The dissolution of $Al(OH)_3$ by a solution of $NaOH$ results in the formation of
- A$[Al(H_2O)_4(OH)_2]^+$
- B$[Al(H_2O)_3(OH)_3]$
- C$[Al(H_2O)_2(OH)_4]^-$
- D$[Al(H_2O)_6(OH)_3]$
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View SolutionIn which of the following reactions is hydrogen one of the products?
$i$. $2NaBH_4 + I_2 \longrightarrow$
$ii$. $2BF_3 + 6NaH \xrightarrow{450 \ K}$
$iii$. $4BF_3 + 3LiAlH_4 \longrightarrow$
$iv$. $3B_2H_6 + 6NH_3 \xrightarrow{\text{heat}} 2B_3N_3H_6 + 12H_2$
$i$. $2NaBH_4 + I_2 \longrightarrow$
$ii$. $2BF_3 + 6NaH \xrightarrow{450 \ K}$
$iii$. $4BF_3 + 3LiAlH_4 \longrightarrow$
$iv$. $3B_2H_6 + 6NH_3 \xrightarrow{\text{heat}} 2B_3N_3H_6 + 12H_2$
In the reaction $B(OH)_{3} + 2 H_{2}O \rightarrow [B(OH)_{4}]^{-} + H_{3}O^{+}$,$B(OH)_{3}$ functions as:
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