Question

Aluminium reacts with $NaOH$ and forms compound �$X$�. If the coordination number of aluminium in �$X$ � is $6$, the correct formula of $X$ is

• A. $\left[ Al \left( H _{2} O \right)_{4}( OH )_{2}\right]^{+}$
• B. $\left[ Al \left( H _{2} O \right)_{3}( OH )_{3}\right]$
• C. $\left[ Al \left( H _{2} O \right)_{2}( OH )_{4}\right]^{-}$
• D. $\left[ Al \left( H _{2} O \right)_{6}\right]( OH )_{3}$

Answer 1

Answer: (C)

$\left[ Al \left( H _{2} O \right)_{2}( OH )_{4}\right]^{-}$

Answer 2

$2 Al +2 NaOH +2 H _{2} O \longrightarrow \underset{\text { sodium meta aluminate }}{2 NaAlO} +3 H _{2}$
Sodiummetaaluminate, thus formed, is soluble in water and changes into the complex $\left[ Al \left( H _{2} O \right)_{2}( OH )_{4}\right]^{-}$, in which coordination
number of $Al$ is $6 .$