Question

The divalent metal whose oxide is reduced to metal by electrolysis of its fused salt is ……………………………. $([Al/Na/Mg/K])$.

Answer 1

Hint: The extraction of metal from its ore is done by using a variety of methods. Ore is basically a rock, containing one or more minerals of metals, which can be mined, extracted and processed to obtain metals in the pure form. While extracting metals from their ores either they are reduced using carbon or subjected to electrolytic reduction by passing electric current through the metal oxides, chlorides and hydroxides in its molten form.

Complete Step by step answer: The process by which electric current is passed through ionic compounds in their molten forms to dissociate them in their respective elements. During electrolysis, a positive electrode or anode and a negative electrode or a cathode are used.
If we compare aluminium, sodium, magnesium and potassium, the number of valence electrons in the atoms of these metals are 3, 1, 2 and 1 respectively. Metal atoms lose valence electrons to form cations which are positively charged ions.
When aluminium, odium, magnesium and potassium lose their valence electrons, they form cations. Aluminium on losing its valence electrons, will achieve its octet.
$[Al \to A {l^{ + 3}} + 3{e^ - }]$
Electronic configuration of Al: $[1{s^2}2{s^2}2{p^6}3{s^2}3{p^1}]$
Electronic configuration of $[A {l^{ + 3}}]: [1{s^2}2{s^2}2{p^6}]$
Similarly we can write for sodium, magnesium and potassium. Sodium and potassium each will lose 1 electron, whereas magnesium atom loses 2 electrons to form cation and complete their octet.
$ [Na \to N{a^ + } + {e^ - }]$
Electronic configuration of Na: $[1{s^2}2{s^2}2{p^6}3{s^1}]$
Electronic configuration of $[N{a^ + }]: [1{s^2}2{s^2}2{p^6}]$
$[Mg \to M{g^{ + 2}} + 2{e^ - }]$
Electronic configuration of Mg: $[1{s^2}2{s^2}2{p^6}3{s^2}]$
Electronic configuration of $[M{g^{ + 2}}]$: $[1{s^2}2{s^2}2{p^6}]$
$[K \to {K^ + } + {e^ - }]$
Electronic configuration of K: $[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}]$
Electronic configuration of $[{K^ + }]$: $[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}]$
On losing their valence electrons, Al, Na and Mg attain the nearest noble gas configuration that is neon and K attains the electronic configuration of argon.
As aluminium has to lose 3 electrons in order to achieve its octet, it has +3 valency and hence, it is trivalent. The monovalent sodium and potassium have to lose only electrons in order to achieve its octet.
A divalent or bivalent ion or atom, has a valency of 2. Magnesium has to lose 2 valence electrons for attaining its octet. So, it is divalent.
Magnesium, an alkaline earth metal is highly reactive in nature. Magnesium is one of the most abundant metals in nature.
Therefore, the correct answer is Mg.

The divalent metal whose oxide is reduced to metal by electrolysis of its fused salt is \[Mg\].

Note: Magnesium oxide when subjected to electrolysis in its molten form, it will decompose to form $[M{g^{ + 2}}]$ and $[{O^{ - 2}}]$ ions. These ions move freely in an electrolyte.
At cathode, the $[M{g^{ + 2}}]$ ions accepts 2 electrons forming magnesium metal in its pure form. The magnesium metal will be deposited at the cathode.
$[M{g_{(l)}}^{ + 2} + 2{e^ - } \to M{g_{(s)}}]$
At anode, $[{O^{ - 2}}]$ ions will lose their electrons to form oxygen molecule ($[{O_2}]$)
$[2{O^{ - 2}} \to {O_{2(g)}} + 4{e^ - }]$
At anode, the oxygen gas evolves.
The complete equation at anode and cathode can be shown as:
$[2M{g_{(l)}}^{ + 2} + 2{O_{(l)}}^{ - 2} \to 2M{g_{(s)}} + {O_2}_{(g)}]$