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Covalent Bonds and the Versatile Nature of Carbon

🎓 Class 10 Science CBSE Theory Ch 4 — Carbon and its Compounds ⏱ ~22 min
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This MCQ module is based on: Covalent Bonds and the Versatile Nature of Carbon

[myaischool_lt_science_assessment grade_level="class_10" science_domain="chemistry" difficulty="intermediate"]

Introduction: Carbon — The Element of Life

Look around you. The air you breathe contains carbon dioxide (CO2). The shells of sea creatures and the limestone of the Himalayas are made of calcium carbonate (CaCO3). Coal, petroleum and natural gas — the fuels that power our world — are carbon compounds. Every protein, carbohydrate, fat, and strand of DNA in every living cell is built from a backbone of carbon atoms.

All the other elements together form a certain number of compounds, but carbon alone forms millions. How does this single element, which makes up only about 0.02% of Earth's crust, create such astonishing diversity? The answer lies in the way carbon bonds.

Chapter focus: We will explore why carbon forms covalent bonds, what makes carbon so versatile, and meet its remarkable allotropes — diamond, graphite and fullerene.

4.1 Bonding in Carbon — The Covalent Bond

Carbon has atomic number Z = 6, giving it the electronic configuration 2, 4. It therefore has four valence electrons. To become stable (achieve a noble-gas configuration of 8 electrons in the outer shell), carbon needs four more electrons.

It has two theoretical options:

  • Lose 4 electrons to form C4+ — but this requires an enormous amount of energy (ionisation energy). Removing four electrons from a small atom leaves a highly charged cation, which is energetically unfavourable.
  • Gain 4 electrons to form C4− — but then 6 protons would have to hold on to 10 electrons. This concentration of negative charge is also unstable.

Carbon chooses a third path: sharing. By sharing its four valence electrons with the valence electrons of other atoms, carbon (and its partner) can both achieve an octet without any ion formation. A bond formed by the mutual sharing of electron pairs is called a covalent bond.

Activity 4.1 — Drawing Lewis (Electron-Dot) StructuresL3 Apply
Predict first: Using the idea of octet completion, how many electron pairs must be shared between the two atoms in Cl2, O2, and N2?
  1. Write the electronic configuration of H (1), Cl (2,8,7), O (2,6), N (2,5), C (2,4).
  2. Count the valence electrons of each atom.
  3. Pair up the atoms so each achieves a duplet (H) or octet (others) through sharing.
  4. Represent each shared pair of electrons by a pair of dots between the two atoms — or by a short line.
Lewis structures:
  • H2: H:H or H–H (single bond — 1 shared pair)
  • Cl2: Cl–Cl with 3 lone pairs on each Cl (single bond)
  • O2: O=O (double bond — 2 shared pairs)
  • N2: N≡N (triple bond — 3 shared pairs)
  • H2O: H–O–H, two lone pairs on O
  • NH3: three N–H bonds, one lone pair on N
  • CH4: four C–H single bonds — carbon uses all four valence electrons
H:H H₂ (single) Cl:Cl Cl₂ (single) O::O O₂ (double) N⋮⋮⋮N N₂ (triple) H:O:H H₂O H:N:H H: NH₃ H:C:H H: H: CH₄ — four C–H bonds Bond types Single : (1 pair = 2 e⁻) Double :: (2 pairs = 4 e⁻) Triple ⋮⋮⋮ (3 pairs = 6 e⁻)
Fig 4.1: Electron-dot (Lewis) structures showing single, double, and triple covalent bonds

Methane (CH₄) — Tetrahedral Geometry

In methane, carbon shares one electron with each of four hydrogen atoms. Because the four electron pairs repel one another equally, they spread out into a tetrahedron with H–C–H bond angles of 109.5°.

C H H H H 109.5°
Fig 4.2: Tetrahedral shape of the methane molecule — dashed bond projects behind the page

Properties of Covalent Compounds

PropertyReason
Low melting and boiling pointsWeak intermolecular forces between neutral molecules
Poor conductors of electricityNo free ions or mobile electrons to carry charge
Usually insoluble in waterNon-polar molecules do not interact with polar water
Soluble in organic solvents"Like dissolves like" — non-polar in non-polar

4.2 Versatile Nature of Carbon

Why does a single element create over ten million known compounds? Four special features of carbon — saturation choice, catenation, tetravalency, and the ability to bond with many other elements — explain this unmatched versatility.

4.2.1 Saturated and Unsaturated Compounds

Saturated compound: contains only single bonds between carbon atoms. General formula for open-chain alkanes — \(C_nH_{2n+2}\).
Unsaturated compound: contains at least one double (C=C) or triple (C≡C) bond. Alkenes — \(C_nH_{2n}\); alkynes — \(C_nH_{2n-2}\).

Example: CH4 (methane) is saturated. C2H4 (ethene) is unsaturated because of its C=C bond.

4.2.2 Catenation

Catenation is carbon's remarkable ability to link to other carbon atoms, forming chains of two, three, four — even hundreds — of atoms in a row. The chains may be straight, branched, or closed into rings. Carbon–carbon bonds are exceptionally strong (bond energy ≈ 348 kJ/mol), which is why these large structures remain stable.

4.2.3 Tetravalency of Carbon

Because carbon has four valence electrons, it forms four covalent bonds. This tetravalency means one carbon atom can bond to up to four other atoms — and those atoms can themselves be carbon or many other elements: H, O, N, S, the halogens, P, and more. This is how diverse classes of compounds — alcohols, acids, amines, halides — all arise.

4.2.4 Allotropes of Carbon

Allotropes are different structural forms of the same element. Pure carbon comes in several allotropic forms whose atoms are arranged differently — leading to wildly different properties.

