Comparison of Properties Between Graphite and Diamond

To determine the correct statements regarding graphite and diamond, we analyze each property individually:

1. Hardness (Statement A):

   • Diamond: Each carbon atom is $sp^3$ hybridized and bonded to four other carbon atoms in a rigid, three-dimensional tetrahedral lattice with strong covalent bonds. This makes diamond one of the hardest known natural substances.
   • Graphite: Each carbon atom is $sp^2$ hybridized, forming two-dimensional hexagonal sheets. These individual sheets are held together by weak van der Waals forces, allowing them to slide past one another easily. Thus, graphite is soft and slippery.
   • Therefore, diamond is harder than graphite, so Statement (A) is incorrect.

2. Electrical Conductivity (Statement B):

   • Graphite: Each carbon atom uses three of its four valence electrons to form $\sigma$-bonds in a planar hexagonal network ($sp^2$ hybridization). The fourth valence electron occupies a $p$-orbital perpendicular to the sheet, contributing to a delocalized $\pi$-system. These mobile $\pi$-electrons are free to move along the layers, giving graphite high electrical conductivity.
   • Diamond: All four valence electrons are tightly bound in localized $sp^3$ $\sigma$-bonds, leaving no free electrons for electrical conduction. Hence, diamond is an insulator.
   • Therefore, graphite has higher electrical conductivity than diamond, so Statement (B) is correct.

3. Thermal Conductivity (Statement C):

   • Diamond: Heat transfer in non-metallic solids occurs via lattice vibrations (phonons). The strong, light carbon-carbon $sp^3$ covalent bonds and rigid 3D lattice structure in diamond result in extremely low phonon scattering, giving it an exceptionally high thermal conductivity ($\sim 2000 \text{ W m}^{-1}\text{K}^{-1}$).
   • Graphite: Although its in-plane thermal conductivity is high, the overall thermal conductivity (especially perpendicular to the layers) is lower than that of diamond due to weak interlayer coupling.
   • Therefore, diamond has a higher thermal conductivity than graphite, so Statement (C) is incorrect.

4. C-C Bond Order (Statement D):

   • Diamond: Each C-C interaction is a pure single $\sigma$-bond formed by $sp^3-sp^3$ orbital overlap. Thus, the bond order is: $\text{Bond Order (Diamond)} = 1$
   • Graphite: Each carbon atom forms three $sp^2-sp^2$ $\sigma$-bonds and participates in a delocalized $\pi$-network across the hexagonal rings. Considering resonance structures of a benzene-like hexagonal lattice, the C-C bond order is: $\text{Bond Order (Graphite)} = 1 + \frac{1}{3} = 1.33$
   • Since $1.33 > 1.00$, graphite has a higher C-C bond order than diamond, so Statement (D) is correct.

Thus, the correct options are B and D.