The complex \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) contains five unpaired electrons. Sketch the energy-level diagram for the \(d\) orbitals, and indicate the placement of electrons for this complex ion. Is the ion a high-spin or a low-spin complex?

The complex ion \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) contains an Mn(II) ion with an electron configuration of \([Ar] 3d^5\). In the octahedral complex, the \(d\) orbitals split into two sets: \(\mathrm{t}_{2\mathrm{g}}\) (lower energy) and \(\mathrm{e}_{\mathrm{g}}\) (higher energy). Due to the weak field ligand ammonia, the splitting energy is small, and electrons occupy the higher energy \(\mathrm{e}_{\mathrm{g}}\) orbitals before pairing up in the lower energy \(\mathrm{t}_{2\mathrm{g}}\) orbitals, resulting in a high-spin configuration: t2g: ↓ ↓ ↓ --- --- --- dxy dyz dxz eg: ↓ ↓ --- --- dx^2-y^2 d(z^2) Thus, the complex ion \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) is a high-spin complex.