When metal XX is treated with sodium hydroxide- a white ppt- of AA is obtained which is soluble in excess of NaOHNaOH to form soluble complex BB-Compound AA is soluble in dilute HClHCl to form compound CC-The compound AA when strongly heated gives compound DD which is used to extract metals-Compound C is -

The correct option is A AlCl_3 Al+3NaOH⟶ Al(OH)_3+3Na^+ #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; ...

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Byju's Answer

Standard XII

Chemistry

Chemical Properties of Boron Family

When metal XX...

Question

When metal X is treated with sodium hydroxide, a white ppt. of A is obtained which is soluble in excess of NaOH to form soluble complex B.Compound A is soluble in dilute HCl to form compound C.The compound A when strongly heated gives compound D which is used to extract metals.

Compound $C$ is :

A l C l_{3}

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C u_{2} C l_{2}

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H g_{2} C l_{2}

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M g C l_{2}

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Solution

The correct option is A $A l C l_{3}$
$A l + 3 N a O H ⟶ A l (O H)_{3} + 3 N a^{+}$
(white ppt.)

$A l (O H)_{3} + N a O H ⟶ N a^{+} [A l (O H)_{4}]^{-}$ (Compound B)
(soluble complex)

$A l (O H)_{3} + 3 H C l ⟶ A l C l_{3} + 3 H_{2} O$
(dil) (Compound C)

$2 A l (O H)_{3} S t r o n g - ----- \to h e a t i n g △ A l_{2} O_{3} + 3 H_{2} O$
$↓$
(used to extract metals)
Compound $D = A l_{2} O_{3}$

Compound $A = A l (O H)_{3}$

Compound $B = N a^{+} [A l (O H)_{4}]^{-}$

Compound $C = A l C l_{3}$

Compound $D = A l_{2} O_{3}$

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