Question

In diamond, the numbers of free electrons are:
A.1
B.2
C.0
D.4

Answer 1

To answer this question, you must recall the structure and bonding of diamond. Diamond is a solid allotropic form of carbon. Apart from diamond, another allotropic form of carbon exists as solid at room temperature which is graphite.

Complete step by step answer:
We know that carbon is an element of group 14 of the modern periodic table and it has the atomic number 6. The number of valence electrons is 4. The ground state electronic configuration of carbon can be written as
$C:\left[ {He} \right]2{s^2}2{p^2}$
One electron is excited from the $2s$ orbital to $2p$ orbital to increase the covalency. The electronic configuration in this excited state is given as $C:\left[ {He} \right]2{s^1}2{p^3}$
Diamond is a crystalline solid and it has a crystal lattice consisting of carbon atoms. In diamond, each carbon atom undergoes $s{p^3}$ hybridization and is linked to four other carbon atoms. Each carbon atom is present at the center of a tetrahedron structure and is surrounded by four other carbon atoms bonded to it, present at the corners of this tetrahedron.
All four valence electrons of carbon are used in bonding and each carbon has a complete octet. As a result, diamond forms a rigid three dimensional network. We can conclude that there are no free electrons in diamond.

Thus, the correct option is C.

Note:
Due to the rigid three dimensional network structure formed by the carbon atoms, diamond is the hardest substance. Also due to the rigid network and strong covalent bonds, diamond is chemically inactive. A very high energy is required in order to break the large number of carbon to carbon sigma bonds in the extensive network and thus, it has a high melting point as well.