Why cfse in tetrahedral complexes is less than octahedral complex?
Answers
Answer
In solutions (i.e. as ions), they form transition metal complexes that absorb visible light.
CRYSTAL FIELD SPLITTING DIAGRAMS
It is easiest to show why the ideal octahedral and tetrahedralcomplexes are sometimes colored. Recall their splitting diagrams from crystal field theory?
Well, the correct version (which accounts for stabilization of the dxy, dxz, and dyz orbitals for an octahedral field, or for the dz2 and dx2−y2 for the tetrahedral field), is:
These splitting energies are called Δo for the octahedral complex and Δt=49Δo for the tetrahedral complex.
HOW DO THESE SPLITTING ENERGIES COME IN?
Whenever Δo or Δt are in the range of 400~700 nm (or 25000~14285 cm−1), the compound can absorb visible light to excite an electron from a lower-energy dorbital to the ones right above them.
Then, the relaxation of these electrons would emit light back to your eyes and allow you see a specific color.
You will see the color complementary to the one corresponding to the wavelength absorbed (e.g. absorb blue light, see orange light).
So, you would see colors resulting from eg→t2g relaxations, or t2→erelaxations. These can be fairly weak, however. These have molar absorptivities at around ε≈50 L/mol⋅cm.
Explanation:
Generally speaking, octahedral complexes will be favored over tetrahedral ones because: It is more (energetically) favorable to form six bonds rather than four. The CFSE is usually greater for octahedral than tetrahedral complexes. Remember that Δo is bigger than Δtet (in fact, Δtet is approximately 4/9 Δo).