Question

Write the names of the following compounds, using the standard nomenclature rules for coordination complexes: (a) $\left[\mathrm{Rh}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}$ (b) ${\mathrm{K}}_2\left[{\mathrm{TiCl}}_6\right]$ (c) ${\mathrm{MoOCl}}_4$ (d) $\left[\mathrm{Pt}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4\left({\mathrm{C}}_2{\mathrm{O}}_4\right)\right]{\mathrm{Br}}_2$

Short answer

The names of the coordination complexes are: (a) tetrachlorobis(ammine)rhodium(III) chloride (b) potassium hexachlorotitanate(IV) (c) molybdenum(VI) tetrachlorooxide (d) dibromido-tetraaqua(oxalato)platinum(IV)

Step-by-step solution

Name compound (a)

Compound (a) is $\left[\mathrm{Rh}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}$. The metal is Rhodium (Rh), which means that we should use the name "Rhodium" in the compound's name. The ligands are ${\mathrm{NH}}_3$, named "ammine" and $\mathrm{Cl}$, named "chloro". There are 4 ammine ligands and 2 chloro ligands. The complex carries one negative charge, so the counterion is simply $\mathrm{Cl}$, named "chloride". The name of the complex in alphabetical order is "tetrachlorobis(ammine)rhodium(III) chloride".

Name compound (b)

Compound (b) is ${\mathrm{K}}_2\left[{\mathrm{TiCl}}_6\right]$. The metal in the complex is Titanium (Ti). The ligand attached is 6 chloro ligands ($\mathrm{Cl}$). The complex ion has a charge of -2. The counterions are 2 potassium ions ($\mathrm{K}$) naming "potassium". The name of the compound in alphabetical order is "potassium hexachlorotitanate(IV)".

Name compound (c)

Compound (c) is ${\mathrm{MoOCl}}_4$. It consists of a Molybdenum (Mo) metal with oxygen and chloro ligands. There is no counterion since it's a neutral complex, so there is no need to indicate any cation. We have 1 oxygen atom and 4 chloro ligands. The name of the compound in alphabetical order is "molybdenum(VI) tetrachlorooxide".

Name compound (d)

Compound (d) is $\left[\mathrm{Pt}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4\left({\mathrm{C}}_2{\mathrm{O}}_4\right)\right]{\mathrm{Br}}_2$. The metal is Platinum (Pt). The ligands are ${\mathrm{H}}_2\mathrm{O}$, named "aqua" and ${\mathrm{C}}_2{\mathrm{O}}_4$, named "oxalato". There are 4 aqua ligands and 1 oxalato ligand. The complex carries two positive charges, so the counterion is ${\mathrm{Br}}_2$, each named "bromide". The name of the complex in alphabetical order is "dibromido-tetraaqua(oxalato)platinum(IV)".