Question

Sulfite is similar to hydroxide in its behavior as a ligand: most sulfites are insoluble in water, but excess sulfite can result in the formation of a complex ion. Write the two equilibrium reactions of sulfite with the copper(II) ion that show this behavior.

Answer 1

`$Cu_2+(SO_3)^2 \rightleftharpoons CuSO_3 \ (insoluble)$`

`$Cu_2+2(SO_3)^2 \rightleftharpoons [Cu(SO_3)_2]_2 \ (soluble)$`

Step-by-step explanation:

A ligand may be defined as a molecule or an ion or that binds to the central metal atom in order to form a more coordination complex. Sulfite is one such ligand and it behaves similarly as hydroxide as a ligand.

Now, according to the question, when we react copper with sulfite ion, it forms copper sulfite. The equation is

`$Cu_2+(SO_3)^2 \rightleftharpoons CuSO_3 \ (insoluble)$`

Now when excess of the sulfite ion is used in the reaction, we get a complex formation, which is shown by

`$Cu_2+2(SO_3)^2 \rightleftharpoons [Cu(SO_3)_2]_2 \ (soluble)$`

The way the sulfite reacts is quite similar to hydroxide ion where they form a complex ion when hydroxide ion in excess is used in the reaction with metal cation.