Which of the following complex ions is most likely to be colourless?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → [Ag(NH3)2]+
In crystal field theory a complex is coloured when a d electron can jump from a lower to a higher d level (a d–d transition) by absorbing visible light. This needs a partly filled d subshell; with d10 there is no vacant d level to go to. For each ion:
[Co(H2O)6]2+: water is neutral, so Co is +2; Co (Z = 27) is [Ar]3d74s2, so \( \ce{Co^2+} \) is 3d7 → d–d transition possible, coloured.
[Mn(CN)6]3−: x + 6(−1) = −3, so Mn is +3; Mn (Z = 25) is [Ar]3d54s2, so \( \ce{Mn^3+} \) is 3d4 → coloured.
[CrCl3(H2O)3]: x + 3(−1) = 0, so Cr is +3; Cr (Z = 24) is [Ar]3d54s1, so \( \ce{Cr^3+} \) is 3d3 → coloured.
[Ag(NH3)2]+: ammonia is neutral, so Ag is +1; Ag (Z = 47) is [Kr]4d105s1, so \( \ce{Ag^+} \) is 4d10 → no d–d transition, colourless.
NCERT (2026-27 edition):
5.5.5 Colour in Coordination Compounds – “… The crystal field theory attributes the colour of the coordination compounds to d-d transition of the electron.”