[elt2112]

Could anyone help me with this question please? I thought that the crystal field splitting at which there would be a changeover would just be 21000cm^-1 but wouldn't that stay the same for a d⁷ ion?

[mjc123]

How many spin-pairs are there in low-spin d⁶? How many in high-spin d⁶? What is the difference in spin-pairing energy? What is the difference in crystal field energy if the crystal field splitting is Δ?
Do the same for d⁷.

[elt2112]

Thanks but I'm still arriving at the same answer, could you tell me where I've gone wrong please?
High spin d⁶:
No spin pairs = 1
Crystal field energy = (4x-2/5Δ) - (2 x 3/5Δ) = -2/5Δ

Low spin d⁶:
No spin pairs = 3
Crystal field energy = 6 x -2/5Δ = -12/5Δ + 2P

Difference in spin-pairing energy = (3x21000) - (21000) = 42000cm^-1
Difference in crystal field energy = -10/5Δ = -2Δ
Therefore 2P must be greater than 2Δ

d⁷ high spin
CFSE = -4/5Δ

d⁷ low spin
Crystal field energy = -9/5Δ + P

Difference in CFE = -1Δ
So it loses 1Δ but gains 1P therefore P must be greater than Δ

[mjc123]

You're not wrong, I agree with you, but now you have given reasons for your answer, which is what the question asked for.
How about part iii?

[elt2112]

Oh okay, thanks!
Tetrahedral d⁶ high spin:
CFE =  -3/5Δtet

Tetraheadral d⁶ low spin:
CFE = -8/5Δtet + P

So again if Δ is greater than P the complex will be low spin.

Δtet is around 4/9ths of Δoct so it's unlikely for the tetrahedral splitting to be larger than the pairing energy. Therefore tetrahedral complexes are always high spin as it is more energetically favourable to put an electron in a t2 orbital than to pair it in an e orbital.

[mjc123]

Looks right to me.