[mcrdotcom]

Obviously octahedral is by far the most likely answer to any question asked about CN6 complexes, but how would you tell if it was trigonal prismatic?

My notes say that [Zr(Me)6]2- is trig. pris. and I'm answering an exam question asking for [Ni(CN)6]3- geometry and to explain in a few words... As it asks to explain, I kinda wonder if its an unusual answer, can't see why octahedral would need explaining...

Anyway just wondering if there are any rule of thumbs for this.

Posted this in undergrad section but no replies, so posting here now

[grs35]

Obviously octahedral is by far the most likely answer to any question asked about CN6 complexes, but how would you tell if it was trigonal prismatic?

My notes say that [Zr(Me)6]2- is trig. pris. and I'm answering an exam question asking for [Ni(CN)6]3- geometry and to explain in a few words... As it asks to explain, I kinda wonder if its an unusual answer, can't see why octahedral would need explaining...

Anyway just wondering if there are any rule of thumbs for this.

Posted this in undergrad section but no replies, so posting here now

I think you mean trigonal bipyramidal. For a geometry like that, you need 5 ligands not 6. That zirconium complex can't be trigonal bipyramidal. May be there's a mistake. It's octahedral.

Nickel complex is octahedral too due to those 6 -CN ligands.

[Big-Daddy]

I think you mean trigonal bipyramidal.

I thought that octahedral or trigonal prismatic corresponded to coordination number of 6, trigonal bipyramidal or square pyramidal corresponds to coordination number 5?

Sorry that this doesn't help the OP's question in any way. But I don't see how [Zr(Me)6]^2- can be trigonal bypramidal?  Probably I'm thinking about it wrong ...