Chemistry · Coordination Compounds · NEET
In an octahedral complex the five d-orbitals split into a lower set (t2g) and a higher set (eg). When light hits the complex, a d-electron in the lower set absorbs energy and jumps to the higher set. This jump between two d-orbitals is called a d-d transition. The energy needed for the jump equals the crystal field splitting, called delta (Δo).
White light is made of all colours. The complex absorbs one colour (the one whose energy matches Δo). The colours that are NOT absorbed pass through or reflect, and your eye mixes them into one colour. This seen colour is the complementary colour of the absorbed one. Example: if it absorbs green, it looks red.
Use E = hc/λ. Here E is the energy absorbed, which equals Δo. So Δo and wavelength (λ) are inversely related. A LARGER Δo means MORE energy is needed, so a SHORTER wavelength is absorbed. A smaller Δo means less energy, so a longer wavelength is absorbed. This one line is the key to almost every NEET colour question.
Strong-field ligands (from the spectrochemical series: I- < Br- < Cl- < F- < H2O < NH3 < en < CN-) give a bigger Δo. A bigger Δo means a shorter wavelength is absorbed. So if you swap a weak ligand for a strong one, the absorbed wavelength gets shorter. More strong ligands (like more 'en') also increase Δo.
Colour needs a d-d transition, so you need d-electrons that can jump AND empty higher d-orbitals to jump into. Sc3+ and Ti4+ are d0 (no d-electrons to jump). Zn2+ and Cu+ are d10 (all d-orbitals full, no empty orbital to jump into). With no possible d-d transition, no visible light is absorbed, so these ions are colourless/white.
No. NEET tricks students here. The wavelength/colour ABSORBED is what matches Δo. The colour we SEE is the complementary (opposite) colour. Read the question carefully: 'wavelength of light absorbed' relates directly to Δo, while 'colour observed' is the complement of that.
Correct increasing order for the wavelengths of absorption in the visible region for the complexes of Co3+ is:
The order of energy absorbed which is responsible for the colour of complexes: (A) [Ni(H2O)2(en)2]2+ (B) [Ni(H2O)4(en)]2+ (C) [Ni(en)3]2+
The correct order of the wavelength of light absorbed by the following complexes is: A. [Co(NH3)6]3+ B. [Co(CN)6]3- C. [Cu(H2O)4]2+ D. [Ti(H2O)6]3+
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Because a d-electron absorbs visible light and jumps between split d-orbitals (a d-d transition), and we see the leftover complementary colour.
E = hc/λ, where the absorbed energy E equals the crystal field splitting Δo. Bigger Δo means shorter absorbed wavelength.
The spectrochemical series (increasing field strength): I- < Br- < Cl- < F- < OH- < H2O < NH3 < en < CN- < CO. Stronger field = bigger Δo = shorter wavelength absorbed.
No. Only ions with a partly filled d-subshell (d1 to d9) can be coloured. d0 (like Sc3+, Ti4+) and d10 (like Zn2+, Cu+) ions are colourless because no d-d transition is possible.
No. The seen colour is the complementary (opposite) colour of the absorbed one. Questions about 'wavelength absorbed' link straight to Δo; questions about 'colour seen' need the complement.
