Aluminium hydroxide is soluble in excess of $N a O H$ forming the ion :

A l {O_{2}}^{3 +}

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A l {O_{2}}^{- 3}

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A l {O_{2}}^{-}

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A l {O_{3}}^{-}

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Solution

The correct option is C $A l {O_{2}}^{-}$

Alluminium hydroxide reacts with excess of $N a O H$ and forms $A l O_{2}^{-}$ .
$A l (O H)_{3} + N a O H \to N a A l O_{2} + 2 H_{2} O$
So the $N a A l O_{2}$ forms $A l O_{2}^{-}$ .
Hence option C is correct.

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{A l O}_{2}^{-}

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