To solve the problem, we need to determine the type of isomerism shown by the complex \([Co(NH₃)_6][Cr(CN)_6]\).
1. Understanding the Complex:
The given complex is a coordination compound with two parts: \([Co(NH₃)_6]^{3+}\) and \([Cr(CN)_6]^{3-}\). The first part, \([Co(NH₃)_6]^{3+}\), is a complex of cobalt with six ammonia (NH₃) molecules as ligands, while the second part, \([Cr(CN)_6]^{3-}\), is a complex of chromium with six cyanide (CN⁻) ions as ligands.
2. Types of Isomerism:
Coordination compounds can exhibit several types of isomerism, including:
- Ionization isomerism: Occurs when the complex ions in the compound interchange their positions in the formula, leading to different ions in solution.
- Linkage isomerism: Occurs when a ligand can coordinate to the metal through two different atoms, giving rise to different isomers.
- Geometrical isomerism: Occurs when ligands are arranged in different spatial orientations around the metal center.
- Optical isomerism: Occurs when the complex can exist in non-superimposable mirror image forms (chirality).
3. Identifying the Type of Isomerism in \([Co(NH₃)_6][Cr(CN)_6]\):
This complex can exhibit ionization isomerism. Ionization isomerism occurs when two different compounds with the same molecular formula produce different ions in solution due to the exchange of ligands between the metal centers. In this case, if the cyanide ion (CN⁻) were to replace one of the ammonia (NH₃) ligands in the \([Co(NH₃)_6]^{3+}\) complex, and vice versa, two different compounds would result. One complex would have the \([Co(NH₃)_5(CN)]^{3+}\) ion and the \([Cr(CN)_6]^{3-}\) ion, while the other complex would have the \([Co(CN)_6]^{3-}\) ion and the \([Cr(NH₃)_6]^{3+}\) ion. These would give rise to different ions in solution, showing ionization isomerism.
4. Final Answer:
The complex \([Co(NH₃)_6][Cr(CN)_6]\) shows ionization isomerism.
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) \([Cr(NH_3)_3Cl_3] \)