Question

The carbon-carbon bond length in graphite is
A.$1.34A^\circ $
B.$1.54A^\circ $
C.$1.42A^\circ $
D.$1.20A^\circ $

Answer 1

Hint: The usual $\,C - C\,$ single bond length is $1.54A^\circ $. The carbon-carbon bond length in graphite will be less than this usual bond length due to some reasons which will be discussed further in this solution. Graphite is a hexagonal layered structure in which each carbon has $s{p^2}$ hybridization.

Complete step by step answer:
The actual $\,C - C\,$ single bond has a length of $1.54A^\circ $ but in graphite each carbon atom is bonded to three other carbon atoms that form layers in a hexagonal manner.
Now, you might wonder why its bond length is not the same as a single $\,C - C\,$ bond, the answer is that each carbon in graphite has a non-bonded electron which can delocalize through the structure. The hybridization of carbon in graphite is $s{p^2}$ so one $\,p\,$ orbital is available for the delocalization of electrons. So, the bond length will be less than that of usual $\,C - C\,$ single bond $\,(1.54A^\circ )\,$ and becomes $1.42A^\circ $.
So, the right option is C that is $1.42A^\circ $.


Additional information:
The delocalized electron allows a free movement of electrons through the entire structure which makes it able to conduct electricity. So graphite is a good conductor. And also the layers in graphite structure can slide over due to its weak forces of attraction. This makes graphite a good lubricant because of its slipping nature.
The graphite structure is as follows,

Note:
Note that the bond length of diamond is $1.54A^\circ $ which is a usual $\,C - C\,$ single bond length. Here, in graphite due to delocalization it gets decreased, So, don’t be confused with the two values. And $1.34A^\circ $ is the $\,C = C\,$ double bond length so the real answer will be also more than this value.