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. $[Fe(H_2O)_6]^{3+}$ is an octahedral complex. In octahedral field, d-orbitals split into $t_{2g} (d_{xy}, d_{xz}, d_{yz})$ and $e_g (d_{x^2-y^2}, d_{z^2})$. The $e_g$ set is higher in energy. The pattern given ($d_{xy} = d_{xz} = d_{yz} < d_{x^2 - y^2} = d_{z^2}$) is correct.
2. $[Ni(Cl)_4]^{2-}$ is a tetrahedral complex (due to weak field $Cl^-$ ligand). In tetrahedral field, d-orbitals split into $e (d_{x^2-y^2}, d_{z^2})$ and $t_2 (d_{xy}, d_{xz}, d_{yz})$. The $t_2$ set is higher in energy. The second pattern given ($d_{x^2 - y^2} = d_{z^2} = d_{xy} < d_{xz} = d_{yz}$) is incorrect because $d_{xy}$ belongs to the $t_2$ set, not $e$.
Thus, Statement II is Incorrect.
Step 3: Final Answer:
Statement I is correct but Statement II is incorrect.
Identify A in the following reaction. 

Identify A in the following reaction. 