Chemistry of Carbon: Practice Questions

Multiple Choice Questions (MCQs)

Question 1

Which of the following statements about the allotropes of carbon is INCORRECT?

A) Diamond has a 3D network structure with sp³ hybridized carbon atoms. B) Graphite has a layered structure with sp² hybridized carbon atoms. C) Fullerenes are cage-like molecules with pentagonal and hexagonal rings of carbon atoms. D) Carbon nanotubes are 2D structures formed by rolling up graphene sheets.

Solution 1

Correct Answer: D

Explanation: Carbon nanotubes are one-dimensional (1D) cylindrical structures formed by rolling up single or multiple sheets of graphene. Graphene itself is a 2D material, but the nanotubes derived from it are inherently 1D along their axis. The other statements (A, B, C) accurately describe the structures of diamond, graphite, and fullerenes, respectively.

Question 2

Which of the following is true for Carbon Monoxide (CO)?

A) It is an acidic oxide. B) It is a strong oxidizing agent. C) It is highly poisonous due to its affinity for hemoglobin. D) It turns moist blue litmus red.

Solution 2

Correct Answer: C

Explanation: Carbon monoxide (CO) is a highly poisonous gas. It binds to the iron in hemoglobin in red blood cells about 200-250 times more strongly than oxygen, forming carboxyhemoglobin. This stable complex reduces the oxygen-carrying capacity of blood, leading to oxygen deprivation in tissues and potentially death.

• CO is a neutral oxide, not acidic (A is incorrect).
• CO is a strong reducing agent, not an oxidizing agent (B is incorrect).
• Being a neutral oxide, it does not change the color of litmus paper (D is incorrect).

Question 3

Which type of carbide is typically formed when carbon reacts with highly electropositive elements like Calcium (Ca) or Aluminium (Al)?

A) Covalent carbides B) Interstitial carbides C) Ionic carbides D) Polymeric carbides

Solution 3

Correct Answer: C

Explanation: Highly electropositive elements, especially from Group 1 (alkali metals), Group 2 (alkaline earth metals), and Group 13 (like Aluminium), form ionic carbides. These carbides contain discrete carbide anions such as C₂²⁻ (acetylide ion, e.g., CaC₂) or C⁴⁻ (methanide ion, e.g., Al₄C₃). Upon hydrolysis, acetylides produce acetylene, and methanides produce methane.

• Covalent carbides (e.g., SiC, B₄C) are formed with elements of similar electronegativity.
• Interstitial carbides (e.g., Fe₃C, WC) are formed with transition metals where carbon atoms occupy the interstitial sites in the metal lattice.

Assertion-Reason Questions

Question 1

Assertion (A): Graphite is a good conductor of electricity, while diamond is an electrical insulator. Reason (R): In graphite, each carbon atom is sp² hybridized, and one valence electron is delocalized, forming a mobile electron cloud. In diamond, each carbon atom is sp³ hybridized, and all valence electrons are involved in sigma bonds.

A) Both A and R are true, and R is the correct explanation of A. B) Both A and R are true, but R is NOT the correct explanation of A. C) A is true, but R is false. D) A is false, but R is true. E) Both A and R are false.

Solution 1

Correct Answer: A

Explanation: Assertion (A) is true. Graphite’s layered structure allows it to conduct electricity, while diamond’s rigid covalent network makes it an insulator. Reason (R) is also true and correctly explains the assertion. In graphite, each carbon atom is sp² hybridized and bonded to three other carbon atoms in the same plane, leaving one unhybridized p-orbital perpendicular to the layers. These p-orbitals overlap to form a delocalized electron cloud, allowing electrons to move freely, thus making graphite a good conductor. In diamond, each carbon atom is sp³ hybridized and bonded to four other carbon atoms in a tetrahedral arrangement. All valence electrons are tightly held in strong sigma bonds, with no free electrons available for conduction, hence it acts as an insulator.

Question 2

Assertion (A): Carbon monoxide is highly toxic. Reason (R): Carbon monoxide forms a stable complex with hemoglobin, called carboxyhemoglobin, which is about 200-250 times more stable than oxyhemoglobin.

A) Both A and R are true, and R is the correct explanation of A. B) Both A and R are true, but R is NOT the correct explanation of A. C) A is true, but R is false. D) A is false, but R is true. E) Both A and R are false.

Solution 2

Correct Answer: A

Explanation: Assertion (A) is true. Carbon monoxide is indeed a highly toxic gas. Reason (R) is also true and provides the correct explanation for its toxicity. CO has a much stronger affinity for the heme iron in hemoglobin than oxygen. When CO binds to hemoglobin, it forms carboxyhemoglobin, which is significantly more stable than oxyhemoglobin. This prevents hemoglobin from transporting oxygen to the body’s tissues, leading to cellular hypoxia and potentially death.

