Question

Using IUPAC norms write the formula for the following:
a. Potassium trioxalatoaluminate(III)
b. Dichlorobis(ethane-$1,2$-diamine)cobalt(III) chloride

Answer 1

Hint We can write the formula of given coordination compounds by using the IUPAC norms as per compounds with complex anion and complex cation respectively.

Complete step by step solution:
Let’s have a brief look at the IUPAC norms that need to be followed for writing chemical formula for coordination compounds:
1. In case a compound contains both the cation and anion, we will write the formula of cation followed by formula of anion
2. Now, for the complex entity, first of all, we will list the central metal
3. Alphabetical order is followed for the ligands while we ignore the prefix or charge
4. For polyatomic or abbreviated ligands, we use parenthesis for enclosing their formulae
5. Then, we will enclose the coordination entity with square brackets.
6. Finally, for any charge on the complex ion, we will write that as the right superscript or we will use the counter ions as per formula.
Now, let’s write the formula for the given compounds:

a. We are given the name of the coordination compound to be potassium trioxalatoaluminate(III). As we can see that in the given coordination compound, potassium ions are cations whereas anion is a complex one. Let’s deduce the formula for complex anion first by following the above listed rules as follows:
1. The central metal is aluminum: ${\rm{Al}}$
2. The ligand is oxalate ions with a prefix tri-
i. The formula of the oxalate is ${{\rm{C}}_2}{\rm{O}}_4^{2 - }$
ii. The anion is polyatomic, so we will use the parenthesis: $\left( {{{\rm{C}}_2}{\rm{O}}_4^{2 - }} \right)$
iii. Now we will write the number as per the prefix (tri-three): ${\left( {{{\rm{C}}_2}{\rm{O}}_4^{2 - }} \right)_3}$
3. We will enclose the complex anion: $\left[ {{\rm{Al}}{{\left( {{{\rm{C}}_2}{\rm{O}}_4^{2 - }} \right)}_3}} \right]$
4. Now, we will determine the charge on the complex anion by using the given oxidation state of aluminum $\left( { + 3} \right)$ and the charge on the oxalate as follows:
$x = \left( { + 3} \right) + \left( {3 \times \left( { - 2} \right)} \right)\\\ = + 3 - 6\\\ = - 3$
5. So, three potassium ions will be used to counter this charge.
6. Finally, we can write the formula to be ${{\rm{K}}_{\rm{3}}}\left[ {{\rm{Al}}{{\left( {{{\rm{C}}_2}{\rm{O}}_4^{2 - }} \right)}_3}} \right]$
Hence, the formula of Potassium trioxalatoaluminate(III) is ${{\rm{K}}_{\rm{3}}}\left[ {{\rm{Al}}{{\left( {{{\rm{C}}_2}{\rm{O}}_4^{2 - }} \right)}_3}} \right]$.

b. We are given the name of the coordination compound to be Dichlorobis(ethane-$1,2$-diamine)cobalt(III) chloride. As we can see that in the given coordination compound, cation is a complex entity whereas anions are chlorides. Let’s deduce the formula for complex cation first by following the above listed rules as follows:
1. The central metal is cobalt: ${\rm{Co}}$
2. The first ligand is chlorido with a prefix di-:
i. Chlorido is used for chlorine with formula as ${\rm{Cl}}$
ii. The anion is monoatomic, so we will write the number as per the prefix (di-two): ${\rm{C}}{{\rm{l}}_2}$
3. The next ligand is ethane-$1,2$-diamine with a prefix bis-:
i. The formula for ethane-$1,2$-diamine is: ${{\rm{H}}_{\rm{2}}}{\rm{NC}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{N}}{{\rm{H}}_{\rm{2}}}$
ii. The anion is polyatomic, so we will use the parenthesis: $\left( {{{\rm{H}}_{\rm{2}}}{\rm{NC}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{N}}{{\rm{H}}_{\rm{2}}}} \right)$
iii. Now, we will write the number as per the prefix (bis-two): ${\left( {{{\rm{H}}_{\rm{2}}}{\rm{NC}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{N}}{{\rm{H}}_{\rm{2}}}} \right)_2}$
4. We will enclose the complex anion: $\left[ {{\rm{CoC}}{{\rm{l}}_2}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{NC}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{N}}{{\rm{H}}_{\rm{2}}}} \right)}_2}} \right]$
5. Now, we will determine the charge on the complex anion by using the given oxidation state of cobalt $\left( { + 3} \right)$ and the charge on the ligands as follows:
$x = \left( { + 3} \right) + \left( {2 \times \left( { - 1} \right)} \right) + \left( {2 \times 0} \right)\\\ = + 3 - 2\\\ = + 1$
6. So, one chloride ion will be used to counter this charge.
Finally, we can write the formula to be $\left[ {{\rm{CoC}}{{\rm{l}}_2}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{NC}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{N}}{{\rm{H}}_{\rm{2}}}} \right)}_2}} \right]{\rm{Cl}}$
7. Hence, the formula of Dichlorobis(ethane-$1,2$-diamine)cobalt(III) chloride is $\left[ {{\rm{CoC}}{{\rm{l}}_2}{{\left( {{{\rm{H}}_{\rm{2}}}{\rm{NC}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{N}}{{\rm{H}}_{\rm{2}}}} \right)}_2}} \right]{\rm{Cl}}$.

Note:
For ethane-$1,2$-diamine, its abbreviated form (en) can also be used in the formula in place of complete formula.