Q. 23 Addition of a few drops of concentrated HCl to an aqueous solution of CoCl2
forms a dark blue complex X.
The CORRECT statement(s) for this reaction is(are)
(Given: Atomic number of Co: 27)
(A) X is a centrosymmetric complex.
(B) The oxidation state of cobalt does not change in this reaction.
(C) The number of unpaired electrons on cobalt in X and in CoCl2 (aqueous solution)
are the same.
(D) The spin only magnetic moment value for X is 3.87 BM.
CoCl₂ with HCl forms the dark blue tetrahedral complex [CoCl₄]²⁻, where cobalt remains in the +2 oxidation state with 3 unpaired electrons, matching the spin-only magnetic moment of 3.87 BM.
Reaction Overview
Aqueous CoCl₂ exists primarily as the pink [Co(H₂O)₆]²⁺ octahedral complex. Adding concentrated HCl provides excess Cl⁻ ions, which displace H₂O ligands to form the dark blue tetrahedral [CoCl₄]²⁻ (complex X) via ligand exchange: [Co(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CoCl₄]²⁻ + 6H₂O. The color change arises from tetrahedral geometry causing smaller crystal field splitting and different d-d transitions compared to octahedral.
Option Analysis
-
(A) Incorrect: [CoCl₄]²⁻ adopts tetrahedral geometry (sp³ hybridization), lacking a center of symmetry, unlike square planar or octahedral complexes.
-
(B) Correct: Cobalt stays at +2 oxidation state in both [Co(H₂O)₆]²⁺ (x + 0 = +2) and [CoCl₄]²⁻ (x + 4(-1) = -2 → x = +2); no redox occurs.
-
(C) Correct: Co²⁺ (3d⁷) shows high-spin configuration in both cases due to weak-field ligands (H₂O, Cl⁻). Each has 3 unpaired electrons: octahedral t₂g⁵e_g² and tetrahedral e⁴t₂³.
-
(D) Correct: Spin-only magnetic moment μ = √[n(n+2)] BM with n=3 unpaired electrons gives √[3(5)] = √15 ≈ 3.87 BM.
Correct options: (B), (C), (D)
Introduction to CoCl2 HCl Dark Blue Complex
The classic reaction where aqueous CoCl₂ (pink) forms a dark blue complex upon adding concentrated HCl fascinates coordination chemistry students. This ligand substitution creates [CoCl₄]²⁻, a tetrahedral species with Co in +2 oxidation state, 3 unpaired electrons, and spin-only magnetic moment of 3.87 BM – key for CSIR NET Life Sciences and JEE preparation.
Reaction Mechanism and Complex Formation
In water, CoCl₂·6H₂O dissociates to [Co(H₂O)₆]²⁺ (octahedral, pink due to d-d transitions). Concentrated HCl floods Cl⁻ (weak field ligand), driving equilibrium: [Co(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CoCl₄]²⁻ + 6H₂O. The tetrahedral [CoCl₄]²⁻ appears deep blue from smaller Δ_t (t₂ < e splitting).
Electronic Configuration and Geometry
Co (Z=27: [Ar] 3d⁷4s²) as Co²⁺ is d⁷.
-
[Co(H₂O)₆]²⁺ (octahedral, high-spin): t₂g⁵e_g² (3 unpaired e⁻).
-
[CoCl₄]²⁻ (tetrahedral, sp³, high-spin): e⁴t₂³ (3 unpaired e⁻).
Both maintain 3 unpaired electrons; oxidation state unchanged (+2).
| Property | [Co(H₂O)₆]²⁺ | [CoCl₄]²⁻ |
|---|---|---|
| Geometry | Octahedral | Tetrahedral |
| Unpaired e⁻ | 3 | 3 |
| Magnetic Moment | ~3.87-4.4 BM | 3.87 BM |
| Color | Pink | Dark Blue |
Magnetic Properties: Spin-Only Moment Calculation
μ = √[n(n+2)] BM (n=3): √15 ≈ 3.87 BM confirms paramagnetism in X. Matches experimental values; orbital contributions minimal in tetrahedral field.
CSIR NET Exam Relevance
This question tests crystal field theory, VBT, and magnetism. Remember: tetrahedral [CoCl₄]²⁻ lacks centrosymmetry, retains Co(II), same unpaired electrons as aqua complex.
