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Conductivity MP BP and Electronegativity (Group 2)

A metal X d...

Question

A metal $X$ dissolves both in dilute $H C l$ and dilute $N a O H$ to liberate $H_{2}$ . Addition of $N H_{4} C l$ and excess $N H_{4} O H$ to an $H C l$ solution of $X$ produce $Y$ as a precipitate. $Y$ is also produced by adding $N H_{4} C l$ to the $N a O H$ solution of $X$ . The species $X$ and $Y$ , respectively, are:

A

Z n

and

Z n (O H)_{2}

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B

A l

and

A l (O H)_{3}

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C

Z n

and

N a_{2} Z n O_{2}

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D

A l

and

N a A l O_{2}

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Solution

The correct option is B $A l$ and $A l (O H)_{3}$
$Z n + N a O H \to N a_{2} Z n O_{2} A l + N a O H \to N a C l O_{2}$
$Z n + H C l \to Z n C l_{2} + H_{2} A l + H C l \to A l C l_{3} + H_{2}$
$Z n C l_{2}$ gives no ppt with $N H_{4} C l + N H_{4} O H$
But $A l C l_{3}$ gives gelatinious white ppt of $A l (O H)_{3}$ with $N H_{4} C l + N H_{4} O H$ .

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Similar questions

A metal (X) dissolves both in dilute HCl and dilute NaOH to liberate $H_{2}$ . Addition of $N H_{4} C l$ and excess $N H_{4} O H$ to an HCl solution of X produce Y as a precipitate. Y is also produced by adding $N H_{4} C l$ to the NaOH solution of X. The species X and Y, respectively, are

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If true enter 1, else enter 0.

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