6
$\begingroup$

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?

References:

$\endgroup$
1
  • 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

11
$\begingroup$

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 $\ce{[Ar]}3d^9$

and

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.