Science (086) — AI-generated practice question

(i) Structure, Hardness, and Conductivity:

In diamond, each carbon atom is bonded to four others in a rigid tetrahedral 3D network. All four valence electrons are used in bonding, so no free electrons exist → diamond is a non-conductor and extremely hard.

In graphite, each carbon atom bonds to three others in flat hexagonal layers. The fourth electron is free to move between layers → graphite is a good conductor. The layers are held by weak forces and can slide over each other → graphite is soft.

(ii) Combustion behaviour:

Despite structural differences, both diamond and graphite are allotropes of carbon. On combustion, both burn in oxygen to produce only carbon dioxide with release of heat and light:
$$\text{C} + \text{O}_2 \rightarrow \text{CO}_2 + \text{heat and light}$$

Their chemical behaviour during combustion is identical because the product depends on the element (carbon), not its structural form.

(iii) Disagreement with the student:

No, the student is incorrect. Hardness is a physical property related to structure, not a chemical property. Since diamond and graphite are both pure carbon, they produce the same product — CO₂ — on combustion. The textbook states: "Carbon, in all its allotropic forms, burns in oxygen to give carbon dioxide."

Source: Chapter 4, Section 4.3.1 (Combustion)

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• Examiners expect you to clearly contrast diamond (3D network, non-conductor, hard) and graphite (layered, conductor, soft) using the concept of free electrons and bonding.
• The key textbook line for parts (ii) and (iii) is: "Carbon, in all its allotropic forms, burns in oxygen to give CO₂." Quote or paraphrase it — it directly answers the student's argument.
• Part (iii) tests whether you can distinguish physical properties (hardness) from chemical properties (combustion products). Always clarify this distinction.
• Keep the equation balanced and include "heat and light" — examiners often check for this.