Question

Write the formula of tetraammineaquachlorocobalt(III) chloride.

Answer 1

Hint:
We can deduce the formula of a given coordination compound by following the guidelines provided by IUPAC.

Complete step by step solution:
We have nomenclature and formula writing rules for different types of compounds recommended by IUPAC. Here, we will take up the IUPAC set rules for writing chemical formulas for coordination compounds. We can enlist the rules as follows:

- First of all, we will write the formula of cation followed by that of anion
- In the complex, we will list the central metal to which ligands are attached
- For, ligands, alphabetical order is followed ignoring the charge or any prefix in the name
- In case, any ligand is polyatomic or we are using an abbreviation, we will use parenthesis for enclosing these and their number would be written outside
- We don’t leave any space between the formulae
- After writing the formula for central metal and the ligands, we will enclose this coordination entity by square brackets.
- Now, if we have complex ion then its charge is written as right superscript otherwise counter ions are written

Here, we are given the name as tetraammineaquachlorocobalt(III) chloride. We can see that the given coordination compound has complex cation and chlorides as anions. So, let’s write the formula for complex cation first by following the above listed rules as follows:

- The central metal is cobalt: $\mathrm{C}\mathrm{o}$
- First listed ligand is amine with a prefix tetra-:
a. The formula of the amine is $\mathrm{N}{\mathrm{H}}_3$
b. It is polyatomic, so we will enclose this: $\left(\mathrm{N}{\mathrm{H}}_3\right)$
c. Now we will write the number derived from the prefix (tetra-four): ${\left(\mathrm{N}{\mathrm{H}}_3\right)}_4$
- The next listed ligand is aqua:
a. we use this name for water molecule
b. the formula of water is: ${\mathrm{H}}_2\mathrm{O}$
c. It is polyatomic, so we will enclose this: $\left({\mathrm{H}}_2\mathrm{O}\right)$
- The next listed ligand is chlorido:
a. we use this name for anionic chloride ion
b. the formula for this is: $\mathrm{C}{\mathrm{l}}^{-}$

- From all these, we can write the formula of complex cation to be $\mathrm{C}\mathrm{o}\mathrm{C}\mathrm{l}\left({\mathrm{H}}_2\mathrm{O}\right){\left(\mathrm{N}{\mathrm{H}}_3\right)}_4$
- We will enclose it: $\left[\mathrm{C}\mathrm{o}\mathrm{C}\mathrm{l}\left({\mathrm{H}}_2\mathrm{O}\right){\left(\mathrm{N}{\mathrm{H}}_3\right)}_4\right]$
- Now, to determine the number of counter ions we will find out the charge on the complex ion by using the given oxidation state of cobalt $\left(+3\right)$ along with charge on the ligands as follows:
$$\begin{gathered} x=\left(+3\right)+\left(1\times \left(-1\right)\right)+\left(1\times 0\right)+\left(4\times 0\right) \\ =+3-1 \\ =+2 \end{gathered}$$
- It means that chlorides that are counter ions will counter $\left(+2\right)$ charge. We can find the number of chlorides by using the charges in accordance with charge neutrality principle as follows
$$\begin{gathered} \left\{n\times \left(-1\right)\right\}+2=0 \\ -n+2=0 \\ n=2 \end{gathered}$$
- Finally, we can write the formula for the given compound to be $$\left[\mathrm{C}\mathrm{o}\mathrm{C}\mathrm{l}\left({\mathrm{H}}_2\mathrm{O}\right){\left(\mathrm{N}{\mathrm{H}}_3\right)}_4\right]\mathrm{C}{\mathrm{l}}_2$$

- Hence, the formula of tetraammineaquachlorocobalt(III) chloride is $$\left[\mathrm{C}\mathrm{o}\mathrm{C}\mathrm{l}\left({\mathrm{H}}_2\mathrm{O}\right){\left(\mathrm{N}{\mathrm{H}}_3\right)}_4\right]\mathrm{C}{\mathrm{l}}_2$$.

Note:
We have to follow all the recommended rules while writing the formula.Missing any of the steps will result in wrong naming of the compound.