Answer to Question #4671 in Inorganic Chemistry for Steven
pairing energy is 280 kJ/mol.
μ = √(n(n+2), n is
Theelectron pairing energy is P = 280 kJ/mol = 23406 cm-1
Δ0> P, thus thecomplex is high-spin, the number of unpaired electrones is 1.
μ = √(n(n+2) = √(1(1+2) = √3 = 1.73 mB
Thespin-only magnetic moment of Ru3+(aq) is 1.73 mB.
Need a fast expert's response?Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!