Question

In Graphite electrons are:
A.Localized on each $C$ atom
B.Localized on every third $C$ atom.
C.Spread out between the structure
D.Present in antibonding orbital

Answer 1

Hint: Graphite carbon atoms have $s{p}^2$ hybridization. They have electrons that are delocalized. These electrons are part of the$\pi$bond orbitals that are present in Graphite.

Complete step by step answer:
Graphite is an allotrope of carbon. It is different from diamond in that it has $s{p}^2$ hybridization whereas diamond has $s{p}^3$ hybridization.
Being an $s{p}^2$ hybrid, graphite is known to have $\pi$ bonds. These $\pi$ bonds are formed by the overlapping of p-p orbitals in a plane perpendicular to the internuclear axis and the overlapping is indirect and sideways.
This means that the atoms are free to move around. In addition to that, since it is $s{p}^2$ hybridized we understand that the carbon only uses three of its four orbitals for bonding leaving one electron unpaired thus causing a lone electron to move around.
In general, using the above information, we can say that graphite has lone electrons that are moving freely in the interstitial spaces present within the lattice of the graphite.
Because of this phenomenon graphite is one of the few nonmetals present in the periodic table that can conduct electricity. Graphite can do so because of the presence of these unpaired electrons that are floating around. They can propagate electric current.
Therefore, the answer to the above question is option C, that is, the electrons are spread out between the structures.

Note: It is important to remember that graphite is an $s{p}^2$ hybridized carbon atom and hence it contains delocalized electrons.
The electrons present in the graphite lattice float freely.
Graphite is generally a soft substance when compared to the other allotropes of carbon like diamond due to its hybridization.