Statement-1 : $C u {(O H)}_{2}$ is soluble in excess $N H_{4} O H (a q .)$ but not soluble in excess $N a O H (a q .)$ but reverse is true for $A l {(O H)}_{3} .$
Statement-2 : $A l {(O H)}_{3}$ is amphoteric but $C u {(O H)}_{2}$ is basic

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Statement-1 is False, Statement-2 is True.

Statement -1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

Statement-1 is True, Statement-2 is False

Statement -1 is True, Statement-2 is True; Statement-2 is a NOT correct explanation for Statement-1.

answer is B.

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Detailed Solution

$A l {(O H)}_{3} =$ Amphoteric hydroxide
$C u {(O H)}_{2} =$ Basic hydroxide

$Al {(OH)}_{3} + NaOH \to {NaAlO}_{2} (soluble) + 2 H_{2} O Cu {(OH)}_{2} + 4 {NH}_{4} OH \to \underset{soluble complex}{(Cu {({NH}_{3})}_{4}) {(OH)}_{2}} + 4 H_{2} O$

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