JEE Main 2026 — Chemistry Inorganic Chemistry

Statement I: The enthalpy of atomisation of transition metals depends on the strength of metallic bonding. A greater number of unpaired electrons in the d-orbitals leads to stronger metallic bonding, which results in a higher enthalpy of atomisation. Thus, Statement I is correct.

Statement II: $[Fe(H_2O)_6]^{3+}$[Fe(H2​O)6​]3+ is an octahedral complex. In an octahedral crystal field, the d-orbitals split such that the $t_{2g}$t2g​ set of orbitals ($d_{xy}, d_{xz}, d_{yz}$dxy​,dxz​,dyz​) has lower energy than the $e_g$eg​ set of orbitals ($d_{x^2-y^2}, d_{z^2}$dx2−y2​,dz2​). Therefore, $d_{xy} = d_{xz} = d_{yz} < d_{x^2-y^2} = d_{z^2}$dxy​=dxz​=dyz​<dx2−y2​=dz2​.

$[Ni(Cl)_4]^{2-}$[Ni(Cl)4​]2− is a tetrahedral complex because $Cl^{-}$Cl− is a weak field ligand and the coordination number is $4$4. In a tetrahedral crystal field, the splitting is reversed compared to the octahedral field. The $e$e set of orbitals ($d_{x^2-y^2}, d_{z^2}$dx2−y2​,dz2​) has lower energy than the $t_2$t2​ set of orbitals ($d_{xy}, d_{xz}, d_{yz}$dxy​,dxz​,dyz​). Therefore, $d_{x^2-y^2} = d_{z^2} < d_{xy} = d_{xz} = d_{yz}$dx2−y2​=dz2​<dxy​=dxz​=dyz​. Thus, Statement II is also correct.

Both Statement I and Statement II are correct.

Answer: Both Statement I and Statement II are correct