Question

Write the formulas for following coordination compounds.
(a)Tetraammineaquachloro cobalt (III) chloride
(b) Potassium tetrahydroxozincate (II)
(c) Potassium trioxalatoaluminate (III)
(d) Tetracarbonylnickel (0)

Answer 1

First part of the name represents the name of the cation and the second part represents the name of the anion. Each entity name contains the name of ligands in alphabetical order followed by the name of metal with its oxidation state.

Complete step by step answer:
The IUPAC rules for the naming of coordination compounds is as follows:
Write the name of the cation part then the name of the anion part.
Write the name of the ligand in alphabetical order.
For neutral ligand, the name of the compound is used as such but for water use ‘aqua’ for ammonia use ‘ammine’
For anionic ligands replace the,
Ide with o.
Ate with ato.
Ite with ito.
Use Greek prefix to denote the number of the ligand. If the name of the ligand has a Greek prefix then use bis for two, tris for three.
Write the name of the metal.
For the cationic complex, write the metal name as such.
For anionic complexes, use the suffix ‘ate’.
(a) Write the formula of Tetraammineaquachloro cobalt (III) chloride as follows:
The name shows the coordination entity has four amine ligands, one aqua (water) ligand, one chloro (chlorine atom) ligand and cobalt metal. The counter ion is chloride ion.
So, the coordination number of cobalt metal six.
The charge on the cobalt metal is + 3 and the charge on the chloro ligand is - 1. Suppose the charge on the complex is x so,
$\left( { + 3 \times 1} \right) + \left( {0 \times 4} \right) + \left( {0 \times 1} \right) + \left( { - 1 \times 1} \right) = x$
x = + 2
Charge on the complex is + 2 so, the chlorine ions (counter ions) will be two.
So, the formula of the coordination compound is $\left[ {{\text{Co}}{{\left( {{\text{N}}{{\text{H}}_{\text{3}}}} \right)}_4}\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right){\text{Cl}}} \right]{\text{C}}{{\text{l}}_{\text{2}}}$.

(b) Write the formula of Potassium tetrahydroxozincate (II) as follows:
The name shows the coordination entity has four hydroxo ligands, and zinc metal.
So, the coordination number of zinc metal four.
The charge on the zinc metal is + 2 and the charge on the hydroxo ligand is - 1. Suppose the charge on the complex is x so,
$\left( { + 2 \times 1} \right) + \left( { - 1 \times 4} \right) = x$
$x = - 2$
Charge on the complex is - 2 so, the potassium ions (counter ions) will be two.
So, the formula of the coordination compound is ${{\text{K}}_2}\left[ {{\text{Zn}}{{\left( {{\text{OH}}} \right)}_4}} \right]$.

(c) Write the formula of Potassium trioxalatoaluminate (III) as follows:
The name shows the coordination entity has three oxalate ligands, and aluminium metal.
The Density of oxalate ions is two.
So, the coordination number of aluminium metal six.
The charge on the aluminium metal is + 3 and the charge on the oxalate ligand is - 2. Suppose the charge on the complex is x so,
$\left( { + 3 \times 1} \right) + \left( { - 2 \times 3} \right) = x$
$x = - 3$
Charge on the complex is - 3 so, the potassium ions (counter ions) will be three.
So, the formula of the coordination compound is ${{\text{K}}_{\text{3}}}\left[ {{\text{Al}}{{\left( {{\text{ox}}} \right)}_{\text{3}}}} \right]$.

(d) Write the formula of Tetracarbonylnickel (0) as follows:
The name shows the coordination entity has four carbonyl ligands, and nickel metal.
So, the coordination number of aluminium metal four.
The charge on the nickel metal is zero and the charge on the carbonyl ligand is zero.
Charge on the complex is zero.
So, the formula of the coordination compound is $\left[ {{\text{Ni}}{{\left( {{\text{CO}}} \right)}_4}} \right]$.

Note: The molecules that bounds with the central atom are known as ligands. Denticity defines the number of donor atoms in a ligand. The number of counter ions depends upon the charge on the complex.