An octahedral complex of a d^6 transition metal ion is formed under the condition _o < P , where _o is the crystal field splitting energy and P is the pairing energy. The Crystal Field Stabilization Energy (CFSE) ignoring pairing energy terms, and the spin-only magnetic moment (in Bohr Magnetons) for this complex are, respectively :

Step-by-step solution

The condition _o For a d^6 metal ion in a high spin octahedral field, the electrons are distributed as t_ 2g ^4 e_g^2 . The number of unpaired electrons ( n ) is 4 (one in each e_g orbital and two in t_ 2g orbitals). The spin-only magnetic moment is given by = n(n+2) = 4(4+2) = 24 4.90 BM. The Crystal Field Stabilization Energy (CFSE) is calculated as: CFSE = (-0.4 n_ t_ 2g + 0.6 n_ e_g ) _o CFSE = (-0.4 4 + 0.6 2) _o = (-1.6 + 1.2) _o = -0.4 _o . Answer: -0.4 _o and 4.90 BM