A few drops of sodium hydroxide is added to an aqueous solution, a green precipitate is formed. The ion that may be present in the solution is
Cu\(^{2+}\)
Fe\(^{2+}\)
Fe\(^{3+}\)
NH\(_4\)\(^{+}\)
Explanation
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Discussions (1)

Let's check the options:
A. Cu²⁺ (Copper II ion)
With NaOH → blue precipitate (Cu(OH)₂)
Not green
𝗕. 𝗙𝗲²⁺ (𝗜𝗿𝗼𝗻 𝗜𝗜 𝗶𝗼𝗻) 
𝗪𝗶𝘁𝗵 𝗡𝗮𝗢𝗛 → 𝗴𝗿𝗲𝗲𝗻 𝗽𝗿𝗲𝗰𝗶𝗽𝗶𝘁𝗮𝘁𝗲 (𝗙𝗲(𝗢𝗛)₂)
THIS MATCHES!
Reaction (simple form):
Fe²⁺ + 2OH⁻ → Fe(OH)₂ (green precipitate)
C. Fe³⁺ (Iron III ion)
With NaOH → brown/reddish-brown precipitate (Fe(OH)₃)
Not green
D. NH₄⁺ (Ammonium ion)
With NaOH → no precipitate, instead gives off ammonia gas (NH₃) with heating
No precipitate at all
So, our final answer is:
B. Fe²⁺ 
Quick memory trick (this will save you in exam!):
𝗙𝗲²⁺ → 𝗚𝗿𝗲𝗲𝗻 🟢
𝗙𝗲³⁺ → 𝗕𝗿𝗼𝘄𝗻 🟤
𝗖𝘂²⁺ → 𝗕𝗹𝘂𝗲 
Just lock that in your brain and you're good 

