Explanation

[Solution from source] CN⁻ reduces Cu²⁺ to Cu⁺ (2Cu²⁺ + 4CN⁻ → 2CuCN + (CN)₂), and CuCN dissolves in excess cyanide to give the stable tetracyanidocuprate(I) ion, [Cu(CN)₄]³⁻. The Cu(II) complex [Cu(CN)₄]²⁻ is not the product, and CuCN is only the intermediate before excess KCN acts. Answer (2).