On passing ammonia gas through a solution of copper sulphate, a deep blue solution is obtained. The deep blue colour of the solution is due to the formation of:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \([Cu(NH_3)_4]^{2+}\)
When ammonia is passed into copper sulphate solution, a pale blue precipitate of \( \ce{Cu(OH)2} \) forms first; in excess ammonia it dissolves, because \( \ce{NH3} \) displaces the water ligands of \( \ce{Cu^{2+}} \) and forms the tetraamminecopper(II) ion, which is deep blue:
\( \ce{[Cu(H2O)4]^2+ + 4NH3 <=> [Cu(NH3)4]^2+ + 4H2O} \)
\( \ce{NH3} \) is a stronger field ligand than \( \ce{H2O} \), so the colour changes from the light blue of the aqua ion to deep blue. Copper stays in the +2 state and binds four \( \ce{NH3} \) molecules (square planar).
Why the other options are incorrect
Option (1) \([Cu(NH_3)_2]^{2+}\) — Incorrect. \( \ce{Cu^{2+}} \) in excess ammonia takes four \( \ce{NH3} \) ligands; a diammine is not the species that gives the deep blue colour.
Option (3) \([Cu(NH_3)_6]^+\) — Incorrect. It puts copper in the +1 state; \( \ce{NH3} \) does not reduce \( \ce{Cu^{2+}} \), and Cu(I) complexes (3d10) are colourless.
Option (4) \([Cu(NH_3)_6]^{2+}\) — Incorrect. The deep blue ion in aqueous ammonia is the tetraammine; \( \ce{Cu^{2+}} \) does not hold six \( \ce{NH3} \) molecules in this solution.
NCERT (2026-27 edition):
Formation of Complex Compounds
