Question

Which of the following can conduct electricity in the solid state?
(A) Sucrose, \[{C_{12}}{H_{22}}{O_{11}}(s)\]
(B) Graphite, C(s)
(C) Sodium hydroxide, NaOH(s)
(D) Iron(III) chloride, \[FeC{l_3}(s)\]
(E) Iodine, \[{I_2}(s)\]

Answer 1

Hint: We already know that metals like gold, platinum are extremely good conductors of electricity in solid state.
But there are some elemental non metal which too are good conductors of electricity. They have free mobile electrons which make them such good conductors.

Complete step by step answer:
Let us first analyse all the options given in the question to decide which conducts electricity in solid state.
We know that sucrose is a nonelectrolyte. This means that sucrose cannot conduct electricity even in solution. In the solid state, the molecules of sucrose are bonded by strong covalent bonds, as there are no charged particles present it does not conduct electricity.
Next, we know that Sodium hydroxide and Iron chloride are ionic compounds; they conduct electricity in aqueous state or in molten state. But in the solid state they are unionized, they lack mobile electrons hence cannot conduct electricity.
Coming to solid iodine molecules, it has two iodine atoms joined by strong covalent bonds. Hence, this too cannot conduct electricity.
In the case of Graphite, it has delocalized electrons very much like metals. This enables them to move between the layers and hence conduct electricity.

Hence, the correct answer is Option (B) Graphite.

Additional information:
In graphite, each atom is surrounded by three other carbon atoms in a trigonal planar arrangement. There are four possible bonding points but only three are occupied so the fourth electron is free to move.

Note: We should remember that while electricity is conducted in a parallel manner, the plates of graphite are bonded by intermolecular forces where there is no orbital overlap and no delocalized electrons. Therefore, there is no conduction perpendicular to the plates.