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Chemistry Question on p -Block Elements

Orthoboric acid -

A

donate proton to form \ceH2BO3−\ce{H_2BO_3^-}

B

accept proton of form \ceH4BO3+\ce{H_4BO_3^+}

C

donate \ceOH−\ce{OH^-} to form \ceH2BO2+\ce{H_2BO_2^+}

D

accept \ceOH−\ce{OH^-} to form \ce[B(OH)4]−\ce{[B(OH)_4]^-}

Answer

accept \ceOH−\ce{OH^-} to form \ce[B(OH)4]−\ce{[B(OH)_4]^-}

Explanation

Solution

H2BO3 {H_2BO_3} acts as a Lewis acid and accepts an OH− {OH^-} ions to form [B(OH)4]− {[B(OH)_4]^-}