To calculate the magnetic moment of a chromium III complex, we first need to determine the number of unpaired electrons in the complex. Chromium III has a +3 oxidation state, and the electronic configuration of chromium is [Ar] 3d5 4s1. When it loses 3 electrons, its electronic configuration becomes [Ar] 3d3.In an octahedral complex, the d-orbitals split into two sets: the lower energy t2g orbitals dxy, dyz, and dxz and the higher energy eg orbitals dz2 and dx2-y2 . In a high-spin complex, the electrons will fill the orbitals according to Hund's rule, which states that electrons will occupy orbitals singly before pairing up.For the [Cr en Cl ClO4 2] complex, the 3d3 configuration of Cr III will result in the following distribution of electrons in the d-orbitals:- t2g orbitals: dxy , dyz , dxz - eg orbitals: dz2 empty , dx2-y2 empty There are 3 unpaired electrons in the high-spin state of the chromium III complex.Now, we can calculate the magnetic moment using the formula: = n n + 2 Bwhere n is the number of unpaired electrons, and B is the Bohr magneton approximately 9.27 10^-24 J/T . = 3 3 + 2 B = 15 B = 3.87 BThe magnetic moment of the [Cr en Cl ClO4 2] complex in its high-spin state at room temperature is approximately 3.87 Bohr magnetons.