The reagents $N H_{4} C l$ and aqueous $N H_{3}$ will precipitate :

C a^{2 +}

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A l^{3 +}

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M g^{2 +}

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Z n^{2 +}

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Solution

The correct option is B $A l^{3 +}$
The reagents $N H_{4} C l$ and aqueous $N H_{3}$ will precipitate $A l^{3 +}$ .

$A l^{3 +}$ belongs to the third group and its group reagents are $N H_{4} O H$ in presence of $N H_{4} C l$ .

So, the corret option is $B$

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0.1

A g^{+}, C a^{2 +}, M g^{2 +}

A l^{3 +}

[P O_{4}^{3 -}]

N a_{3} P O_{4}

B e^{2 +}

M g^{2 +},

S r^{2 +}

C a^{2 +}

A l^{3 +}

A l {(O H)}_{3}

{N a}^{+}, {C a}^{2 +}, {M g}^{2 +}, {A l}^{3 +}

{C a}^{2 +} < {N a}^{+} < {M g}^{2 +} < {A l}^{3 +}

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