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Which of the following statements is incorrect?
A
Aluminium reacts with excess NaOH to give $Al(OH)_3$
B
$NaHCO_3$ on heating gives $Na_2CO_3$
C
Pure sodium metal dissolves in liquid ammonia to give blue solution
D
NaOH reacts with glass to give sodium silicate
Detailed Solution
Aluminium is amphoteric. With excess aqueous NaOH it dissolves to form soluble sodium tetrahydroxoaluminate(III) and hydrogen: $2Al + 2NaOH + 6H_2O \rightarrow 2Na[Al(OH)_4] + 3H_2$
$Al(OH)_3$ is not the product, because it is itself soluble in excess NaOH. So this statement is incorrect.
$2NaHCO_3 \xrightarrow{\Delta} Na_2CO_3 + H_2O + CO_2$ (true).
Sodium dissolves in liquid ammonia to give a deep blue solution due to ammoniated electrons: $Na + (x + y)NH_3 \rightarrow [Na(NH_3)_x]^+ + [e(NH_3)_y]^-$ (true).
NaOH attacks the silica of glass: $2NaOH + SiO_2 \rightarrow Na_2SiO_3 + H_2O$ (true).
Hence the incorrect statement is that aluminium reacts with excess NaOH to give $Al(OH)_3$.
$Al(OH)_3$ is not the product, because it is itself soluble in excess NaOH. So this statement is incorrect.
$2NaHCO_3 \xrightarrow{\Delta} Na_2CO_3 + H_2O + CO_2$ (true).
Sodium dissolves in liquid ammonia to give a deep blue solution due to ammoniated electrons: $Na + (x + y)NH_3 \rightarrow [Na(NH_3)_x]^+ + [e(NH_3)_y]^-$ (true).
NaOH attacks the silica of glass: $2NaOH + SiO_2 \rightarrow Na_2SiO_3 + H_2O$ (true).
Hence the incorrect statement is that aluminium reacts with excess NaOH to give $Al(OH)_3$.