A $d^{6}$ octahedral ion has $Δ_{o} = 19800 c m^{- 1}$ and mean pairing energy $P = 17600 c m^{- 1}$ . The energy of each spin state is measured from the barycentre, adding $P$ for each electron pair formed beyond those in the high-spin arrangement. The lower-energy spin state and its energy are

Answer: B

High spin

t_{2 g}^{4} e_{g}^{2}

E = - 0.4 (19800) = - 7920 c m^{- 1}

with no extra pairs. Low spin

t_{2 g}^{6}

has 2 extra pairs:

E = - 2.4 (19800) + 2 (17600) = - 47520 + 35200 = - 12320 c m^{- 1}

. This is lower, as expected because

Δ_{o}

is greater than

P