5
$\begingroup$

Why is tetracyanonickelate, $\ce{[Ni(CN)4]^{2-}}$ square planar?

  • $\ce{CN}$, for being a strong field, would not follow Hund's rule.
  • the central atom $\ce{Ni}$ would have a +2 charge so its configuration would change from $\ce{3d^8 4s^2}$ to $\ce{3d^8 4s^0}$
  • Using the statements above and using crystal field theory I get its hybridization as $\ce{dsp^3}$, which suggests that its shape must be trigonal bipyramidal while it is not.
|improve this question
$\endgroup$
  • 2
    $\begingroup$ Crystal field theory and hybridisation does not really mix well. $\endgroup$ – Martin - マーチン♦ Dec 22 '14 at 7:52
  • $\begingroup$ Alright i made a mistake.. 3d8 4s2 would change to 3d8 4s0 .But even after this correction and on applying crystal field i get the hybridisation as dsp3 and not dsp2 $\endgroup$ – Infi Dec 22 '14 at 8:00
  • 1
    $\begingroup$ You should not apply hybridisation. The splitting in CFT refers only to the d-orbitals. $\endgroup$ – Martin - マーチン♦ Dec 22 '14 at 8:08
  • $\begingroup$ So you are saying that CFT cannot be applied to this compound (to predict its shape)? $\endgroup$ – Infi Dec 22 '14 at 9:47
  • $\begingroup$ No, I am saying hybridisation should not be applied to this compound. $\endgroup$ – Martin - マーチン♦ Dec 22 '14 at 10:04
7
$\begingroup$

This is the square planar splitting diagram

enter image description here

Many d8 metal complexes are usually square planar because a very stable electron configuration is achieved if you put the electrons on this splitting diagram -6 electrons stabilizing the compound and only two slightly destabilizing electrons-

Now you could ask, why then not d6 and d7 adopt this cinfiguration? Because for 6 or 7 electrons, an even more stable state is achieved with octahedral fields.

About hybridization... Forget about it when talking about coordination complexes. It's completely useless. It can't explain or predict anything that CFT/LFT can't

|improve this answer
$\endgroup$
  • $\begingroup$ $\ce{[NiCl_4]^{2-}}$, $\ce{[NiBr_4]^{2-}}$, and $\ce{[NiI_4]^{2-}}$ are tetrahedral $\endgroup$ – DavePhD Dec 22 '14 at 14:56
  • $\begingroup$ Yeah, of course they are not all sqp $\endgroup$ – Altered State Dec 22 '14 at 14:59
  • 2
    $\begingroup$ You can rationalize this in terms of the spectrochemical series; the halogens are very weak field ligands so they encourage the formation of high spin tetrahedral complexes. $\endgroup$ – J. LS Dec 22 '14 at 19:16
3
$\begingroup$

The use of hybridisation for transition metal complexes is extremely outdated and often leads to incorrect predictions and rationalisations of transition metal chemistry. If you are using it routinely, please stop using it.

You may wish to first refer to this other answer of mine for a discussion of a very similar case:

Why is [PdCl4]2- square planar whereas [NiCl4]2- is tetrahedral?

There, the comparison is between a 3d and a 4d metal. However, the factors that affect the geometry of a $\mathrm{d^8}$ complex are the same. They are described in much more detail in that link. To summarise:

  1. Adoption of a square planar geometry is favoured due to occupation of four low-energy orbitals formed.

  2. Adoption of a tetrahedral geometry is favoured due to (a) relief of steric repulsions; and (b) the relief of electron-electron repulsions due to adoption of a high-spin configuration.

When the ligand is $\ce{CN-}$, a strong sigma donor and pi acceptor, the corresponding splitting of the orbitals in $D_\mathrm{4h}$ symmetry will be very large. Consequently, the electronic stabilisation derived upon adoption of a square planar geometry in $\ce{[Ni(CN)4]^2-}$ is much larger than for other complexes, e.g. $\ce{[NiCl4]^2-}$.

It's quite possible that the linear ligand $\ce{CN-}$ also introduces less steric repulsions in a square planar geometry than a halide, like $\ce{Cl-}$ or $\ce{Br-}$, would. However, the main factor is likely to simply be the large electronic stabilisation, which in turn derives from the fact that $\ce{CN-}$ is very high in the spectrochemical series.

|improve this answer
$\endgroup$
-3
$\begingroup$

It's tetrahedral; as not 6 lone pairs are coming from $\ce{CN}$, only $\ce{CN}$ are attached. Therefore, only 4 lone pairs will affect the configuration.

So by now applying CFT you will get its hybridization to be $\ce{dsp^2}$ which will make the shape of tetrahedral.

|improve this answer
$\endgroup$
  • $\begingroup$ Please make use of the punctuation marks. It is difficult to understand what you've stated. $\endgroup$ – Yashas Mar 24 '17 at 7:50
  • 1
    $\begingroup$ Sorry, this is incorrect. It is square planar as is common for d8 complexes for the reasons given above, see e.g. en.wikipedia.org/wiki/Cyanonickelate . Also hybridization is one of those tools you learn earlyish in your chemical career and then spend a lot of time trying to avoid using it. $\endgroup$ – Ian Bush Mar 24 '17 at 8:26

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.