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(a) The actinoids exhibit more number of oxidation states and give their common oxidation states. (b)A wellknown orange crystalline compound (A) when burnt impart violet colour to flame. (A) on treating (B) and conc. $H_2SO_4$ gives red gas (C) which gives red yellow solution (D) with alkaline water. (D) on treating with acetic acid and lead acetate gives yellow p. pt. (E). (B) sublimes on heating. Also on heating (B) with NaOH gas (F) is formed which gives white fumes with HCl. What are (A) to (F) ?

1 Answer

(a) As the distance between the nucleus and 5 f orbitals (actinoides) is more than the distance between the nucleus and 4 f (lanthanoids) hence the hold of the nucleus on valence electrons decrease in actinoids. For this reason the actinoids exhibit more number of oxidation states in general. Common O. N. exhibited are + 3 (similar to Canthanoids) besides + 3 state, also show + 4, maximum oxidation state in middle of series i. e. Pu and Np. have anoidation state upto + 7.
(b) $(i) K_2Cr_2O_7 + 4 NH_4Cl + 3H_2SO_4 \Rightarrow K_2SO_4 + 2Cr_2O_2Cl_2 + 2(NH_4 )2SO_4 + 3H_2O$
$\;\;\;\;\;\;(A)\;\;\;\;\;\;\;\;\;\;\;\;\; (B) Sublime\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$Chromyl Chloride red gas (C)
$(ii) CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(D) Yellow Soln.$
$(iii) Na_2CrO_4 + (CH_3COO)_2Pb \rightarrow PbCrO_4 + 2CH_3COONa$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;Yellow p. pt. (E)$
answered Dec 29, 2016 by sharmaaparna1

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