Question

What is the end product when $Sn$ is fused with $NaOH$ ?

Answer 1

Here, we are asked to find the end product when $Sn$ is fused with$NaOH$ . The symbol ‘ $Sn$ ’ denotes tin and ‘ $NaOH$ ’ is sodium hydroxide. When tin is fused with sodium hydroxide, it leads to the formation of colourless salt of an inorganic compound.

Complete answer:
When tin ‘ $Sn$ ’ reacts with excess sodium hydroxide ‘ $NaOH$ ’, it leads to the formation sodium stannite $(N{a_2}Sn{O_2})$ , sodium stannate $(N{a_2}Sn{O_3})$ and hydrogen gas$({H_2})$.
The following reaction are involved during the reaction:
$Sn\, + \,2NaOH \to N{a_2}Sn{O_2}\, + {H_2}$
$Sn\, + \,2NaOH\, + {H_2}O \to N{a_2}Sn{O_3}\, + \,2{H_2}$
The IUPAC name of sodium stannate is Sodium hexahydroxostannate (IV). It is a colourless salt and it forms by dissolving metallic tin or tin (IV) oxide in sodium hydroxide. It is an inorganic compound and has the chemical formula $N{a_2}[Sn{(OH)_6}]$ . Stannates is the simple oxyanion which is represented by$Sn{O_3}^{2 - }$ . In some cases, this compound is named as sodium stannate-3-water and represented with the chemical formula $N{a_2}Sn{O_3}.\,3{H_2}O$ . It is a hydrate with three waters of crystallization.
The anhydrous form of sodium stannate is $N{a_2}Sn{O_3}$ .
So, alkali metal stannate compounds are prepared by dissolving element tin in a suitable metal hydroxide as in the case of formation of sodium stannate.
$Sn\,\, + \,\,2NaOH\,\, + \,\,4{H_2}O \to N{a_2}[Sn{(OH)_6}]\,\, + \,\,2{H_2}$
A similar type of reaction also occurs when tin dioxide is dissolved in base.
$Sn{O_2}\,\, + \,\,\,2NaOH\,\, + \,\,2{H_2}O \to N{a_2}[Sn{(OH)_6}]\,$
The anhydrous form can also be prepared by the fusion of tin dioxide with sodium carbonate in a mixed carbon monoxide / carbon dioxide environment.
$Sn{O_2}\,\, + \,N{a_2}C{O_3}\,\,\, \to \,N{a_2}Sn{O_3}\,\, + C{O_2}$
The anion of sodium stannate is a coordination complex and it is octahedral in shape just like hexachlorostannate anion ${[SnC{l_6}]^{2 - }}$

Note:
Sodium hydroxide $(NaOH)$ is also known as lye and caustic soda. Remember that when $Sn$ is fused with $NaOH$ , sodium stannite $(N{a_2}Sn{O_2})$ is formed. But in the presence of water, an anhydrous form of sodium stannite i.e. sodium stannate $(N{a_2}Sn{O_3})$ is formed.