Question

Which one of the following will not conduct electricity?
(A). Solid $NaCl$
(B). $CuS{O_4}$solution
(C). Graphite
(D). Acidified water

Answer 1

- Hint: You can start by describing why current flows in a medium or material. Then move on to explain how $CuS{O_4}$ has free ions, graphite has free electrons and acidified water has $H +$ions that can make the flow of current possible. Then explain how solid $CuS{O_4}$ does not conduct electricity.

Complete step-by-step answer:
The concept of flow of current – Current flows through a material or medium when charged carries like electrons or ions move in the medium due to the presence of a potential difference (Potential difference leads to the formation of electric field).
$CuS{O_4}$ - The solution of $CuS{O_4}$ solution contains one $C{u^{2 + }}$ and one $SO_4^{2 - }$ ion. Under the effect on an electric field the positive ions $(C{u^{2 + }})$ move towards the cathode and negative ions $(SO_4^{2 - })$ move towards the anode. This movement of ions results in the flow of current.
Graphite – Graphite is the only non-metal that can conduct electricity. Graphite is made-up of long layers of C-atom chains. The carbon atom can form 4 bonds, but in graphite carbon only forms three bonds and the last electron helps in the flow of current.
Acidified water – Acidified water contains $H +$ (hydrogen ion). These form ${H_3}{O^ + }$ (hydronium ion) when they combine with water molecules. These hydronium ions can move under the influence of an electric field and result in the flow of current.
Solid $NaCl$- Solid $NaCl$ does not have free ions, nor does it have free electrons. So it cannot conduct electricity.
Hence, option A is the correct choice.

Note: Unlike solid $NaCl$, a solution of $NaCl$ or $NaCl$ in its molten state has free ions that can carry under the influence of an electric field. $NaCl$ solution in the process of electrolysis conducts electricity through the movement of ions $N{a^ + }$ moves towards the cathode and $C{l^ - }$ moves towards the anode. Thus $N{a^ + }$ and $C{l^ - }$ are separated.