A) III and IV
B) I and III
C) I only
D) II and III
E) III only
Correct Answer: E
Solution :
(i)\[Be\]and\[Al\]both react with\[HCl\]to liberate \[{{H}_{2}}\]. \[Be+2HCl\xrightarrow[{}]{{}}BeC{{l}_{2}}+{{H}_{2}}\uparrow \] \[2Al+6HCl\xrightarrow[{}]{{}}2AlC{{l}_{3}}+3{{H}_{2}}\uparrow \] (ii) Be renders passive with cone.\[HN{{O}_{3}}\]and so does not react. Cone.\[HN{{O}_{3}}\]is strong oxidizing agent and forms very thin layer of oxide on the surface of metal. Similarly,\[Al\] become passive in cone.\[HN{{O}_{3}}\]due to formation of oxide layer on the surface. (iii) Their carbides gives methane on treatment with water. \[B{{e}_{2}}C+4{{H}_{2}}O\xrightarrow[{}]{{}}2Be{{(OH)}_{2}}+C{{H}_{4}}\] \[A{{l}_{4}}{{C}_{3}}+12{{H}_{2}}O\xrightarrow[{}]{{}}4Al{{(OH)}_{3}}+3C{{H}_{4}}\] (iv) Their oxides,\[BeO\]and\[A{{l}_{2}}{{O}_{3}}\]are amphoteric.
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