To solve the problem, we need to explain why low spin tetrahedral complexes are not known.
1. Understanding Tetrahedral Complexes:
A tetrahedral complex has a central metal ion surrounded by four ligands arranged at the corners of a tetrahedron. The crystal field splitting in a tetrahedral complex results in the splitting of the metal ion's d-orbitals into two sets: one set with lower energy (called the eₓₓ orbital) and the other with higher energy (called the t₂ₓₓ orbitals).
2. High and Low Spin Complexes:
In coordination chemistry, a complex is considered "high spin" if the electrons tend to occupy higher energy orbitals to maximize unpaired electrons, and "low spin" if the electrons tend to pair up in the lower energy orbitals to minimize the number of unpaired electrons. The tendency to form high or low spin complexes depends largely on the ligand field strength.
3. Crystal Field Splitting in Tetrahedral Geometry:
In tetrahedral geometry, the crystal field splitting is relatively small compared to octahedral geometry. This means that the energy difference between the eₓₓ and t₂ₓₓ orbitals is not large enough to favor pairing of electrons in the lower energy orbitals. In tetrahedral complexes, the ligands create a weaker field, and as a result, the electrons are more likely to occupy higher energy orbitals (t₂ₓₓ), leading to high spin complexes.
4. Why Low Spin Tetrahedral Complexes Are Not Known:
For low spin complexes to form, there needs to be a significant crystal field splitting to overcome the pairing energy of electrons, which typically occurs in strong ligand fields like those seen in octahedral complexes (e.g., with ligands such as CN⁻). In tetrahedral complexes, the splitting is too small to favor low spin configurations, making low spin tetrahedral complexes extremely rare or not observed.
5. Final Answer:
Low spin tetrahedral complexes are not known because the crystal field splitting in tetrahedral geometry is too small to favor electron pairing, and thus high spin complexes are more commonly observed.
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) \([Cr(NH_3)_3Cl_3] \)