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)

$2 Al +2 NaOH +2 H _{2} O \longrightarrow \underset{\text{sodium meta aluminate}}{2 NaAlO _{2}}+3 H _{2}$

Sodium metaaluminate, 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 $6 .$