Question

The conductivity measurement of a coordination compound of Cobalt (III) shows that it dissociates into 3 ions in solution. The compound us:
A) Hexamminecobalt(III) chloride
B) Pentaammine Sulphato Cobalt(III) chloride
C) Pentaamminechloridocobalt(III)sulphate
D) Pentaamminechloridocobalt(III) chloride

Answer 1

Determine the formulas of all complexes. Ions are charged species. Species having a negative or positive charge on it is known as an ion. Ammine is a neutral ligand. Chloride and sulfates are anionic ligands. Roman number (III) indicates the charge on cobalt is +3. From the formulas of complexes determine which complex dissociates into 3 ions in solution.

Complete step-by-step answer:
We have given names of different complexes. So using the names of complexes we will determine the chemical formulas of all complexes. Also, we will determine the number of ions produced by each complex after dissociation.

The oxidation state of cobalt is +3 in all complexes.

Ammine is neutral ligand its formula is ${\text{N}}{{\text{H}}_{\text{3}}}$.

Chlorido is an anionic ligand of ion chloride and its formula is ${\text{C}}{{\text{l}}^{\text{ - }}}$.

Sulphato is an anionic ligand of anion sulphate its formula is ${\text{S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}$.

A) Hexamminecobalt (III) chloride: Here, the ligand is ammine and Hexa indicates that there are 6 amine ligands. So, the formula of Hexamminecobalt (III) ion is ${[Co{({\text{N}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{6}}}]^{3 + }}$.
Thus, we can say that 3 chloride ions will associate with cation to neutralize the +3 charge on it. Hence, the formula of Hexamminecobalt (III) chloride complex is $[Co{({\text{N}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{6}}}]C{l_3}$.

Now we will write the dissociation reaction for this complex.
$[Co{({\text{N}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{6}}}]C{l_3} \to {[Co{({\text{N}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{6}}}]^{3 + }} + {\text{3C}}{{\text{l}}^{\text{ - }}}$

Thus, Hexamminecobalt (III) chloride complex dissociates into 4 ions in solution. Hence, option (A) Hexamminecobalt (III) chloride is an incorrect answer.

B) Pentaammine Sulphato Cobalt(III ) chloride

Here, ligands are ammine and sulphato. Penta indicates that there are 5 amine ligands. So, the formula of Pentaammine Sulphato Cobalt (III ) ion is ${[Co(S{O_4}){({\text{N}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{5}}}]^ + }$. Thus, we can say that 1 chloride ion will associate with cation to neutralize the +1 charge on it. Hence, the formula of Pentaammine Sulphato Cobalt (III) chloride is$[Co(S{O_4}){({\text{N}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{5}}}{\text{]Cl}}$.

Now we will write the dissociation reaction for this complex.
$[Co(S{O_4}){({\text{N}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{5}}}{\text{]Cl}} \to {[Co(S{O_4}){({\text{N}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{5}}}]^ + } + {\text{C}}{{\text{l}}^{\text{ - }}}$

Thus, Pentaammine Sulphato Cobalt (III) chloride complex dissociates into 2 ions in solution. Hence, option (B) Pentaammine Sulphato Cobalt (III) chloride is an incorrect answer.

C) Pentaamminechloridocobalt(III)sulphate: Here, ligands are ammine and chlorido. Penta indicates that there are 5 amine ligands. So, the formula of Pentaamminechlorocobalt(III) ion is ${[CoCl{({\text{N}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{5}}}]^{2 + }}$. Thus, we can say that 1 sulphate ion will associate with cation to neutralize the +2 charge on it. Hence, the formula of Pentaamminechlorocobalt (III) sulphate is$[CoCl{({\text{N}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{5}}}]S{O_4}$.

Now we will write the dissociation reaction for this complex.
$[CoCl{({\text{N}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{5}}}]S{O_4} \to {[CoCl{({\text{N}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{5}}}]^{2 + }} + {\text{S}}{{\text{O}}_{\text{4}}}^{2 - }$

Thus, Pentaamminechloridocobalt (III) sulphate complex dissociates into 2 ions in solution. Hence, option (C) Pentaamminechlorocobalt (III) sulphate is an incorrect answer.

D) Pentaamminechloridocobalt(III) chloride: Here, ligands are ammine and chlorido. Penta indicates that there are 5 amine ligands. So, the formula of Pentaamminechlorocobalt (III) ion is ${[CoCl{({\text{N}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{5}}}]^{2 + }}$.
Thus, we can say that 2 chloride ions will associate with cation to neutralize the +2 charge on it. Hence, the formula of Pentaamminechlorocobalt (III) chloride is$[CoCl{({\text{N}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{5}}}]C{l_2}$.
Now we will write the dissociation reaction for this complex.
$[CoCl{({\text{N}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{5}}}{\text{]C}}{{\text{l}}_{\text{2}}} \to {[CoCl{({\text{N}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{5}}}]^{2 + }} + {\text{2C}}{{\text{l}}^{\text{ - }}}$
Thus, Pentaamminechloridocobalt (III) chloride complex dissociates into 3 ions in solution.

Hence, option (D) Pentaamminechloridocobalt (III) chloride is the correct answer.

Note: Sum of the oxidation state of metal ions and total charge of ligand gives the charge on the complex ion. Using the charge on the complex we can determine the number of ions associated with the complex ion.