To solve the problem, we need to explain why Co\(^{2+}\) is easily oxidized to Co\(^{3+}\) in the presence of a strong ligand.
1. Understanding the Oxidation States of Cobalt:
Cobalt (Co) has two common oxidation states: Co\(^{2+}\) and Co\(^{3+}\). Co\(^{2+}\) has a [Ar] 3d⁷ electron configuration, while Co\(^{3+}\) has a [Ar] 3d⁶ electron configuration. Co\(^{3+}\) is more highly charged, making it more electronegative, and it has a greater tendency to attract electrons from its environment.
2. Role of Strong Ligands:
In coordination chemistry, ligands are molecules or ions that surround a metal ion, forming a coordination complex. Strong ligands are those that form stable complexes with metal ions, leading to significant changes in the metal's electronic structure. A strong ligand is a good electron donor, and it stabilizes the metal ion by interacting strongly with it, often leading to oxidation or reduction of the metal ion.
3. Oxidation of Co\(^{2+}\) to Co\(^{3+}\):
Co\(^{2+}\) is more easily oxidized to Co\(^{3+}\) in the presence of a strong ligand because the ligand stabilizes the Co\(^{3+}\) ion. The presence of a strong ligand such as ammonia (NH₃), cyanide (CN⁻), or ethylenediamine (en) creates a stronger field around the metal, which can lower the energy of the Co\(^{3+}\) ion and make its formation more favorable. In the absence of a strong ligand, Co\(^{2+}\) is relatively stable, but the strong ligand induces an oxidation state change by stabilizing the higher oxidation state (Co\(^{3+}\)).
4. Effect of Ligand Field on Oxidation Potential:
The oxidation potential for Co\(^{2+}\) to Co\(^{3+}\) is relatively low in the absence of a strong ligand. However, the presence of a strong ligand increases the stability of Co\(^{3+}\), thus making the oxidation of Co\(^{2+}\) more favorable. This is because strong field ligands cause a large splitting of the metal's d-orbitals, which stabilizes the higher oxidation state and makes the oxidation process energetically more favorable.
5. Final Answer:
Co\(^{2+}\) is easily oxidized to Co\(^{3+}\) in the presence of a strong ligand because the strong ligand stabilizes the Co\(^{3+}\) ion, making the oxidation of Co\(^{2+}\) to Co\(^{3+}\) more thermodynamically favorable.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3 )2CHNH2 (ii) CH3 (CH2 )2NH2 (iii) CH3NHCH(CH3 )2
(iv) (CH3 )3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2 )2NCH3 (vii) m–BrC6H4NH2
Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylaniline
Account for the following:
(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
\(FeSO_4\) solution mixed with\( (NH_4)_2SO_4 \) solution in\( 1:1\) molar ratio gives the test of \(Fe^{2+}\) ion but \(CuSO_4\) solution mixed with aqueous ammonia in \( 1:4 \) molar ratio does not give the test of \(Cu^{2+}\) ion. Explain why?
Specify the oxidation numbers of the metals in the following coordination entities:
(i)\( [Co(H_2O)(CN)(en)_2] ^{2+}\)
(ii) \([CoBr_2(en)_2]^{+}\)
(iii)\( [PtCl_4]^{ 2–} \)
(iv) \(K_3[Fe(CN)_6]\)
(v) \([Cr(NH_3)_3Cl_3] \)