Answer
Tetraamminediaquairidium(III) nitrate.
Work Step by Step
In $[Ir(NH_{3})_{4}(H_{2}O)_{2}](NO_{3})_{3}$, $[Ir(NH_{3})_{4}(H_{2}O)_{2}]$ is the cation whereas $(NO_{3})_{3}$ is the anion. Cations are named first, then are anions.
2. In this complex, $Ir$ is bonded to two different ligands $(NH_{3})_{4}$ and $(H_{2}O)_{2}$ The ligands are named first then is the metal. Prefixes are added and Ligands are named alphabetically. $(NH_{3})_{4}$ is tetrammine and $(H_{2}O)_{2}$ is diaqua.
3. Ir is Iridium.
4. In order to find the oxidation number on the metal, we add the charges and equal it to the charge on the complex, in this case zero. $(NH_{3})_{4}$ and $(H_{2}O)_{2}$ have no charge and $(NO_{3})_{3}$ is $3\times( -1)$.
Thus,
$x+3\times(-1)=0$
$x-3=0$
$x=+3$
The name is:
Tetraamminediaquairdium(III) nitrate.
