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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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$\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}(\mathrm{OH})_{4}\right]^{-}$
$2 \mathrm{Al}+2 \mathrm{NaOH}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow \underset{\begin{array}{c}\text { sodum meta } \\ \text { aluminata }\end{array}}{\mathrm{NaAlO}_{2}}+3 \mathrm{H}_{2}$ Sodiummetaaluminate, 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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