Answer
See the explanation
Work Step by Step
a. Neutral Na has 11 electrons → $Na^+$ has 10 electrons. A negative ion with 10 electrons is $F^⁻$ (fluoride ion, atomic number 9 + 1 extra e⁻ = 10).
b. Neutral Ca has 20 electrons → $Ca^{2+}$ has 18 electrons. A negative ion with 18 electrons is $S^{2-}$ (sulfide ion, atomic number 16 + 2 = 18).
c. Neutral Al has 13 electrons → $Al^{3+}$ has 10 electrons. A negative ion with 10 electrons is $F^⁻$ (fluoride ion, same as $Na^⁺$).
d. Neutral Rb has 37 electrons → $Rb^⁺$ has 36 electrons. A negative ion with 36 electrons is $Se^{2-}$ (selenide ion, atomic number 34 + 2 = 36).
