Some Applications Of D And F Block Elements

(a) Cu is unstable in aqueous solution because it undergoes disproportionation: This is thermodynamically favored as Cu has a tendency to be both oxidised to Cu and reduced to Cu.
This instability arises due to the positive value of the standard electrode potential for the disproportionation reaction.
(b) Out of Cr and Fe , which is a stronger reducing agent and why?
Solution:
(b) Cr is a stronger reducing agent than Fe because:
- Cr readily loses an electron to form the more stable Cr (having half-filled t configuration).
- Fe does not gain as much stability on oxidation.
(c) Actinoid contraction is greater from element to element than lanthanoid contraction. Why?
Solution:
(c)Actinoid contraction is more pronounced because:
- Poor shielding of 5f electrons compared to 4f electrons in lanthanoids.
- Greater increase in effective nuclear charge across the series.
(d) KMnO acts as an oxidising agent in acidic medium. Write the ionic equation to support this.
Solution:\ (d) In acidic medium, KMnO gets reduced from Mn to Mn . The ionic redox equation is: (e) Name the metal in the first transition series which exhibits oxidation state most frequently.
Solution:
(e) The metal is **Copper (Cu)**. Cu shows +1 oxidation state frequently due to configuration: A completely filled d-orbital provides extra stability. (f) Transition metals and their compounds are good catalysts. Justify.
(f) Solution:
Transition metals act as good catalysts due to:
- Variable oxidation states → allow redox cycling
- Ability to form complexes with reactants → intermediate formation
- Provide surface area for adsorption
(g) Scandium forms no coloured ions, yet it is regarded as a transition element. Why?
Solution:
(g) Scandium forms Sc with 3d configuration → no unpaired electrons → no d-d transitions → hence, no colour.
However, it is a transition element because:
- It has an incompletely filled d-subshell in its ground state (Sc: 3d 4s )