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The dissolution of stannous hydroxide in excess base produces

$\begin{array}{1 1}Sn(OH)_4\\Sn(OH)_6^{2-}\\SnO_3^{2-}\\Sn(OH)_3^-\end{array} $

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Answer : $Sn(OH)_3^-$
$Sn(OH)_2+OH^-\rightarrow Sn(OH)_3^-$
answered Jul 24, 2014 by sreemathi.v
 
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