1.

The dissolution of `Al(OH)_(3)` by a solution of `NaOH` results in the formation ofA. `[Al(H_(2)O)_(2)(OH)_(4)]^(-)`B. `[Al(H_(2)O)_(4)(OH)]^(2+)`C. `[Al(H_(2)O)_(3)(OH)_(3)]`D. `[Al(H_(2)O)_(6)(OH)_(3)]`

Answer» Correct Answer - A
`Al(OH)_(3)+OH^(-)to[Al(OH)_(4)^(-)]`.
Coordinate number is `6`, thus, it exist as
`[Al(H_(2)O)_(2)(OH)_(4)]^(-)`


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