Check statement (A): Covalent radius decreases down the group from $\mathrm{C}$ to $\mathrm{Pb}$ in a regular manner.

This is false.

In group 14, atomic/covalent radius generally increases down the group, not decreases.

Order is roughly: $\mathrm{C} < \mathrm{Si} < \mathrm{Ge} \approx \mathrm{Sn} < \mathrm{Pb}$

So statement (A) is incorrect.

Check statement (B): Electronegativity decreases from $\mathrm{C}$ to $\mathrm{Pb}$ down the group gradually.

This is also false.

In group 14, electronegativity does not decrease regularly down the group. Because of poor shielding by $d$ and $f$ electrons, the values are somewhat irregular.

Approximate trend: $\mathrm{C} > \mathrm{Si} < \mathrm{Ge} \approx \mathrm{Sn} < \mathrm{Pb}$

Hence it does not decrease gradually from $\mathrm{C}$ to $\mathrm{Pb}$.

Check statement (C): Maximum covalence of $\mathrm{C}$ is 4 whereas other elements can expand their covalence due to presence of d orbitals.

This is true.

• Carbon has no vacant $d$ orbitals in its valence shell, so maximum covalency is 4.
• Heavier group 14 elements like Si, Ge, Sn, Pb can show covalency greater than 4 in some compounds due to availability of vacant orbitals (traditionally explained using $d$ orbitals in JEE-level chemistry).

Therefore, (C) is correct.

Check statement (D): Heavier elements do not form $\mathrm{p}\pi-\mathrm{p}\pi$ bonds.

This is true.

Carbon readily forms multiple bonds like: $\mathrm{C=C}, \quad \mathrm{C\equiv C}, \quad \mathrm{C=O}$ because effective sideways overlap of small 2p orbitals is possible.

Heavier elements (Si, Ge, Sn, Pb) have larger and more diffuse orbitals, so effective $\mathrm{p}\pi-\mathrm{p}\pi$ overlap is poor. Thus they generally do not form such bonds.

Hence (D) is correct.

Check statement (E): Carbon can exhibit negative oxidation states.

This is true.

Carbon shows negative oxidation states in compounds with less electronegative elements, for example:

• In methane, $\mathrm{CH_4}$, oxidation state of carbon is $x + 4(+1) = 0 \Rightarrow x = -4$
• In carbides also carbon may have negative oxidation state.

So (E) is correct.

Collect the correct statements

• (A) False
• (B) False
• (C) True
• (D) True
• (E) True

Therefore the correct set is: $\boxed{(C), (D) \text{ and } (E)\text{ only}}$

Match with options

This corresponds to Option C.