To determine which substance is most likely to form in a precipitation reaction, we need to analyze each option using the provided solubility rules:
1. Calcium nitrate ([tex]$\text{Ca(NO}_3\text{)_2}$[/tex]):
- According to Rule 2, nitrates ([tex]$\text{NO}_3{^-}$[/tex]) are soluble regardless of the cation. Therefore, calcium nitrate is soluble.
2. Copper(I) nitrate ([tex]$\text{CuNO}_3$[/tex]):
- Again, Rule 2 states that nitrates ([tex]$\text{NO}_3{^-}$[/tex]) are soluble regardless of the cation. Therefore, copper(I) nitrate is soluble.
3. Iron(I) chloride ([tex]$\text{FeCl}$[/tex]):
- According to Rule 3, chlorides ([tex]$\text{Cl}^-$[/tex]) are generally soluble, except when combined with silver ([tex]$\text{Ag}^+$[/tex]), mercury(I) ([tex]$\text{Hg}_2^{2+}$[/tex]), and lead ([tex]$\text{Pb}^{2+}$[/tex]). Since iron is not one of these exceptions, iron(I) chloride is soluble.
4. Iron(II) sulfide ([tex]$\text{FeS}$[/tex]):
- Rule 1 and 2 do not apply here. According to general solubility rules for sulfides ([tex]$\text{S}^{2-}$[/tex]), they are usually insoluble except when paired with group 1 alkali metals or ammonium ([tex]$\text{NH}_4{^+}$[/tex]). Since iron does not fall into these exceptions, iron(II) sulfide is insoluble.
5. Potassium sulfate ([tex]$\text{K}_2\text{SO}_4$[/tex]):
- According to Rule 4, sulfates ([tex]$\text{SO}_4^{2-}$[/tex]) are soluble except with calcium ([tex]$\text{Ca}^{2+}$[/tex]), strontium ([tex]$\text{Sr}^{2+}$[/tex]), barium ([tex]$\text{Ba}^{2+}$[/tex]), and lead ([tex]$\text{Pb}^{2+}$[/tex]). Potassium is a group 1 metal, so potassium sulfate is soluble.
Among these options, the substance that is insoluble and thus most likely to form in a precipitation reaction is:
D. iron(II) sulfide
