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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
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$\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_2(\mathrm{OH})_4\right]^{-}$
$2 \mathrm{Al}+2 \mathrm{NaOH}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\substack{\text { sodium } \\ \text { meta } \\ \text { aluminate }}}{2 \mathrm{NaAlO}_2}+3 \mathrm{H}_2$
Sodium metaaluminate, thus formed, is soluble in water and changes into the complex $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_2(\mathrm{OH})_4\right]^{-}$, in which coordination number of $\mathrm{Al}$ is 6 .
Sodium metaaluminate, thus formed, is soluble in water and changes into the complex $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_2(\mathrm{OH})_4\right]^{-}$, in which coordination number of $\mathrm{Al}$ is 6 .
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