation state of the central metal. A higher positive charge pulls ligands closer, causing greater splitting. If oxidation states are the same, use the spectrochemical series to compare ligand strength. A stronger ligand causes greater splitting and higher CFSE.
A simplified spectrochemical series to remember is: I\(^-\) < Br\(^-\) < Cl\(^-\) < F\(^-\) < OH\(^-\) < H\(_2\)O < NH\(_3\) < en < CN\(^-\) < CO.

Answer 1

The Correct Option is D

Step 1: Understanding the Question
We need to arrange four cobalt complexes in increasing order of their Crystal Field Stabilization Energy (CFSE).
Step 2: Key Factors Affecting CFSE
CFSE depends on three main factors:
1. Oxidation state of the central metal ion: Higher oxidation state leads to greater crystal field splitting (\(\Delta_o\)) and thus higher CFSE.
2. Nature of the ligand: Strong field ligands cause greater splitting than weak field ligands. The order is given by the spectrochemical series.
3. Geometry of the complex: All given complexes are octahedral.
Step 3: Detailed Analysis of Each Complex
Let's analyze each complex:
Complex A: [CoF\(_6\)]\(^{3-}\)
Oxidation state of Co: x + 6(-1) = -3 \(\Rightarrow\) x = +3.
Ligand: F\(^-\) is a weak field ligand.
Complex B: [Co(H\(_2\)O)\(_6\)]\(^{2+}\)
Oxidation state of Co: x + 6(0) = +2 \(\Rightarrow\) x = +2.
Ligand: H\(_2\)O is a weak field ligand.
Complex C: [Co(NH\(_3\))\(_6\)]\(^{3+}\)
Oxidation state of Co: x + 6(0) = +3 \(\Rightarrow\) x = +3.
Ligand: NH\(_3\) is a strong field ligand.
Complex D: [Co(en)\(_3\)]\(^{3+}\)
Oxidation state of Co: x + 3(0) = +3 \(\Rightarrow\) x = +3.
Ligand: 'en' (ethylenediamine) is a strong field ligand, stronger than NH\(_3\).
Step 4: Comparing the CFSE Values
Comparing based on oxidation state: Complex B has Co in a +2 oxidation state, while A, C, and D have Co in a +3 oxidation state. A higher oxidation state leads to higher CFSE. Therefore, Complex B will have the lowest CFSE among the four.
Order so far: B < (A, C, D)
Comparing A, C, and D: All these complexes have Co\(^{3+}\). So, the CFSE will depend on the strength of the ligands.
The spectrochemical series gives the order of ligand strength:
F\(^-\) < H\(_2\)O < NH\(_3\) < en
Based on this, the order of CFSE for these three complexes will be:
[CoF\(_6\)]\(^{3-}\) (A) < [Co(NH\(_3\))\(_6\)]\(^{3+}\) (C) < [Co(en)\(_3\)]\(^{3+}\) (D)
Order: A < C < D
Step 5: Final Order
Combining the two comparisons, we get the final increasing order of CFSE:
\[ \text{[Co(H}_2\text{O)}_6\text{]}^{2+}<\text{[CoF}_6\text{]}^{3-}<\text{[Co(NH}_3\text{)}_6\text{]}^{3+}<\text{[Co(en)}_3\text{]}^{3+} \] \[ \text{B < A < C < D} \]