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Beryllium hydroxide dissolves in excess of alkali to give – ion

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${[Be {(OH)}_{2}]}^{-}$

${[Be {(OH)}_{2}]}^{- 2}$

${[Be {(OH)}_{4}]}^{- 2}$

${[Be {(OH)}_{4}]}^{3 -}$

answer is C.

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Detailed Solution

${[Be {(OH)}_{4}]}^{- 2}$

$2NaOH (aq) + Be {(OH)}_{2} (s) \to {Na}_{2} Be {(OH)}_{4} (aq)$

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