Question

Name each of the compounds or ions given in \(Exercise 19.28,\)including the oxidation state of the metal.

Short answer

Coordination compounds are the compound in which atoms or molecules are attached to the central metal ion through Lewis acid-Lewis base interactions.

Step-by-step solution

Rules for naming compounds

• There are few rules for naming these compounds:
• If we have the ionic compound, the nomenclature will be the same as the name of cation followed by name of the anion
• In all other conditions, the name of the ligands is written alphabetically.
• We also need to put prefix before name if we have certain number of molecules: \(2 = d{\rm{ }}i{\rm{ }}.3 = {\rm{ }}tri...\)
• Finally, we need to put the oxidation number of metal.
• If the ligand has negative charge, it has sufix \( - o\) to its name.
• If the complex has negative charge, it has sufix \( - ate\) added to the metal name.

Explanation

(a)

• For\({\left( {{\rm{Co}}{{\left( {{\rm{C}}{{\rm{O}}_3}} \right)}_3}} \right)^{3 - }}\), We can use \(x\)for the oxidation state of \(Co\). As we know that general oxidation state of carbonate ion \( - 2,\) we can determine the oxidation state of \(Co\)like this:

\(\begin{aligned}{l}x + 3 \cdot - 2 = - 3\\x - 6 = - 3\\x = + 3\end{aligned}\)

• The oxidation state of\(Co\) is \( + 3.\)
• As per the rules, we can see that we need to write first the ligand and then the central metal. In this case the central metal is \(Co\) which is linked to the \(3\)carbonate ions. Also, we can see that this complex is an anion.
• The name of the complex is \(tricarbonatocobaltate{\rm{ }}\left( {III} \right){\rm{ }}ion.\)

(b)

• For \({\left( {{\rm{Cu}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_4}} \right)^{2 + }}\), We can use \(x\)for the oxidation state of \(Cu.\)We know that general oxidation state of ammonia is \(0\) , so we can determine the oxidation state of \(Cu.\)like this:

\(\begin{aligned}{l}x + 4 \cdot 0 = + 2\\x = + 2\end{aligned}\)

• It means that the oxidation state of \(Cu.\)is \( + 2.\)
• As per the rules, we can see that we need to write first the ligand and then the central metal. In this case the central metal is \(Cu.\)which is linked to the \(4\)ammonia molecules. Also, we can see that this complex is a cation.
• The name of the complex is:\(\;tetraaminecopper{\rm{ }}\left( {II} \right){\rm{ }}ion.\)

(c)

• For \({\left( {{\rm{Co}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_4}{\rm{B}}{{\rm{r}}_2}} \right)_2}{\left( {{\rm{S}}{{\rm{O}}_4}} \right)_3}\), We can use \(x\)for the oxidation state of \(Co\).
• As we know that general oxidation state of ammonia molecule is\(0\), \(B{r^ - }\)ion is \( - 1\)and sulfate ion is \( - 2\).
• Hence, we can determine the oxidation state of \(Co\)like this:

\(\begin{aligned}{l}2 \cdot x + 2 \cdot (4 \cdot 0) + 2 \cdot (2 \cdot - 1) = + 6\\2x - 4 = + 6\\2x = 10\\x = + 5\end{aligned}\)

• It means the oxidation state of \(Co\)is \( + 5.\)
• As per the rules, we need to write first the ligand and then the central metal.
• In this case the central metal is \(Co\) which is linked to the \(4\)ammonia molecules and \(2\)bromide ions. Also, we can see that this complex is a cation.
• The name of the complex is: \(tetraaminedibromocobalt{\rm{ }}\left( {III} \right){\rm{ }}sulfate\)

(d)

• For\(\left( {{\rm{Pt}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_4}} \right)\left( {{\rm{PtCl}}{l_4}} \right)\), We can use \(x\)for the oxidation state of \(Pt\).
• As we know that general oxidation state of ammonia is \(0\) and of \(Cl\)ion is \( - 1,\)
• Hence, we can determine the oxidation state of \(Pt\)like this:

\(\begin{aligned}{l}2x + (4 \cdot 0) + (4 \cdot - 1) = 0\\2x - 4 = 0\\2x = 4\\x = + 2\end{aligned}\)

• It means the oxidation state of \(Pt\)is \( + 2.\)
• If we look at the rules we can see that we need to write first the ligand and then the central metal. In this case the central metal is \(Pt\) which is linked to the \(4\)ammonia molecules and other with \(4\) chloride ions. Also, we can see that this complex is a cation and the other is an anion.
• The name of the complex is: \(tetraamineplatinum{\rm{ }}\left( {II} \right){\rm{ }}tetrachloroplatinate{\rm{ }}\left( {II} \right)\)

