Crystal Field Theory

In general, low-spin complexes are formed by transition metals with higher oxidation states and/or ligands that create strong crystal field splitting (such as ).

Based on this: - and are low-spin because is a strong field ligand.

- and are high-spin due to weaker ligands or lower oxidation states.

Thus, the correct complexes that are low-spin are and .

To determine the correct order of the complexes in terms of their Crystal Field Stabilization Energies (CFSE), we need to analyze each complex based on the following factors:

1. Nature of the ligand: The ability of the ligand to split the d-orbitals is called the ligand's field strength. Ligands like en (ethylenediamine) are stronger than NH₃ according to the spectrochemical series.
2. Oxidation state of the metal: Higher oxidation states increase the splitting of the d-orbitals.
3. Geometry of the complex: The complexes are given to have octahedral geometry unless mentioned otherwise.

Now, let's consider each complex individually:

• : Cobalt is in a +2 oxidation state, and NH₃ is a moderate field ligand.
• : Cobalt is in a +2 oxidation state, and with six ammonia molecules, the ligand field slightly increases compared to four ligands due to enhanced coordination.
• : Cobalt is in a +3 oxidation state, significantly increasing the field splitting of the d-orbitals because of the higher positive charge.
• : Cobalt in a +3 oxidation state with en ligands, which are stronger field ligands than NH₃, results in maximum stability of the complex.

Considering the above analyses and the spectrochemical series, we conclude the increasing order of CFSE:

1. [\text{Co(NH}_3)_4]^{2+}
2. [\text{Co(NH}_3)_6]^{2+}
3. [\text{Co(NH}_3)_6]^{3+}
4. [\text{Co(en)}_3]^{3+}

This is because the combination of the stronger ligand and higher oxidation state results in the greatest CFSE.

Thus, the correct order is:

The complex is low spin.

Why?

• is a strong field ligand (large ), so it causes big splitting of the d-orbitals.
• For iron in this complex the large splitting favours electron pairing in the lower set of d-orbitals, giving a low-spin configuration.

Note (comparison)

• : F⁻ is a weak field ligand → typically high spin.
• : NH₃ is a stronger field than F⁻ and can give low spin for some metal/oxidation states (e.g. Co(III) d⁶ often low spin), but the clearest low-spin example among the list is due to CN⁻ being a very strong field ligand.

The crystal field stabilization energy (CFSE) depends on the oxidation state and the ligand field strength.

Generally:
- The higher the oxidation state of the metal, the stronger the ligand field and the higher the CFSE.
- is a stronger field ligand than (ethylenediamine).

Thus: - will have the lowest CFSE, as it is in a lower oxidation state.
- has a higher CFSE compared to .
- has a higher oxidation state, leading to higher CFSE.
- has the highest CFSE due to the strong ligand field of .

Hence, the correct order is:

Step 1: Understanding the Concept:
Statement I relates the strength of metallic bonding to electronic configuration. Statement II concerns the Crystal Field Splitting patterns for octahedral and tetrahedral complexes.
Step 2: Detailed Explanation:
Analysis of Statement I:
Transition metals have unpaired d-electrons which participate in interatomic metallic bonding. The greater the number of unpaired electrons, the stronger the metallic bond, and hence higher is the enthalpy of atomisation. This is True.

Analysis of Statement II:
1. is an octahedral complex. In octahedral field, d-orbitals split into and . The set is higher in energy. The pattern given ( ) is correct.
2. is a tetrahedral complex (due to weak field ligand). In tetrahedral field, d-orbitals split into and . The set is higher in energy. The second pattern given ( ) is incorrect because belongs to the set, not .
Thus, Statement II is Incorrect.
Step 3: Final Answer:
Statement I is correct but Statement II is incorrect.