Be(OH)2 and AI(OH)3 dissolve in NaOH to produce

BeOH2+2OH−→ BeOH42−Beryllate ionAlOH3+OH−→ AlOH4−Aluminate ion

Be(OH)2 and AI(OH)3 dissolve in NaOH to produce

BeOH2+2OH−→ BeOH42−Beryllate ionAlOH3+OH−→ AlOH4−Aluminate ion

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Be(OH)₂ and AI(OH)₃ dissolve in NaOH to produce

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

${[Be {(OH)}_{4}]}^{2 -} and {[Al {(OH)}_{6}]}^{3 -}$

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

${[Be {(OH)}_{6}]}^{4 -} and {[Al {(OH)}_{6}]}^{3 -}$

answer is C.

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$Be {(OH)}_{2} + 2 O H^{-} \overset{}{\to} \underset{Beryllate ion}{{[Be {(OH)}_{4}]}^{2 -}}$

$Al {(OH)}_{3} + O H^{-} \overset{}{\to} \underset{Aluminate ion}{{[Al {(OH)}_{4}]}^{-}}$

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Be(OH)2 and AI(OH)3 dissolve in NaOH to produce

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Be(OH)₂ and AI(OH)₃ dissolve in NaOH to produce