Assertion: Among the carbon allotropes, diamond is an insulator, whereas, graphite is a good conductor of electricity.
Reason: Hybridization of carbon in diamond and graphite are sp$$^3$$ and sp$$^2$$ respectively.

First, let us read the two statements with care.

Assertion: “Among the carbon allotropes, diamond is an insulator, whereas graphite is a good conductor of electricity.”

Reason: “Hybridization of carbon in diamond and graphite are $$sp^3$$ and $$sp^2$$ respectively.”

We want to judge two things: (i) whether each sentence is factually correct, and (ii) whether the second sentence is the real, complete cause for the first sentence. We shall proceed slowly, checking all the chemical facts step by step.

Step 1 - Verifying the assertion. In diamond every carbon is covalently bonded to four other carbons in a three-dimensional network. No free or mobile electrons are left. Therefore the electrical conductivity $$\sigma$$ is practically zero; diamond is an insulator.
In graphite each carbon is covalently bonded to three other carbons in a two-dimensional hexagonal sheet. One valence electron of every carbon atom remains unbonded in a $$p_z$$ orbital. These $$p_z$$ orbitals overlap sideways over the whole sheet, producing a delocalised $$\pi$$-electron cloud. These delocalised electrons can move under an electric field, so graphite conducts electricity well, especially along the planes. Hence the assertion is true.

Step 2 - Verifying the reason. The carbon in diamond has four $$\sigma$$-bonds arranged tetrahedrally. According to valence bond theory, tetrahedral geometry corresponds to $$sp^3$$ hybridisation. In graphite the carbon forms only three $$\sigma$$-bonds lying in one plane at $$120^{\circ}$$. Planar trigonal geometry is obtained from $$sp^2$$ hybridisation. Thus the reason is also true.

Step 3 - Testing whether the reason explains the assertion.
We now ask: “Does merely stating that the hybridisations are $$sp^3$$ and $$sp^2$$ automatically explain the difference in conductivity?”

• Hybridisation describes how atomic orbitals mix to give bond geometry and bond type; it does not, by itself, mention free or delocalised electrons.
• The actual cause of conductivity in graphite is the presence of one unhybridised $$p_z$$ orbital per carbon whose electrons become delocalised to form a conduction band. Saying “graphite is $$sp^2$$” certainly tells us there is an unhybridised $$p_z$$ orbital, but the key phrase “delocalised $$\pi$$ electrons that are mobile” is missing. Similarly, the statement “diamond is $$sp^3$$” only conveys tetrahedral bonding; it does not explicitly point out that all electrons are localised in $$\sigma$$-bonds and no conduction band exists.
• Therefore, while the difference in hybridisation is related to the electronic structures, the reason as written is not a complete, explicit explanation of why one allotrope conducts and the other does not.

Because the assertion is correct, the reason is also correct, yet the reason does not sufficiently and directly explain the assertion, we select the option that says exactly this.

Option A states: “Both assertion and reason are correct, but the reason is not the correct explanation for the assertion.”

Hence, the correct answer is Option A.