Aluminium reacts with $\mathrm{NaOH}$ and forms compound ' $X$ '. If the coordination number of aluminium in…

Aluminium reacts with $\mathrm{NaOH}$ and forms compound ' $X$ '. If the coordination number of aluminium in ' $X$ ' is 6 , the correct formula of $X$ is
  1. $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}ight)_4(\mathrm{OH})_2ight]^{+}$
  2. $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}ight)_3(\mathrm{OH})_3ight]$
  3. $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}ight)_2(\mathrm{OH})_4ight]^{-}$
  4. $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}ight)_6ight](\mathrm{OH})_3$

Solution

$2 \mathrm{Al}+2 \mathrm{NaOH}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\substack{\text { sodium } \\ \text { meta } \\ \text { aluminate }}}{2 \mathrm{NaAlO}_2}+3 \mathrm{H}_2$ Sodium metaaluminate, thus formed, is soluble in water and changes into the complex $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}ight)_2(\mathrm{OH})_4ight]^{-}$, in which coordination number of $\mathrm{Al}$ is 6 . .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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