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Why is the geometry of $\ce{[Ni(CN)4)]^{2-}}$ square planar given that it has a coordination number of 4? Shouldn't its geometry be tetrahedral as it would minimise the bond angle and hence repulsion?

Also, why is it colorless? I saw that it is related to how $\ce{CN^{-}}$ is a relatively strong field ligand and can cause "pairing" of the electrons which would not allow the electrons to be transferred between d-orbitals, but what does it mean by pairing?

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  • $\begingroup$ Because it has two lone pairs? $\endgroup$ – DHMO Oct 7 '16 at 14:15
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    $\begingroup$ This has nothing to do with lone pairs... I think, you can start by reading my answer here: chemistry.stackexchange.com/a/40881/16683 If you have any questions, tell me. $\endgroup$ – orthocresol Oct 7 '16 at 14:22
  • $\begingroup$ that is a comparison between two different metal centres so does not answer this specific question. $\endgroup$ – gamma1 Jun 4 '17 at 12:12
  • $\begingroup$ @gamma1 Nowhere did I claim that it answered this specific question. However, the factors that affect the adoption of a square planar geometry vs tetrahedral geometry are the same, regardless of whether it is the metal centre or the ligand that is being compared. Hence, I felt it was related, but of course not a duplicate for the reason you mentioned. If you want to, you're more than welcome to write up an answer that actually compares the ligands. $\endgroup$ – orthocresol Jun 4 '17 at 12:44

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