Question

Give the chemical formula that corresponds to each of the following compounds:

(a) Sodium tetrahydroxozincate(II)

(b) Dichlorobis(ethylenediamine)cobalt(III) nitrate

(c) Triaquabromoplatinum(II) chloride

(d) Tetra-amminedinitroplatinum(IV) bromide

Short answer

$\begin{array}{l}{\text{a)Na}}_2\left[\text{Zn}{\left(\text{OH}\right)}_{\text{4}}\right]\\ \mathrm{b})\left[{\text{CoCl}}_2{\left(\text{en}\right)}_{\text{2}}\right]{\text{NO}}_{\text{3}}\\ \mathrm{c})\left[\text{Pt}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{2}}\text{Br}\right]\text{Cl}\\ \mathrm{d})\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{NO}}_{\text{2}}\right]{\text{Br}}_{\text{2}}\end{array}$

Step-by-step solution

Step 1:Coordination compound

Coordination compounds are “complex molecules” in which metal ions (mainly transition metal) are bounded to ligands that can be neutral or possess a negative charge.

a) Chemical formula of Sodium tetrahydroxozincate(II)

In the coordination compound, the Sodium tetrahydroxozincate(II) anion has four hydroxide ligand (with -1 charges) are coordinated to zinc atom in +2 oxidation state complex will have overall of -2 charge, two sodium cation are required thus the chemical formula of Sodium tetrahydroxozincate(II) becomes:

${\text{Na}}_2\left[\text{Zn}{\left(\text{OH}\right)}_{\text{4}}\right]$

b) Chemical formula of Dichlorobis(ethylenediamine)cobalt(III) nitrate

In the coordination compound dichlorobis(ethylenediamine), the cobalt(III)nitrate anion part has two chloride ligands and two ethylenediamine ligands which are coordinated to cobalt metal which is in a +3 oxidation state; ethylenediamine is neutral ligand two chloride ligand contribute to $2\times (-1)=-2$charge that sum with +3 charge of cobalt giving complex +1 charge one nitrate anion is required to balance this charge thus the formula of dichlorobis(ethylenediamine)cobalt(III)nitrate becomes:

$\left[{\text{CoCl}}_2{\left(\text{en}\right)}_{\text{2}}\right]{\text{NO}}_{\text{3}}$

c) Chemical formula of Triaquabromoplatinum(II) chloride

In the coordination compound triaquabromoplatinum(II) chloride anion part has one bromide ligand and three water ligand (aqua) which are coordinated to platinum metal which is in +2 oxidation state, water is neutral ligand one bromide ligand contribute to $1\times (-1)=-1$charge that sum with +2 charge of platinum giving complex +1 charge one chloride anion is required to balance this charge thus the formula of triaquabromoplatinum(II) chloride becomes:

$\left[\text{Pt}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{2}}\text{Br}\right]\text{Cl}$

d) Chemical formula of Tetra-amminedinitroplatinum(IV) bromide

In the coordination compound, tetra-amminedinitroplatinum(IV) bromide anion part has four ammine ligands and two nitro ligands which are coordinated to platinum metal which is in +4 oxidation state; the amine is neutral ligand two nitro ligand contribute to $2\times (-1)=-2$charge that sum with +4 charge of platinum giving complex +2 charge two bromide anion is required to balance this charge thus the formula of tetra-amminedinitroplatinum(IV) bromide becomes:

$\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{NO}}_{\text{2}}\right]{\text{Br}}_{\text{2}}$