Crystal field stabilization energy equals pairing energy what happens?

Question

Crystal field stabilization energy equals pairing energy what happens?

Answer 1

I'm assuming you mean "what happens when a compound has an orbital splitting equal to its pairing energy?", not CFSE = P.

(You can't have CFSE = P.  CFSE is defined as the total energy of the d electrons after splitting, as compared to five degenerate d orbitals, e.g. –12 Dq for an octahedral d3 complex.  It's always a negative value, at worst it's zero.  Pairing energy is always a positive quantity, and there's nothing to learn by comparing the two values in any case.  So your question only makes sense if you mean "orbital splitting", not "crystal field stabilization energy".)

Consider an equilibrium problem.  If the left side is lower energy (more stable), the K < 1, and the system stays on the left.  If the right side is lower energy, K > 1, and the system shifts to the right.  If the two sides are equal energy, K = 1, and you get equal population of both sides, with a mixture of reactant and products.  You can change K and so drive the system to one side or the other by changing temperature.

So you're trying to decide if a compound is LS or HS.  If orbital splitting (Delta) is greater than pairing energy (P), Delta wins and the system is LS.  That is, K for the process LS <--> HS is less than 1.  If Delta < P, then P wins and the system is HS, so K > 1.

So what if Delta = P?  Then K = 1, and you get an equal distribution of the two sides of the equilibrium: some is HS and some is LS, because there's no energetic preference for either form!  There are lots of compounds, called "spin crossover" complexes, where that's the case.  (Many octahedral Fe(II) compounds, for example.)  You can drive it to either side with T: low temperature, favours low spin (K < 1), high temperature favours high spin (K > 1), and there's a temperature at which the spin state changes and you get a population of both states, called the "crossover temperature" (K = 1).