How would you account for the following:
(i) Cr2+ is strongly reducing in nature. It has a d4 configuration. While acting as a reducing agent, it gets oxidized to Cr3+ (electronic configuration, d3 ). This d3 configuration can be written as \(t^3_2\)g configuration, which is a more stable configuration. In the case of Mn3+ (d4), it acts as an oxidizing agent and gets reduced to Mn2+ (d5). This has an exactly half-filled d-orbital and is highly stable.
(ii) Co(II) is stable in aqueous solutions. However, in the presence of strong field complexing reagents, it is oxidized to Co (III). Although the 3rd ionization energy for Co is high, but the higher amount of crystal field stabilization energy (CFSE) released in the presence of strong field ligands overcomes this ionization energy.
(iii) The ions in d1 configuration tend to lose one more electron to get into stable d0 configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present in the d-orbital of these ions. Therefore, they act as reducing agents.
(i) The standard electrode potential for Cr³⁺/Cr²⁺ is negative (-0.41 V), while Mn³⁺/Mn²⁺ has a positive value (+1.57 V). Cr²⁺ can undergo oxidation and serves as a reducing agent, whereas Mn³⁺ can undergo reduction and acts as an oxidizing agent.
(ii) Due to the greater stability of Co(III) compared to Co(II), the transition of cobalt from the +2 to +3 oxidation state is facilitated in the presence of a complexing agent.
(iii) Once ns electrons are lost, the removal of a d¹ electron leads to a stable electronic configuration. Elements exhibiting this behavior either act as reducing agents or undergo disproportionation.
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.
Write down the electronic configuration of:
(i) Cr3+ (iii) Cu+ (v) Co2+ (vii) Mn2+
(ii) Pm3+ (iv) Ce4+ (vi) Lu2+ (viii) Th4+
Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state?
What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms : 3d3 ,3d5 ,3d8 and 3d4?
The d-block elements are placed in groups 3-12 and F-block elements with 4f and 5f orbital filled progressively. The general electronic configuration of d block elements and f- block elements are (n-1) d 1-10 ns 1-2 and (n-2) f 1-14 (n-1) d1 ns2 respectively. They are commonly known as transition elements because they exhibit multiple oxidation states because of the d-d transition which is possible by the availability of vacant d orbitals in these elements.
They have variable Oxidation States as well as are good catalysts because they provide a large surface area for the absorption of reaction. They show variable oxidation states to form intermediate with reactants easily. They are mostly lanthanoids and show lanthanoid contraction. Since differentiating electrons enter in an anti-penultimate f subshell. Therefore, these elements are also called inner transition elements.
Read More: The d and f block elements