To calculate the magnetic moment of a cobalt II ion Co2+ in an octahedral field, we need to determine the electron configuration of the ion and then find the number of unpaired electrons. Finally, we will use the spin-only formula to calculate the magnetic moment.Cobalt has an atomic number of 27, so its ground state electron configuration is [Ar] 4s 3d. When cobalt loses two electrons to form the Co ion, its electron configuration becomes [Ar] 3d [Ar] 3d.In an octahedral field, the five d orbitals split into two energy levels: the lower-energy tg orbitals dxy, dyz, and dxz and the higher-energy e_g orbitals dz and dx-y . The ligand field strength 800 cm does not affect the electron configuration in this case, as it only influences the energy gap between the tg and e_g orbitals.The electron configuration of Co in an octahedral field is:tg 3 electrons in tg orbitals e_g 2 electrons in e_g orbitals There are a total of 5 unpaired electrons in the Co ion.Now, we can use the spin-only formula to calculate the magnetic moment : = n n+2 _Bwhere n is the number of unpaired electrons and _B is the Bohr magneton approximately 9.27 10 Am . = 5 5+2 _B = 5 7 _B = 35 _BThe magnetic moment of the Co ion in an octahedral field is approximately 35 _B.