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On Wikipedia, it says that

$\ce{NiCl2·6H2O}$ consists of separated trans-$\ce{[NiCl2(H2O)4]}$ molecules linked more weakly to adjacent water molecules. Only four of the six water molecules in the formula are bound to the nickel, and the remaining two are water of crystallisation.[4] Cobalt(II) chloride hexahydrate has a similar structure.

So nickel chloride hexahydrate has an octahedral symmetry and the ligands are two chloride ions and 4 water molecules? Or is $\ce{NiCl2}$ a solid in water, but separates into $\ce{Ni^2+}$ and $\ce{Cl-}$ ions in water since they are ionic?

Does this also indicate that there are 8 electrons in the $d$ orbitals and that it will form a high-spin complex (I know that the low-spin orbitals look the same though) so that there are two unpaired electrons? I think I'm thrown off by the chloride ions because my textbook only explains complex ions like $\ce{[CoCl4]^4-}$ or $\ce{Fe(CN)6}$ which are easier to understand for me.

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  • $\begingroup$ I think I could answer this, but there is a part of the question I’m not yet understanding. Second paragraph ‘or is $\ce{NiCl2}$ a solid in water, but it probably separates into $\ce{Ni^2+}$ and $\ce{Cl-}$ ions in water since they are ionic, right?’ <- that sentence. $\endgroup$
    – Jan
    Commented Jun 9, 2015 at 19:29

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The hexahydrate has octahedral symmetry, and it is very probable that this is the case for the solid and dissolved state of the complex.

The compound will not separate into the ions, because the solvation of three ions vs the solvation of one neutral complex is higher in energy, especially considering the fact that opposite charges attract. So in solution state it is a $\ce{[NiCl2(OH2)4]}$ complex with octahedral symmetry.

The $d$-electron count is 8, as it starts with 10 and has a formal oxidation state of II. This is always the case for $\ce{Ni^{II}}$ in any medium but vacuum.

Whether it's a high-spin or low-spin complex depends on the splitting energy of the $e_\text{g}$ and $t_\text{2g}$ molecular orbitals that arise from the $d$ orbitals in combination with the ligand group orbitals, colloquially known as $\Delta_\text{o}$ or $\Delta_\text{oct}$. Have a look at this Wikipedia article on crystal field theory for more details.

In short, there are several factors that influence the magnitude of the splitting:

  • Oxidation state of central atom
  • Geometry
  • Ligand types

The octahedral geometry usually gives good energy splitting.$^\text{[citation needed]}$ In conjunction with a moderate oxidation state compared to Ni(0), and four mid-range splitting ligands (water), it is a safe bet to say that we will have a low-spin octahedral $d^8$ complex: Three pairs of electrons in the $t_\text{2g}$ orbitals and two unpaired electrons in the $e_\text{g}$ orbitals, that is.

Note at this point that the terms high-spin and low-spin are superfluous. The electrons in the $e_\text{g}$ orbitals will always be in a triplet configuration according to Hund's Rule. So it will always be in a "high-spin" configuration from an experimental standpoint.

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  • $\begingroup$ Is there a reason for the downvote? I'd be glad to improve this answer. $\endgroup$
    – tschoppi
    Commented Jan 23, 2016 at 22:59
  • $\begingroup$ low-spin octahedral d8 equals square planar. Or, at least, strongly distorted octahedral. Nickel has quite an interesting chemistry here, though AFAIK chloride has not strong enough field to make it low spin. Cianide, on the other hand, does. That said, nickel chloride does dissociate in water, as its solutions have measurable conductance. I suggest you to go for the sources and edit your post accordingly with references and links. I'm not posting the sources here, but they are pretty easily googleable. $\endgroup$
    – permeakra
    Commented Jan 23, 2016 at 23:53
  • $\begingroup$ @permeakra Quote from the wikipedia page: "In some derivatives, the chloride remains within the coordination sphere, whereas chloride is displaced with highly basic ligands." Water is not highly basic. Distilled water also has measurable conductance. And I don't understand at all what you mean by low-spin octahedral d8 equalling square planar... Maybe check this out for reference? $\endgroup$
    – tschoppi
    Commented Jan 24, 2016 at 0:22
  • $\begingroup$ But your point is noted, I'll scour the interwebs for some information whether the chloride ligands dissociate and update my answer accordingly. $\endgroup$
    – tschoppi
    Commented Jan 24, 2016 at 0:23
  • $\begingroup$ distilled water has low enough conductance to be utilized as dielectric in some applications. $\endgroup$
    – permeakra
    Commented Jan 24, 2016 at 7:13

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