Question

Determine the number of unpaired electrons, magnetic moment, and LFSE for following complexes:

f. MnO4−

g. Cu(H2O)6]2+

Answer 1

MnO_4^- The cation here is Mn^+ 7: it has 18 electrons Its electronic configuration is: - 1s^2 2s^2 2p^6 3s^2 3p^6 It has no unpaired electron: its magnetic electron: its magnetic moment is: - 0 and LFSE (Ligand field stabilization energy) = 0 (as it has no d electron) [cu CH_2O_6]^+ 2 here the cation is Cu^+ 2 it has a total 27 electron The electronic configuration of Cu^+ 2 is 1s^2 2s^2 2p^6 3s^2 3p^6 3d^9 if we draw the box diagram for outer most orbitals (3[, 3d) we get So: it has one unpaired electron The magnetic moment of thye ion is mu = squareroot 4s(s + 1) BM = squareroot 4 times 1/2 times (1/2 + 1) BM As it has one electron unpaired: S = 1/2: mu = squareroot BM mu = 1. 73BM (it is spin only magnetic moment) Cu^+ 2 is 3d^9 system so its electronic configuration under