Short Answer Questions

Question 1

Explain why carbon exhibits the maximum tendency for catenation among Group 14 elements.

Model Answer 1

Catenation is the ability of an element to form bonds with atoms of itself, creating long chains or rings. Carbon exhibits the maximum tendency for catenation due to the following reasons:

1. Small Size: Carbon atoms are relatively small, allowing them to form strong, stable covalent bonds with other carbon atoms without significant steric repulsion.
2. High C-C Bond Energy: The carbon-carbon single bond (C-C) is very strong (bond energy ≈ 348 kJ/mol), making the chains and rings stable. As we move down Group 14 (Si, Ge, Sn, Pb), the atomic size increases, and the E-E (element-element) bond strength decreases significantly (e.g., Si-Si bond energy ≈ 222 kJ/mol). This weaker bond strength makes catenated structures less stable for heavier elements.
3. Ability to Form Multiple Bonds: Carbon can form stable single (C-C), double (C=C), and triple (C≡C) bonds with itself and with other elements. This versatility allows for a vast array of stable catenated structures and isomers. Heavier elements in Group 14 show a much lesser tendency to form stable multiple bonds due to the larger size and more diffuse p-orbitals, which leads to poor pπ-pπ overlap.

Question 2

Describe the structure and bonding in carbon dioxide (CO₂).

Model Answer 2

Carbon dioxide (CO₂) is a simple molecular compound with a distinct structure and bonding:

1. Hybridization: The central carbon atom in CO₂ is sp hybridized. It forms two sigma (σ) bonds and two pi (π) bonds.
2. Bonding:
   • The carbon atom forms two C=O double bonds, one with each oxygen atom.
   • Each C=O double bond consists of one sigma (σ) bond formed by the overlap of an sp hybrid orbital of carbon with a p orbital of oxygen, and one pi (π) bond formed by the lateral overlap of an unhybridized p orbital of carbon with another p orbital of oxygen.
3. Geometry: Due to sp hybridization, the molecule has a linear geometry with a bond angle of 180°.
4. Polarity: Although the C=O bonds are polar due to the electronegativity difference between carbon and oxygen, the linear symmetrical arrangement of the two C=O dipole moments results in their cancellation. Therefore, the CO₂ molecule as a whole is nonpolar.
5. Physical State: At room temperature and pressure, CO₂ is a gas. In its solid state (dry ice), it is a molecular solid held together by weak intermolecular forces (London dispersion forces).

High-Order Thinking Skills (HOTS) Question

Question 1

Why is carbon dioxide (CO₂) a gas at room temperature, while silicon dioxide (SiO₂) is a hard, high-melting solid? Explain the fundamental differences in their bonding and structure.

Model Answer 1

The stark difference in the physical states of CO₂ and SiO₂ at room temperature (CO₂ as a gas, SiO₂ as a hard solid) arises from fundamental differences in their bonding and molecular structures.

1. Carbon Dioxide (CO₂):

   • Structure: CO₂ exists as discrete, simple triatomic molecules (O=C=O). The central carbon atom forms two C=O double bonds with sp hybridization, resulting in a linear geometry.
   • Bonding: Within each CO₂ molecule, strong covalent bonds exist between carbon and oxygen. However, the individual CO₂ molecules are held together in the solid or liquid state only by weak intermolecular forces, specifically London dispersion forces (van der Waals forces).
   • Consequence: These weak intermolecular forces require very little energy to overcome. Hence, CO₂ has a very low melting point (-56.6 °C at 5.1 atm, sublimes at -78.5 °C at 1 atm) and is a gas at room temperature.

2. Silicon Dioxide (SiO₂):

   • Structure: SiO₂ does not exist as discrete molecules like CO₂. Instead, it forms a giant covalent network structure (macromolecular structure), similar to diamond. In this structure, each silicon atom is tetrahedrally bonded to four oxygen atoms, and each oxygen atom bridges two silicon atoms. This results in a continuous 3D network of strong covalent Si-O-Si linkages.
   • Bonding: The entire crystal of SiO₂ is held together by a vast network of strong covalent single bonds. Silicon, being larger than carbon, prefers to form four strong Si-O single bonds rather than Si=O double bonds. The formation of pπ-pπ double bonds (Si=O) is not favorable for silicon due to the larger size of its 3p orbitals, which are too diffuse for effective lateral overlap with oxygen’s 2p orbitals.
   • Consequence: To melt or boil SiO₂, a large amount of energy is required to break these extensive and strong covalent bonds throughout the entire network. Therefore, SiO₂ is a hard, high-melting solid (melting point ≈ 1710 °C) at room temperature.

In summary, the key difference lies in carbon’s ability to form stable pπ-pπ multiple bonds (C=O) leading to discrete molecular units, while silicon’s preference for forming strong σ bonds and its inability to form stable pπ-pπ multiple bonds with oxygen leads to an extended giant covalent network structure.