Question

Which of the following is not a conductor of electricity?

• A. Solid NaCl
• B. Cu
• C. Fused NaCl
• D. Brine

Answer 1

The correct option is A

Solid NaCl

The explanation for the correct option

In the case of option A,

1. $\mathrm{NaCl}$ is commonly known as sodium chloride.
2. It is the salt that we use on a regular basis in our homes.
3. Ionic compounds such as $\left(\mathrm{NaCl}\right)$ do not conduct electricity in the solid state because the ions are fixed in place and cannot move, but they do so in an aqueous solution and in the molten state because the ions are free to move.
4. Sodium chloride $\left(\mathrm{NaCl}\right)$ is an ionic compound that contains sodium and chloride ions.

The explanation for the incorrect options

In the case of option B,

1. Copper is a chemical element with the atomic number 29 and the symbol $\left(\mathrm{Cu}\right)$.
2. It is one of the few metals that has the ability to conduct electricity in its pure state.
3. Copper is used mostly in electrical wiring as well as industrial machines.
4. Copper is utilized for electrical wiring because it has the highest conductivity of all metals.

In the case of option C,

1. Fused $\mathrm{NaCl}$ is sodium chloride salt that has been converted to a liquid state by heating.
2. Sodium chloride does not conduct electricity in a free state since it does not have any free mobile ions in its solid-state, which are responsible for electricity conduction, but it does so in its fused or watery state, where it does have free mobile ions.
3. It is a good conductor of electricity because fused $\mathrm{NaCl}$ contains mobile Sodium and Chloride ions obtained by breaking the lattice with heat.

In the case of option D,

1. Brine is a concentrated sodium chloride solution in water.
2. Because it includes oppositely charged ions, brine solution conducts electricity.
3. In water, the sodium and chloride ions split, changing solid $\mathrm{NaCl}$ into ${\mathrm{Na}}^{+}$ and ${\mathrm{Cl}}^{-}$ ions that can freely travel through solutions with oppositely charged ions.

Hence, option A is the correct answer.