Question

Which of the following statements is incorrect?
A) Pure sodium metal dissolves in liquid ammonia to give a blue solution
B) $\text{ NaOH }$reacts with glass to give sodium silicate
C) Aluminium reacts with excess $\text{ NaOH }$ to give $\text{ Al(OH}{{\text{)}}_{\text{3}}}\text{ }$
D) $\text{ NaHC}{{\text{O}}_{\text{3}}}\text{ }$ On heating gives $\text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$

Answer 1

The sodium is an alkali metal. In aqueous solutions, it loses its electrons. The various compounds of sodium are known to have wide application. The sodium hydroxide $\text{ NaOH }$ is a strong base. It can readily react with the glass. However the reaction between the sodium hydroxide and some metals is exothermic and liberates the hydrogen ion. The $\text{ NaHC}{{\text{O}}_{\text{3}}}\text{ }$ is another very well-known compound of sodium. It has a wide application as a raising agent in bakeries.

Complete step by step answer:
A) The pure sodium metal reacts with the ammonia to form a sodium amide. Sodium metal is reduced to the $\text{ N}{{\text{a}}^{\text{+}}}\text{ }$ . The reduction of metal liberates an electron. In the ammoniacal solution, the sodium ion forms a fairly soluble sodium-ammonia complex. The complex has a structure$\text{ }{{\left[ \text{Na(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{6}}} \right]}^{\text{+}}}\text{ }$. Each liberated electron is solvated by the 6 ammonia molecules. Solvated electrons is a free electron. This is responsible for the good conduction of electricity. Since, this is free electrons they can easily excite and impart a colour to the solution.
Thus, the reaction of sodium metal with the ammonia results in the blue colour solution due to the presence of solvated electrons. The reaction between sodium metal and ammonia solvent is as shown below,
$\text{ Na + ( x + y ) N}{{\text{H}}_{\text{3}}}\text{ }\to \text{ Na(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{x}}}\text{ + }{{\text{e}}^{-}}{{\text{(N}{{\text{H}}_{\text{3}}}\text{)}}_{\text{y}}}\text{ }$
Or in simpler words, the reaction is as,
$\text{ }\begin{matrix} \text{Na} & \text{+} & \text{liq}\text{.NH3} & \rightleftharpoons & \text{N}{{\text{a}}^{\text{+}}}\text{(ammoniated)} & \text{+} & {{\text{e}}^{-}}\text{(ammoniated)} \\\ {} & {} & {} & {} & {} & {} & \text{(Blue)} \\\ \end{matrix}\text{ }$

B) The sodium hydroxide $\text{ NaOH }$ slowly reacts with the glass and forms a sodium silicate. The glass is made of silicon dioxide$\text{ Si}{{\text{O}}_{\text{2}}}\text{ }$. Silicon dioxide is a weak acid and thus it reacts with a strong base such as sodium hydroxide $\text{ NaOH }$to give the salt. When it reacts with sodium hydroxide it generates a sodium silicate. The reaction is as shown below,
$\text{ Si}{{\text{O}}_{\text{2}}}_{(s)}\text{ + 2NaO}{{\text{H}}_{\text{(aq)}}}\text{ }\to \text{ N}{{\text{a}}_{\text{2}}}\text{Si}{{\text{O}}_{\text{3}}}_{\text{(aq)}}\text{ + }{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{(l)}}}\text{ }$
Where, $\text{N}{{\text{a}}_{\text{2}}}\text{Si}{{\text{O}}_{\text{3}}}$ is sodium silicate.

C) The aluminium reacts with the sodium hydroxide and forms a sodium aluminate. The reaction between the aluminium and sodium hydroxide is exothermic and this causes the rapid evolution of hydrogen gas. The reaction can be written as,
$\text{ 2Al + 2NaOH + 2}{{\text{H}}_{\text{2}}}\text{O }\to \text{ 2NaAl}{{\text{O}}_{\text{2}}}\text{ + 3}{{\text{H}}_{\text{2}}}\text{ }$
Thus, unlike other aluminium do not aluminium hydroxide when treated with sodium hydroxide.

D) The thermal heating of sodium bicarbonate $\text{ NaHC}{{\text{O}}_{\text{3}}}\text{ }$ , release the carbon dioxide $\text{ C}{{\text{O}}_{\text{2}}}\text{ }$
And water. The thermal decomposition $\text{ NaHC}{{\text{O}}_{\text{3}}}\text{ }$is as shown below,
$\text{ 2NaHC}{{\text{O}}_{\text{3}}}\text{ }\xrightarrow{\Delta }\text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ + }{{\text{H}}_{\text{2}}}\text{O + C}{{\text{O}}_{\text{2}}}\text{ }$
It releases the carbon dioxide and thus acts as a raising agent in the baking.
So, the correct answer is “Option C”.

Note: i) The solution has a high number of the electron , thus conducts electricity, but when all electrons are utilized the colour of the solution changes to blue and it becomes diamagnetic from paramagnetic.
ii) The sodium hydroxide when exposed to the glass joints has a tendency to ‘freeze’. However, the glassware may be damaged by the long exposure to the hot sodium hydroxide.
Iii) Note that, like aluminum, tin and zinc also react rapidly with the sodium hydroxide.