Ncert Exercises & Solutions

When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, which is soluble in excess of NaOH to give soluble complex (B). Compound (A) is soluble in dilute HCl to form compound (C). The compound (A) when heated strongly gives (D), which is used to extract metal. Identify (X), (A), (B), (C) and (D). Write suitable equations to support their identities.

The given metal X gives a white precipitate with sodium hydroxide and the precipitate dissolves in excess of sodium hydroxide. Hence, X must be aluminium.

The white precipitate (compound A) obtained is aluminium hydroxide. The compound B formed when an excess of the base is added is sodium tetrahydroxoaluminate(III).

$2 A I + 3 N a O H \to A I {(O H)}_{3} ↓ + 3 N a^{+}$
$A l u m i n i u m (X) S o d i u m h y d r o x i d e W h i t e p p t . (A)$

$A I {(O H)}_{3} + N a O H \to N a^{+} [A I {(O H)}_{4}] (A) S o d i u m t e r a h y d r o x o a l u m i n a t e (I I I)$
$(S o l u b l e c o m p l e x B)$

Now, when dilute hydrochloric acid is added to aluminium hydroxide, aluminium chloride (compound C) is obtained.

$A I {(O H)}_{3} + 3 H C I \to A I C I_{3} + 3 H_{2} O$
$(A) (C)$

Also, when compound A is heated strongly, it gives compound D. This compound is used to extract metal X. Aluminium metal is extracted from alumina. Hence, compound D must be alumina.

$2 A I {(O H)}_{3} \overset{∆}{\to} A I_{2} O_{3} + 3 H_{2} O$
$(A) (D)$

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