The correct answer is: 1. \(Cl^–\) (chloride ion) is the ligand most likely to form high spin complexes in octahedral fields.
Chloride ion \(Cl^–\) is a weak field ligand, and it causes small crystal field splitting in octahedral complexes. As a result, there is a small energy difference between the \(t_{2g}\) and \(e_g\) orbitals, making it energetically favorable for electrons to occupy the \(e_g\) orbitals before pairing in the \(t_{2g}\) orbitals. This leads to the formation of high-spin complexes in octahedral fields.
On the other hand, the other ligands (\(OH^–\), \(C_2O_4^–\), and \(CN^–\)) are strong field ligands, and they cause large crystal field splitting, leading to a significant energy difference between the \(t_{2g}\) and \(e_g\) orbitals. This favors the pairing of electrons in the \(t_{2g}\) orbitals before occupying the \(e_g\) orbitals, resulting in the formation of low spin complexes.
Therefore, the correct answer is 1. \(Cl^–\) (chloride ion). |
