A) \[Al{{O}_{3}}^{-}\]
B) \[Al{{O}_{2}}^{3+}\]
C) \[Al{{O}_{2}}^{3-}\]
D) \[Al{{O}_{2}}^{-}\]
Correct Answer: D
Solution :
Aluminium hydroxide is soluble in excess of sodium hydroxide forming the ion \[AlO_{2}^{-}\]. \[Al{{(OH)}_{3}}+\underset{excess}{\mathop{NaOH}}\,\xrightarrow{{}}\underset{\begin{smallmatrix} sodium\text{ }meta \\ aluminate \end{smallmatrix}}{\mathop{NaAl{{O}_{2}}}}\,+2{{H}_{2}}O\] \[NaAl{{O}_{2}}N{{a}^{+}}+AlO_{2}^{-}\] Note - \[ZnC{{l}_{2}}+2NaOH\xrightarrow{{}}\underset{white\,\,ppt.}{\mathop{Zn{{(OH)}_{2}}}}\,\] \[\xrightarrow{NaOH}\underset{sodium\,\,zincate}{\mathop{N{{a}_{2}}Zn{{O}_{2}}}}\,\] \[N{{a}_{2}}Zn{{O}_{2}}2N{{a}^{+}}+ZnO_{2}^{2-}\] \[SnC{{l}_{2}}+2NaOH\xrightarrow{{}}\underset{white\,ppt.}{\mathop{Sn{{(OH)}_{2}}}}\,\] \[\xrightarrow{NaOH}\underset{sodium\,\,s\tan nite}{\mathop{N{{a}_{2}}Sn{{O}_{2}}}}\,\]
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