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    How can you tell whether a complex ion/molecule, with a coordination number of 4, is tetrahedral or square planar in shape?
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    (Original post by Clarity Incognito)
    How can you tell whether a complex ion/molecule, with a coordination number of 4, is tetrahedral or square planar in shape?
    square planar is most common for TM complex of d8 configuration, ie Ni2+, Pd2+ and Pt2+.

    For Ni2+, you could still have choice of tetrahedral and square planar depending on size of the ligands. ie NiCl4 and Ni(CN)4 have different shapes.

    But for Pd2+ and Pt2+ complexes, they are all square planar with a single exception of PdF2 which is rutile TiO2 structure.

    Other than d8, you get tetrahedral for those without lone pairs, but for those with one lone pair and 4 bonding pairs, that'd constitute square pyramidal, while if you have two lone pairs and 4 bonding pairs, then you'd get square planar, where the lone pairs hover above and below.

    Just like VSEPR when applied to main group.
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    Square planar is favoured electronically by d8 complexes. They find this configuration particularly favourable as they completely fill the "lower" (i.e. not the dx2-y2 orbital) set of square planar orbitals as opposed to populating the t2 set of tetrahedral orbital energies which are higher in energy. This can be overriden, however, with very large ligands which might prefer the larger 109.5 degree tetrahedral angles. If you're confused by my references to the orbital energies, have a look at the molecular orbital diagrams on this page: http://wwwchem.uwimona.edu.jm:1104/courses/CFT.html
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    Thanks guys, that's really helpful! So it's sort of like a rule of thumb that d8s form square planar molecules?

    (Original post by shengoc)

    Other than d8, you get tetrahedral for those without lone pairs, but for those with one lone pair and 4 bonding pairs, that'd constitute square pyramidal, while if you have two lone pairs and 4 bonding pairs, then you'd get square planar, where the lone pairs hover above and below.

    Just like VSEPR when applied to main group.
    Also, say you have CoCl4, it has four bonding pairs and it has two lone pairs but it forms tetrahedral?
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    (Original post by Kyri)
    Square planar is favoured electronically by d8 complexes. They find this configuration particularly favourable as they completely fill the "lower" (i.e. not the dx2-y2 orbital) set of square planar orbitals as opposed to populating the t2 set of tetrahedral orbital energies which are higher in energy. This can be overriden, however, with very large ligands which might prefer the larger 109.5 degree tetrahedral angles. If you're confused by my references to the orbital energies, have a look at the molecular orbital diagrams on this page: http://wwwchem.uwimona.edu.jm:1104/courses/CFT.html
    kyri, isn't that only true for 3d, not for 4d and 5d where 4-coordinated complexes are exclusively square planar(apart from PdF2 which I mentioned is not)? just a question, not doubting your explanation.
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    (Original post by Clarity Incognito)
    Thanks guys, that's really helpful! So it's sort of like a rule of thumb that d8s form square planar molecules?



    Also, say you have CoCl4, it has four bonding pairs and it has two lone pairs but it forms tetrahedral?
    But then we are talking about Co4+ which is d5, losing two from d and two from s; it uses 4 electrons for bonding, so one is left, that cant be counted as an electron pair. Hence, it would be tetrahedral; to be honest, I am not sure what happens to this one electron; I am sure someone can come up with a better explanation.
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    (Original post by shengoc)
    But then we are talking about Co4+ which is d5, losing two from d and two from s; it uses 4 electrons for bonding, so one is left, that cant be counted as an electron pair. Hence, it would be tetrahedral; to be honest, I am not sure what happens to this one electron; I am sure someone can come up with a better explanation.
    CoCl4 has an overall charge of 2- so we are talking about Co2+, it's only lost two from 4s and then it has 3 and a half pairs, which doesn't help. Baffled.
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    (Original post by Clarity Incognito)
    CoCl4 has an overall charge of 2- so we are talking about Co2+, it's only lost two from 4s and then it has 3 and a half pairs, which doesn't help. Baffled.
    Quoted from an inorganic text:

    Cobalt(II) forms numerous complexes, mostly either octahedral or tetrahedral but 5-coordinate and square planar complexes are also known.,

    There are more TETRAHEDRAL complexes of Co(II) than for other Tm ions, because for a d7 ion, ligand field stabilization energies disfavor the tetrahedral configuration relative to the octahedral one to a smaller extent than for any other dn(1<n<9; meant to be greater than or equal) config. The only d7 ion of common occurrence is Co(II).
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    (Original post by shengoc)
    Quoted from an inorganic text:

    Cobalt(II) forms numerous complexes, mostly either octahedral or tetrahedral but 5-coordinate and square planar complexes are also known.,

    There are more TETRAHEDRAL complexes of Co(II) than for other Tm ions, because for a d7 ion, ligand field stabilization energies disfavor the tetrahedral configuration relative to the octahedral one to a smaller extent than for any other dn(1<n<9; meant to be greater than or equal) config. The only d7 ion of common occurrence is Co(II).
    Haha, funnily enough, I feel that this is beyond my spectrum for now. I'll just have to accept that d8 complexes with 4 ligands favour square planar for now. Thank you so very much though!
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    Yes planar, I simply see no other solution. It must be!
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    (Original post by shengoc)
    kyri, isn't that only true for 3d, not for 4d and 5d where 4-coordinated complexes are exclusively square planar(apart from PdF2 which I mentioned is not)? just a question, not doubting your explanation.
    To be honest, I'm not sure. I had a quick look at the handouts from my first year lecture course which had this stuff in it and they don't mention it only being the case for 3d.

    Also, after refreshing my memory from my notes, tetrahedral geometry comes about more generally from Jahn-Teller distortions. That is lengthening the axial bonds in an octahedral complex until eventually the axial ligands are completely removed. If we were to draw the octahedral set of energies, and the degenerate orbitals are filled with different numbers of electrons, then the orbitals will distort to remove the degeneraacy. This happens to be most prevalent in low spin d8 metals, particularly with ligands that have a high crystal field stabilisation energy such as CN-.
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    (Original post by domino0806)
    Yes planar, I simply see no other solution. It must be!
    I will assume this is an attempt at facetiousness!
 
 
 
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