In the following unbalanced reaction,the product formed is $Al_2O_3 + NaOH_{(aq)} + H_2O \longrightarrow$

  • A
    $Na_3[Al(OH)_6]$
  • B
    $Na_3[Al(OH)_4]$
  • C
    $Na_2[Al(OH)_5]$
  • D
    $Na[Al(OH)_6]$

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$(B)$ all boron atoms in the same plane
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