Correct Answer: Option C

Explanation

Ca\(^{2+}\) (calcium ions) are not soluble in excess NaOH (sodium hydroxide); when mixed, they form a white precipitate of calcium hydroxide (Ca(OH)\(_2\)) which does not dissolve further in excess NaOH, making it a key distinguishing factor in identifying calcium ions in solution compared to other cations like aluminum that can dissolve in excess NaOH. 

Mg\(^{2+}\) is not soluble in excess NaOH; when added to a solution containing Mg\(^{2+}\) ions, sodium hydroxide (NaOH) will produce a white precipitate of magnesium hydroxide (Mg(OH)\(_2\)) which is insoluble in excess NaOH. 

Zn\(^{2+}\) is soluble in excess NaOH, meaning that when you add a solution of sodium hydroxide (NaOH) in excess to a zinc ion Zn\(^{2+}\) solution, the initially formed white precipitate of zinc hydroxide will dissolve due to the formation of a complex ion,. 

Cu\(^{2+}\) (copper ion) is not soluble in excess NaOH (sodium hydroxide); when added to a solution containing Cu\(^{2+}\), NaOH will produce a blue precipitate of copper hydroxide (Cu(OH)\(_2\)) which does not dissolve in excess NaOH unless the NaOH solution is very concentrated. 

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