Question

Name each of the compounds or ions given in $Exercise19.30$.

Short answer

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

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 different circumstances,the name of the ligands is written alphabetically.
• We need to add prefix before name if we have certain number of molecules: $2=di.3=tri..$
• Finally, we need to put the oxidation number of metal.
• If the ligand has negative charge, it has suffix $-o$ to its name.
• If the complex has negative charge, it has suffix $-ate$ added to the metal name.

Explanation

(a)

• For${\left(\mathrm{C}\mathrm{o}{(\mathrm{e}\mathrm{n})}_2\left(\mathrm{N}{\mathrm{O}}_2\right)\mathrm{C}\mathrm{l}\right)}^{+}$, We can use $x$for the oxidation state of $Co$.
• As we know that general oxidation state of ethylenediamine molecule is $0$ and of nitro ion and chloride ion is $-1$, we can determine the oxidation state of $Co$like this:

$\begin{array}{r}lx+(2\cdot 0)-1-1=+1\\ x-2=+1\\ x=+3\end{array}$

• It clearly means that the oxidation state of $Co$is $+3.$
• 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 $2$ ethylenediamine molecules and $1$ nitro and $1$ chloride ions. Also, we can see that this complex is a cation.
• The name of the complex is: $bis-ethylenediaminechloronitrocobalt\left(III\right)ion.$

(b)

• For${\left(\mathrm{C}\mathrm{o}{(en)}_2\mathrm{C}{\mathrm{l}}_2\right)}^{+}$, We can use $x$for the oxidation state of $Co$.
• We know that general oxidation state of ethylenediamine molecule is $0$ and of chloride ion is $-1$, so we can determine the oxidation state of $Co$like this:

$\begin{array}{r}lx+(2\cdot 0)+(2\cdot -1)=+1\\ x-2=+1\\ x=+3\end{array}$

• It means that the oxidation state of $Co$is $+3.$
• According to 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 $2$ ethylenediamine molecules and with $2$ chloride ions. Also, we can see that this complex is a cation.
• The name of the complex is $bis-ethylenediaminedichlorocobalt\left(III\right)ion.$

(c)

• For $\left(Pt{\left(N{H}_3\right)}_{2}C{l}_4\right)$, We can use $x$for the oxidation state of$Pt$. We know that general oxidation state of ammonia is $-2$and of bromide ion is $-1$.so we can determine the oxidation state of $Pt$like this:

$\begin{array}{r}lx+(2\cdot 0)+(4\cdot -1)=0\\ x-4=0\\ x=+4\end{array}$

• It means that the oxidation state of $Pt$is $+4$.
• As per the rule says, 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 $2$ ammonia molecules and with $4$ chloride ions. Also, we can see that this complex is a cation.
• The name of the complex is$diaminetetrachloroplatinum\left(IV\right).$

(d)

• For ${\left(Cr{(en)}_3\right)}^{3+}$, We can use $x$for the oxidation state of $Cr$.We know that general oxidation state of ethylenediamine molecule is $0$ , so we can determine the oxidation state of $Cr$like this:

$\begin{array}{r}lx+3\cdot 0=+3\\ x=+3\end{array}$

• It means that the oxidation state of $Cr$is $+3.$
• 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 $Cr$ which is linked to the $4$ ethylenediamine molecules. Also, we can see that this complex is a cation.
• The name of the complex is: $tris-ethylenediaminechromium\left(III\right)ion.$

(e)

• For $\left(\mathrm{P}\mathrm{t}{\left(\mathrm{N}{\mathrm{H}}_3\right)}_2\mathrm{C}{\mathrm{l}}_2\right)$, We can use $x$for the oxidation state of $Pt$.We know that general oxidation state of ammonia molecule is $0$and of chloride ion is $-1$, so we can determine the oxidation state of $Pt$like this:$\begin{array}{r}lx+(2\cdot 0)+(2\cdot -1)=0\\ x-2=0\\ x=+2\end{array}$
• 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 $2$ ammonia molecules and with $2$ chloride ions. Also, we can see that this complex is a cation.
• The name of the complex is $diaminedichloroplatinum\left(III\right).$

Result

$\begin{array}{r}\mathrm{a})\mathrm{B}\mathrm{i}\mathrm{s}-\mathrm{e}\mathrm{t}\mathrm{h}\mathrm{y}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{c}\mathrm{h}\mathrm{l}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{o}\mathrm{c}\mathrm{o}\mathrm{b}\mathrm{a}\mathrm{l}\mathrm{t}(\mathrm{I}\mathrm{I}\mathrm{I})\mathrm{i}\mathrm{o}\mathrm{n}\\ \mathrm{b})\mathrm{B}\mathrm{i}\mathrm{s}-\mathrm{e}\mathrm{t}\mathrm{h}\mathrm{y}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{c}\mathrm{h}\mathrm{l}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{c}\mathrm{o}\mathrm{b}\mathrm{a}\mathrm{l}\mathrm{t}(\mathrm{I}\mathrm{I}\mathrm{I})\mathrm{i}\mathrm{o}\mathrm{n}\\ \mathrm{c})\mathrm{D}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{c}\mathrm{h}\mathrm{l}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{m}(\mathrm{I}\mathrm{V})\\ \mathrm{d})\mathrm{T}\mathrm{r}\mathrm{i}\mathrm{s}-\mathrm{e}\mathrm{t}\mathrm{h}\mathrm{y}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{c}\mathrm{h}\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{i}\mathrm{u}\mathrm{m}(\mathrm{I}\mathrm{I}\mathrm{I})\mathrm{i}\mathrm{o}\mathrm{n}\\ \mathrm{e})\mathrm{D}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{c}\mathrm{h}\mathrm{l}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{m}(\mathrm{I}\mathrm{I}\mathrm{I}).\end{array}$