The coordination complex [Cr(CO)₆] forms colorless, diamagnetic crystals that melt at 90 °C

c. Given that [Cr(CO)₆] is colorless, would you expect CO to be a weak-field or strong-field ligand?

Video transcript

Hello, everyone. Today, we have the following problem. The following complex is observed to form a light blue solution would water be considered a weak field or strong field ligand, a weak field ligand or B strong field ligand. So this question draws on the crystal field theory. And this theory basically states that the the electronic or the electrons transition from a high energy or they transition, excuse me, from a lower energy to a higher energy. And these consist or these concern D orbitals. And these also help determine the color of coordinator complexes. So this complex here has six L ins, six water L ins. So it has a coordinating number that is equal to six giving it an octahedral geometry for octahedral complexes. The crystal field splitting energy of the metal ion is the following. So we have it these lower energy de orbitals as the following. And then it's followed by two higher energy de orbitals with the following configurations. No, a strong field like in will have a larger energy value. So the higher the energy value will directly correlate or equal a strong filled it like again. And the same can be said for the reverse, a decreased energy value can correlate with a weak field Lian. So the complex is colored that we see it's a light blue solution. And therefore D to D electron transitions are possible. And there must actually be a small gap between the de orbitals and the splitting energy must have a relatively low value, meaning that water is a weak field ligand. And so with that, we have solved the problem overall, I hope this helped and not until next time.