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Zn^(+2) dissolves in excess of NaOH due ...

`Zn^(+2)` dissolves in excess of NaOH due to the formation of

A

Soluble `Zn(OH)_(2)`

B

Soluble `Na_(2)[Zn (OH)_(4)]`

C

Soluble `Na [Zn (OH)_(3)]`

D

`ZnO`

Text Solution

Verified by Experts

The correct Answer is:
2
`Zn^(+2) + 2NaOH to Na_(2)ZnO_(2) + H_(2)`
`Na_(2)ZnO_(2) + 2H_(2)O to Na_(2)[Zn(OH)_(4)]`
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