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When NH₄Cl treated with caustic soda a basic substance (X) is produced. Excess of (X) on reaction with metal ion (Y) in aqueous solution forms a soluble complex. Metal ion (Y) could not be

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Ni⁺²

Zn²⁺

Fe³⁺

Cu²⁺

answer is C.

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Detailed Solution

${[Ni {({NH}_{3})}_{6}]}^{2 +}, {[Zn {({NH}_{3})}_{4}]}^{2 +}, {[Cu {({NH}_{3})}_{4}]}^{2 +}$ are soluble complexes.
But ${Fe}^{3 +}$ ion gives brown precipitate Fe₂O₃.xH₂O

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