# How do transition elements form coordination compounds?

Feb 4, 2017

Transition metal ions are highly Lewis-acidic........

#### Explanation:

Transition metal ions are highly Lewis acidic, and we can typically observe ${M}^{2 +}$ and ${M}^{3 +}$ ions. These ions draft in electron density from some source to stabilize and distribute their charge. In water, we write ${M}^{2 +} \left(a q\right)$ and ${M}^{3 +} \left(a q\right)$. We really mean a coordination complex, i.e. the aquated metal ion: ${\left[M {\left(O {H}_{2}\right)}_{6}\right]}^{2 +}$ or ${\left[M {\left(O {H}_{2}\right)}_{6}\right]}^{3 +}$.

In the presence of other ligands, i.e. other Lewis bases such as $N {H}_{3}$, or ${X}^{-}$, or $H {O}^{-}$, or $P {R}_{3}$, substitution at the metal centre can occur to give an astonishing range of so-called coordination complexes, in which the electron rich Lewis bases, so-called ligands, donate electron density, coordinate to, the metal centre. Typically, 6 ligands around a metal centre give an octahedron, and 4 ligands give a tetrahedron.

As with all chemical reactions, both charge and mass are conserved, so with neutral ligands (such as water above), the charge of the metal ion is preserved, and to isolate this material as a pure substance we would have to introduce a counterion of some sort to give, for instance, $\left[M {\left(O {H}_{2}\right)}_{6}\right] {X}_{2}$ or $\left[M {\left(O {H}_{2}\right)}_{6}\right] {X}_{3}$.

Anyway, there is not much more I can say without leading you down the garden path. You have lecture notes and a text book, which will say the same thing I have said here in detail, and much more eloquently. Read them!