(e)

• For \(\left( {{\rm{Cr}}{{(en)}_3}} \right){\left( {{\rm{N}}{{\rm{O}}_3}} \right)_3}\), We can use \(x\)for the oxidation state of \(Cr\)We know that general oxidation state of ethylenediamine molecule is \(0\) and of nitrate ion is \( - 1,\)so we can determine the oxidation state of \(Cr\)like this:

\(\begin{aligned}{l}x + (3 \cdot 0) + (3 \cdot - 1) = 0\\x - 3 = 0\\x = + 3\end{aligned}\)

• It means the oxidation state of \(Cr\)is \( + 3.\)
• As per the rules, we can see that we need to write first the ligand and then the central metal. In this case the central metal is \(Cr\) which is linked to the \(3\)ethylenediamine molecules and with \(4\)nitrate ions. Also, we can see that this complex is a cation.
• The name of the complex is \(tris - ethylenediaminechromium{\rm{ }}\left( {III} \right){\rm{ }}nitrate\).

(f)

• For \(\left( {Pd{{\left( {N{H_3}} \right)}_2}B{r_2}} \right)\), We can use \(x\)for the oxidation state of \(Pd.\)
• As we know that general oxidation state of ammonia is \(0\) and of bromide \(I\), on is \( - 1,\)so we can determine the oxidation state of \(Pd\)like this:

\(\begin{aligned}{l}x + (2 \cdot 0) + (2 \cdot - 1) = 0\\x = + 2\end{aligned}\)

• It means the oxidation state of \(Pd.\)is\( + 2\).
• As per the rules we can see that we need to write first the ligand and then the central metal. In this case the central metal is \(Pd.\) which is linked to the \(2\) ammonia molecules and with \(2\)bromide ions. Also, we can see that this complex is a cation.
• The name of the complex is\(diaminedibromopalladium{\rm{ }}\left( {II} \right)\).

(g)

• For \({K_3}\left( {Cu{{(Cl)}_5}} \right)\), we can use \(x\)for the oxidation state of \(Cu.\)
• As we know that general oxidation state of chloride ion is \( - 1\)and of Potassium ion is \( + 1,\) so we can determine the oxidation state of \(Cu.\)like this:

\(\begin{aligned}{l}x + (3 \cdot + 1) + (5 \cdot - 1) = 0\\x - 2 = 03\\x = + 2\end{aligned}\)

• It means the oxidation state of \(Cu.\)is \( + 2.\)
• If we look at the rules we can see that we need to write first the ligand and then the central metal. In this case the central metal is \(Cu.\) which is linked to the \(5\)chloride ions. Also, we can see that this complex is an anion.
• The name of the complex is: \(potassium{\rm{ }}pentachlorocuprate{\rm{ }}\left( {II} \right).\)

(h)

• For\(\left( {{\rm{Zn}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_2}{\rm{C}}{{\rm{l}}_2}} \right)\), We can use \(x\)for the oxidation state of \(Zn\).
• As we know that general oxidation state of ammonia is \(0\) and of chloride ion is \( - 1\), we can determine the oxidation state of \(Zn\)like this:
• \(\begin{aligned}{l}x + (2 \cdot 0) + (2 \cdot - 1) = 0\\x = + 2\end{aligned}\)
• Which means the oxidation state of \(Zn\)is \( + 2.\)
• As per the rules, we can see that we need to write first the ligand and then the central metal. In this case the central metal is \(Zn\) which is linked to the \(2\) ammonia molecules and with \(2\) chloride ions. Also, we can see that this complex is a cation.
• The name of the complex is: \(diaminedichlorozinc{\rm{ }}\left( {II} \right).\)

Result

\(\begin{aligned}{}{\rm{a) Tricarbonatocobaltate (III) ion}}\\{\rm{b) Tetraaminecopper (II) ion}}\\{\rm{c) Tetraaminedibromocobalt (III) sulfate}}\\{\rm{d) Tetraamineplatinum (II) tetrachloroplatinate (II)}}\\{\rm{e) Tris - ethylenediaminechromium (III) nitrate}}\\{\rm{f) Diaminedibromopalladium (II)}}\\{\rm{g) Potassium pentachlorocuprate (II)}}\\{\rm{h) Diaminedichlorozinc (II)}}\end{aligned}\)