I'm currently working on a question based on two isomers of a nickel complex, where nickel exists in the oxidation state Ni(II) (i.e. has d-electron configuration d8). It has been deduced that the "normal" isomer is square planar and the "iso" isomer is tetrahedral.

The two isomers.

The first of the two isomers (top) is the "normal" isomer and the second (bottom) is the "iso" isomer.

Apparently steric hindrance causes the "iso" isomer to be tetrahedral rather than square planar, but I am unsure how to explain this. Is it due to the bond angles causing less strain?

  • 4
    $\begingroup$ Welcome to Chemistry StackExchange. Presumably the change in geometry is driven by steric hindrance between the O atom and the $\ce{C(CH3)_2}$ groups, even though there is potentially free rotation about the NC single bonds meaning that the methyl groups will be able to rotate to minimise any steric clashes. $\endgroup$ – porphyrin May 8 '17 at 13:50
  • 1
    $\begingroup$ @polyphyrin What's the IUPAC name of this compound? $\endgroup$ – Mockingbird May 8 '17 at 14:11
  • $\begingroup$ Thank you! I suppose what I really want to know why it is more sterically favourable to take up a tetrahedral geometry then? $\endgroup$ – Emma Louise May 8 '17 at 18:32
  • $\begingroup$ @porphyrin those are not $\ce{C(CH3)2}$ groups. They are $\ce{CH(CH3)2}$ groups. With the bond lengths and angles in the $\ce{O-Ni-N-C}$ loops that additional hydrogen atom can mess you up if you try to tuck both methyl groups away with bond rotation. $\endgroup$ – Oscar Lanzi Nov 12 '19 at 1:35
  • $\begingroup$ @Oscar Lanzi I think we are saying the same thing, the methyl groups will get in the way in a square planar geometry, but can rotate to minimise interaction, but even so they are still bulky enough to force a change to tetrahedral. $\endgroup$ – porphyrin Nov 12 '19 at 13:39

For a given number of ligands surrounding a central atom there is always an arrangement in which the distances between the ligands is maximised and which typically also results in the maximum possible angles between then. You can deduce them logically if you start at a single ligand (which has no steric clashes whatsoever).

A second ligand should approach the same centre from exactly the opposite side leading to a linear arrangement and a bond angle of $180^\circ$.

A third ligand should approach the same centre perpendicular to the existing bond axis. Upon its approach, the two ligands already present will ease away from the approaching one. The final arrangement is trigonal planar with bond angles of $120^\circ$.

A fourth ligand should approach the same centre in a perpendicular manner again; this time relative to the plane of the three existing ligands. Again, they will ease away to make more room for the incoming. The final arrangement is a tetrahedron with bond angles of $109.5^\circ$.

Going from a tetrahedron to a square-planar arrangement means squashing the ligands together i.e. reducing the space between them. Thus, if there are no electronic arguments favouring a different arrangement any tetracoordinated entity should assume a tetrahedral environment to minimise steric strain.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.