Is it possible for a low-spin octahedral Fe(II) complex to be paramagnetic? Explain.

Start by identifying the atomic number of Fe (iron), which is 26. The electron configuration of a neutral iron atom is \([Ar] 3d^6 4s^2\). When an iron atom loses 2 electrons to form the Fe(II) complex, its electron configuration becomes \([Ar] 3d^6\), which has 6 electrons in the d orbitals. 2.

In an octahedral ligand field, the d orbitals split into two sets: \(t_{2g}\) (with \(d_{xy}\), \(d_{yz}\), and \(d_{zx}\) orbitals) and \(e_g\) (with \(d_{z^2}\) and \(d_{x^2-y^2}\) orbitals). The energy gap between these two sets of orbitals (Δ) is relatively large in a low-spin complex. In a low-spin complex, electrons fill the lower energy \(t_{2g}\) orbitals first before occupying the higher energy \(e_g\) orbitals. This is due to strong ligand repulsion in high-spin complexes, which causes electrons to partially fill the \(t_{2g}\) orbitals before moving to \(e_g\) orbitals. For Fe(II), the electron configuration for a low-spin complex is \(t_{2g}^6 e_g^0\), with all 6 d-electrons in the lower-energy \(t_{2g}\) orbitals. 3.

In a low-spin complex, electrons fill up the three \(t_{2g}\) orbitals in pairs, with opposite spins. In the case of the Fe(II) complex, all 6 d-electrons are paired in the \(t_{2g}\) orbitals, resulting in no unpaired electrons. 4.

The magnetic properties of a complex depend on the number of unpaired electrons it possesses. Since there are no unpaired electrons in the low-spin octahedral Fe(II) complex, it exhibits no magnetic moment and is considered diamagnetic. 5.

A low-spin octahedral Fe(II) complex has no unpaired electrons and exhibits diamagnetic behavior rather than paramagnetic behavior. Therefore, it is not possible for a low-spin octahedral Fe(II) complex to be paramagnetic.