Compare the structures of diamond and graphite

• Diamond: Each carbon atom is $sp^3$ hybridised and bonded tetrahedrally to four other carbon atoms by strong single covalent bonds.
• Graphite: Each carbon atom is $sp^2$ hybridised and bonded to three other carbon atoms in planar hexagonal sheets. The fourth electron is delocalised over the layer, giving a $\pi$-electron cloud.

Check statement A: Graphite is harder than diamond

• Diamond has a rigid 3D covalent network.
• Graphite has layered sheets held together by weak van der Waals forces, so layers can slide over one another.

Therefore, graphite is softer than diamond.

A is false.

Check statement B: Graphite has higher electrical conductivity than diamond

• In graphite, delocalised electrons within layers can move freely, so graphite conducts electricity.
• In diamond, all valence electrons are localised in $\sigma$ bonds, so it is an electrical insulator.

Therefore, graphite has higher electrical conductivity than diamond.

B is true.

Check statement C: Graphite has higher thermal conductivity than diamond

• Diamond is known to have exceptionally high thermal conductivity due to its strong 3D covalent lattice and efficient phonon transport.
• Although graphite also conducts heat well along layers, diamond generally has higher thermal conductivity overall.

Therefore, the statement that graphite has higher thermal conductivity than diamond is false.

C is false.

Check statement D: Graphite has higher $C-C$ bond order than diamond

• In diamond, each $C-C$ bond is a single bond, so bond order $=1$.
• In graphite, due to delocalisation in the hexagonal layers, each $C-C$ bond has partial double-bond character. Average bond order is greater than $1$.

Therefore, graphite has higher average $C-C$ bond order than diamond.

D is true.

Final conclusion

The correct statements are: $\boxed{B,\ D}$

Comparison with stored correct answer

• Derived answer: $B, D$
• Stored correct answer: $B, D$

They match.