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When $B_2H_6$ reacts with $H_2O$ or alkalies and produces $x+y$ is liberated.The $x$ is

$\begin{array}{1 1}(a)\;Boric\;acid\\(b)\;Metaboric\;acid\\(c)\;Hydro\;chloric\;acid\\(d)\;Tetraboric\;acid\end{array}$

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The $x$ is Boric acid.
Hence (a) is the correct answer.
answered Nov 28, 2013 by sreemathi.v
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