Question

The pairing energy for the free ion Fe²⁺ is 229 kJ/mol. 10Dq for [Fe(H₂O)₆]²⁺ is 10,400 cm^-1, and the ratio of the Racah parameter B in the complex to the value in the free ion is 0.8 (usually designated as β in the literature). Calculate the energy difference between the high and low spin possibilities for [Fe(H₂O)₆]²⁺ . Use cm^-1 as your energy units.

Answer 1

Fe²⁺ has d⁶ configuration

In high spin it has t_2g⁴ e_g² configuration.

In low spin it has t_2g⁶ configuration

CFSE = x * 4Dq - y* 6Dq. For t_2g^x e_g^y configuration

CFSE for high spin = 4* 4Dq -2* 6Dq + P= 4Dq +P

(P is pairing energy)

P= 229 kJ/mol = 19000cm^-1. (using E= Planck constant*velocity of light* wave number)

10Dq =10400cm^-1

CFSE for low spin(t_2g⁶)= 6* 4Dq +3P =24Dq+3P

Difference in high and low spin energies= 24Dq+ 3P -(4Dq+P)

=20Dq +2P

=20*1040 +2 *19000

= 58000cm^-1