0
$\begingroup$

I'm having trouble understanding how exactly coordination complexes are formed.

Compounds such as $\ce{[Cu(H2O)6]++}$ are the ligands filled into the empty 3 of 4p orbitals+ 1 of 5s orbital + 2 of 4d orbitals or how?

$\endgroup$
1
$\begingroup$

If you want to analyse a compound, it is best to start with the $d$-electron count of the metal. For $\ce{Cu^{2+}}$ we get $d^9$, since $\ce{Cu^0}$ would be $d^{11}$.

Now, consider the following (simplified) molecular orbital diagram for a $d^1$ compound: Molecular orbital diagram for a $d^1$ complex.

As you can clearly see, the ligand electron pairs mix with some of the $d$, the $s$ and $p$ orbitals of the metal.

Now, instead of $d^1$ we have $d^9$, so we just start filling in 8 more electrons into the MOs. This gives us a full $t_\text{2g}$ MO and 3 electrons in the $e_\text{g}$ MO. ($t_\text{2g}$ and $e_\text{g}$ are ways of describing the symmetry of the orbitals. This comes from applied group theory.)

So you see, the $5s$ orbital never comes into play here, only the $3d$, $4s$ and $4p$ orbitals of the metal take part in bonding.

You might want to look at Ligand Field Theory and Molecular Orbital Theory to further your knowledge in inorganic chemistry. Not everything is as simple as the "fill in the boxes" method.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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