Why won't $\ce{Ni^2+}$ form a complex with $\ce{Cl-}$, while the ions $\ce{Cu^2+}$ and $\ce{Co^2+}$ form the complexes $\ce{[CuCl4]^2-}$ and $\ce{[CoCl4]^2-}$?

According to the HSAB theory, $\ce{Cl-}$ is considered a hard/borderline base. So I figured the strongest complex would be the one in which the center ion is the hardest acid. The $\ce{Co^2+}$ ion is the largest and the least electronegative. So it should be the softest acid (although they are all considered borderline). It forms a complex with $\ce{Cl-}$. $\ce{Ni^2+}$ should be the hardest acid, and yet it does not form a complex with $\ce{Cl-}$ at all. Why is that?


  • 1
    $\begingroup$ Are all of the metals in your complexes supposed to be in the $+2$? oxidation state? If so, the charge on your complexes should be $2-$. If not, then all the metals need to be $+3$, which is very unlikely for copper. $\endgroup$
    – Ben Norris
    May 3, 2013 at 10:50

1 Answer 1


The link to your handout explains why $\ce{Fe^{3+}}$ (and it is $\ce{Fe^{3+}}$ in $\ce{FeCl4^{-}}$) is a harder acid than $\ce{Co^{3+}}$ or $\ce{Cu^{2+}}$ or $\ce{Ni^{2+}}$.

Hard acids (in context, HA) are characterized by (s,f blocks, left side of d block in higher OS's)

Iron is the furthest to the left of the four metals you mention. Thus, it has fewer electrons in its $d$ shell to balance the positive charge:

  • $\ce{Fe^{3+}}$ is $\ce{[Ar]}3d^5$
  • $\ce{Co^{3+}}$ is $\ce{[Ar]}3d^6$
  • $\ce{Ni^{2+}}$ is $\ce{[Ar]}3d^8$
  • $\ce{Cu^{2+}}$ is $\ce{[Ar]}3d^9$


Low electronegativity (χ) of the acidic atom. A value in the range 0.7-1.6 is typical of hard acids;

Iron has the lowest electronegativity of the four metals:

  • $\ce{Fe}$ 1.83
  • $\ce{Co}$ 1.88
  • $\ce{Ni}$ 1.91
  • $\ce{Cu}$ 1.90

Iron is the hardest Lewis acid of the metal cations you list.

As for $\ce{Ni^{2+}}$, it does form a complex $\ce{NiCl4^{2-}}$ with $\ce{Cl-}$. Here is a Google search for it.


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