Question

1. pdcl₂ = palladium(ii) chloride
2. ircl₃ = iridium(iii) chloride
3. sbs₃ = antimony(iii) sulfide
4. fe₂(so₄)₃ =
5. cro₄²⁻ =
6. v₂(so₄)₅ =
7. cscl₂ =
8. rhi₃ =

Explanation:

• For \(Fe_2(SO_4)_3\): The cation is \(Fe^{3 +}\) (iron with a +3 charge) and the anion is \(SO_4^{2-}\) (sulfate). So it is iron(III) sulfate.
• For \(CrO_4^{2 -}\): This is a polyatomic ion. The name of \(CrO_4^{2 -}\) is chromate ion.
• For \(V_2(SO_4)_5\): The cation is \(V^{5+}\) (vanadium with a +5 charge) and the anion is \(SO_4^{2-}\) (sulfate). So it is vanadium(V) sulfate.
• For \(CsCl_2\): The cation is \(Cs^{+}\) (cesium with a +1 charge) and the anion is \(Cl^{-}\). But \(CsCl_2\) is not a common stable compound. If we follow the naming rules for ionic compounds (metal + non - metal), it would be cesium chloride (but in reality, cesium usually forms \(CsCl\) as \(Cs^{+}\) and \(Cl^{-}\) combine in a 1:1 ratio. However, if we assume the formula is given as is, we name it as cesium chloride).
• For \(RhI_3\): The cation is \(Rh^{3+}\) (rhodium with a +3 charge) and the anion is \(I^{-}\) (iodide). So it is rhodium(III) iodide.

Answer:

1. iron(III) sulfate
2. chromate ion
3. vanadium(V) sulfate
4. cesium chloride
5. rhodium(III) iodide