Among the following complexes the one which shows zero crystal field stabilization energy (CFSE) is:

Question

What is the chemical formula of glycerol?

Answer 1

The question requires determining which of the given complexes has zero Crystal Field Stabilization Energy (CFSE). Let's analyze each complex to identify the correct answer.

1. [Fe(H₂O)₆]³⁺

Iron in this complex has an oxidation state of +3, which makes it Fe³⁺. The electronic configuration of Fe³⁺ is [Ar] 3d⁵ 4s⁰.
In an octahedral field provided by water ligands (which are weak field ligands), the 3d orbitals of the Fe³⁺ ion split into t_2g and e_g sets.
Since water is a weak field ligand, the splitting is not large enough to pair the electrons, and all five d-electrons remain unpaired, occupying the t_2g and e_g orbitals equally.
This results in no net stabilization from the crystal field splitting, and thus, the CFSE is zero.

2. [Co(H₂O)₆]²⁺

Cobalt in this complex is in the +2 oxidation state, Co²⁺, with an electronic configuration of [Ar] 3d⁷.
With water as a weak field ligand, the d-orbitals will split, and electrons will be distributed among the t_2g and e_g orbitals.
In this case, CFSE is not zero as there is some stabilization due to partial filling of the t_2g level.

3. [Co(H₂O)₆]³⁺

With Co³⁺, we have an oxidation state of +3, having the configuration [Ar] 3d⁶.
This complex will also have partial stabilization due to uneven filling of the t_2g and e_g orbitals, leading to non-zero CFSE.

4. [Mn(H₂O)₆]³⁺

Manganese in the +3 oxidation state has a configuration of [Ar] 3d⁴.
With weak field ligands like water, there will be a splitting, resulting in stabilization due to asymmetric electron distribution.

Based on the above analysis, the complex with zero CFSE is [Fe(H₂O)₆]³⁺, as all its d-electrons are unpaired and evenly distributed between t_2g and e_g orbitals, conferring no net crystal field stabilization.

Answer 2

The question requires determining which of the given complexes has zero Crystal Field Stabilization Energy (CFSE). Let's analyze each complex to identify the correct answer.

1. [Fe(H₂O)₆]³⁺

Iron in this complex has an oxidation state of +3, which makes it Fe³⁺. The electronic configuration of Fe³⁺ is [Ar] 3d⁵ 4s⁰.
In an octahedral field provided by water ligands (which are weak field ligands), the 3d orbitals of the Fe³⁺ ion split into t_2g and e_g sets.
Since water is a weak field ligand, the splitting is not large enough to pair the electrons, and all five d-electrons remain unpaired, occupying the t_2g and e_g orbitals equally.
This results in no net stabilization from the crystal field splitting, and thus, the CFSE is zero.

2. [Co(H₂O)₆]²⁺

Cobalt in this complex is in the +2 oxidation state, Co²⁺, with an electronic configuration of [Ar] 3d⁷.
With water as a weak field ligand, the d-orbitals will split, and electrons will be distributed among the t_2g and e_g orbitals.
In this case, CFSE is not zero as there is some stabilization due to partial filling of the t_2g level.

3. [Co(H₂O)₆]³⁺

With Co³⁺, we have an oxidation state of +3, having the configuration [Ar] 3d⁶.
This complex will also have partial stabilization due to uneven filling of the t_2g and e_g orbitals, leading to non-zero CFSE.

4. [Mn(H₂O)₆]³⁺

Manganese in the +3 oxidation state has a configuration of [Ar] 3d⁴.
With weak field ligands like water, there will be a splitting, resulting in stabilization due to asymmetric electron distribution.

Based on the above analysis, the complex with zero CFSE is [Fe(H₂O)₆]³⁺, as all its d-electrons are unpaired and evenly distributed between t_2g and e_g orbitals, conferring no net crystal field stabilization.

Among the following complexes the one which shows zero crystal field stabilization energy (CFSE) is:

The Correct Option is A

Solution and Explanation

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