The correct statements among I to III are: (I) Valence bond theory cannot explain the color exhibited by transition…
Coordination Compounds · Class 12 · JEE Main Previous Year Question
The correct statements among I to III are: (I) Valence bond theory cannot explain the color exhibited by transition metal complexes. (II) Valence bond theory can predict quantitatively the magnetic properties of transition metal complexes. (III) Valence bond theory cannot distinguish ligands as weak and strong field ones.
- a✓
(I) and (II) only
- b
(II) and (III) only
- c
(I) and (III) only
- d
(I), (II) and (III)
(I) and (II) only
🧠 The "What VBT Can't Do" Question
Three statements about VBT — true/false each:
- (I) VBT cannot explain colour. True. Colour requires a theoretical prediction of d–d transition energies, which CFT/MO theory provides (via ). VBT only assigns hybridisation; it has no orbital-energy gap, so it cannot predict colour.
- (II) VBT can quantitatively predict magnetic properties. True. The unpaired-electron count from VBT, plugged into , gives a quantitative spin-only moment that matches experiment for first-row metals.
- (III) VBT cannot distinguish weak/strong field ligands. False. VBT requires this distinction as an input — you need to know whether the ligand is strong or weak to decide between inner-orbital (d²sp³) and outer-orbital (sp³d²) hybridisation. So VBT depends on this classification rather than failing at it; the statement as worded is misleading.
Correct: (I) and (II) only → option (1).
🗺️ Why VBT Fails on Colour
VBT pictures bonds as hybrid-orbital overlaps. There's no concept of energy splitting between non-bonding d-orbitals — and that splitting is exactly what makes complexes coloured. For colour you need CFT (which gives ) or MO theory.
⚡ The "VBT vs CFT" Cheat
| Property | VBT predicts? | CFT predicts? | |---|---|---| | Geometry | ✓ | ✗ (must assume) | | Magnetism | ✓ (qualitative + quantitative) | ✓ | | Colour | ✗ | ✓ | | Spectrochemical series | ✗ | ✓ | | Stability orders | ~ | ✓ (via CFSE) |
⚠️ The "VBT Predicts Strong/Weak" Trap
Some students misread (III) as "VBT can distinguish" — and then label it true. Read carefully: the original statement says "cannot distinguish". VBT does need this input but doesn't generate it from theory, so the statement is partially defensible — but the JEE keyed answer treats VBT's reliance on the classification as using it, hence (III) is marked false. Always trust the keyed convention on borderline statements.
Practice this question with progress tracking
Want timed practice with adaptive difficulty? Solve this question (and hundreds more from Coordination Compounds) inside The Crucible, our adaptive practice platform.
More JEE Main Coordination Compounds PYQs
Match List-I (Complex) with List-II (Hybridisation & Magnetic characters): (A) [MnBr4]2-, (B) [FeF6]3-, (C) [Co(C2O4)3]3-, (D) [Ni(CO)4] (I) d2sp3 & diamagnetic, (II) sp3d2 & paramagnetic, (III) sp3…
The number of species from the following that are involved in sp3d2 hybridization is: [Co(NH3)6]3+, SF6, [CrF6]3-, [CoF6]3-, [Mn(CN)6]3- and [MnCl6]3-
Match the LIST-I with LIST-II | LIST-I (Complex/Species) | LIST-II (Shape & magnetic moment) | |---|---| | A. [Ni(CO)4] | I. Tetrahedral, 2.8 BM | | B. [Ni(CN)4]2- | II. Square planar, 0 BM | | C.…
The correct statement about [Ni(CO)4] and [Ni(CN)4]2- is:
The hybridisation and magnetic nature of [Ni(CO)4] and [Ni(CN)4]2- respectively are: