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

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BeOH3− and AlOH4−

BeOH42− and AlOH63−

BeOH42−andAlOH4−

BeOH64− and AlOH63−

answer is C.

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Detailed Solution

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

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