Worked Examples

Example 1. Why does carbon not form C4+ or C4− ions readily?
Solution: Losing 4 electrons (to form C4+) requires a very large amount of energy, which is energetically unfavourable. Gaining 4 electrons (to form C4−) is also difficult because 6 protons cannot hold 10 electrons stably. Therefore, carbon completes its octet by sharing electrons — forming covalent bonds.
Example 2. Draw the electron-dot structure of ethane (C₂H₆).
Solution: Two carbons share one electron pair with each other (C–C single bond). Each carbon shares one electron pair with each of three hydrogens.
H3C–CH3, i.e., H:C(:H:H):C(:H:H):H. Total electrons shared = 7 pairs = 14 electrons.
Example 3. Why is diamond very hard while graphite is soft, although both are pure carbon?
Solution: In diamond, every carbon is bonded to four other carbons in a rigid 3D tetrahedral network — no layers can slide. In graphite, carbons are bonded to only three others in flat hexagonal sheets; the sheets are held together by weak van der Waals forces, so they slip easily over each other — making graphite soft and slippery.
Example 4. Why does graphite conduct electricity while diamond does not?
Solution: In graphite, each carbon forms 3 covalent bonds; the 4th valence electron is delocalised and free to move along the hexagonal sheet — hence graphite conducts. In diamond, all 4 valence electrons of every carbon are used up in forming 4 covalent bonds; there are no free electrons, so diamond is a non-conductor.
Example 5. Give the molecular formula of C₆₀ fullerene and explain its shape.
Solution: Molecular formula is C60. The 60 carbon atoms arrange themselves into 20 hexagons and 12 pentagons, forming a closed cage resembling a football (soccer ball). It was named Buckminsterfullerene after the architect who designed similar geodesic domes.

Competency-Based Questions

Priya's chemistry teacher shows her a pencil and a diamond ring and says, "Both are made of pure carbon, yet one marks paper and the other cuts glass." Priya wonders how this is possible.

Q1. L1 Remember What is the electronic configuration of carbon?

  • A. 2, 6
  • B. 2, 8, 4
  • C. 2, 4
  • D. 2, 2
Answer: C. 2, 4. Carbon has atomic number 6, so 2 electrons in K-shell and 4 in L-shell.

Q2. L2 Understand Explain the difference in hardness between diamond and graphite based on their structures. (2 marks)

Answer: In diamond each C is bonded to 4 others in a rigid 3D tetrahedral lattice (hard, rigid). In graphite, C atoms form hexagonal sheets held by weak forces, which slide over each other — making graphite soft and slippery.

Q3. L3 Apply Draw the electron-dot structure of ammonia (NH₃). How many lone pairs are on the nitrogen? (2 marks)

Answer: N shares one electron with each of three H atoms, forming 3 single N–H bonds. The two remaining valence electrons on N form one lone pair. Structure: H–N(lone pair)–H with a third H bonded to N.

Q4. L4 Analyse Why do covalent compounds generally have low melting and boiling points compared to ionic compounds? (3 marks)

Answer: Covalent compounds consist of discrete molecules held together by weak intermolecular forces (van der Waals). Only a small amount of energy is needed to overcome these weak forces, so melting and boiling points are low. Ionic compounds have strong electrostatic forces in a 3D lattice, which require much more energy to break — giving high melting/boiling points.

Q5. L5 Evaluate A student claims that carbon nanotubes and fullerenes are the same thing. Evaluate this claim. (3 marks)

Answer: The claim is incorrect. Both are nano-sized allotropes of carbon derived from graphene-like sheets, but they differ structurally. Fullerenes (e.g., C60) are closed cages of carbon atoms (hexagons + pentagons) resembling footballs. Carbon nanotubes are long hollow cylinders formed by rolling a graphene sheet. They have different shapes, sizes, and applications.

Assertion-Reason Questions

Assertion (A): Carbon forms covalent bonds rather than ionic bonds.

Reason (R): It is energetically unfavourable for carbon to lose or gain four electrons.

  • 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.
Answer: A. Both statements are true. Carbon cannot easily lose 4 e⁻ (too much energy) or gain 4 e⁻ (unstable highly charged anion), so it shares electrons — forming covalent bonds. R correctly explains A.

Assertion (A): Graphite conducts electricity.

Reason (R): In graphite, only 3 of the 4 valence electrons of each carbon are used in bonding; the fourth is free to move.

  • 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.
Answer: A. Both are true. The free, delocalised fourth electron per carbon is exactly what enables graphite's electrical conductivity.

Assertion (A): Fullerene is an allotrope of carbon.

Reason (R): Allotropes are different compounds of the same element.

  • 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.
Answer: C. A is true — C60 fullerene is an allotrope of carbon. R is false — allotropes are different forms of the same element, not different compounds.

Did You Know?

Frequently Asked Questions — Covalent Bonds & Versatile Carbon

What is covalent bonds & versatile carbon in Class 10 Science (CBSE board)?

Covalent Bonds & Versatile Carbon is a key topic in NCERT Class 10 Science Chapter 4 — Carbon and its Compounds. It explains covalent bonding in carbon, tetravalency, catenation and allotropes — the reason for carbon's vast compound family. Core ideas covered include covalent bond, tetravalency, catenation, allotropes. Mastering this subtopic is essential for scoring well in the CBSE Class 10 Science board exam because board papers repeatedly test these concepts through MCQs, short answers and long-answer questions. This part gives a complete, exam-ready explanation with activities, diagrams and competency-based practice aligned to NCERT.

Why is covalent bond important in NCERT Class 10 Science?

Covalent bond is important in NCERT Class 10 Science because it forms the foundation for understanding covalent bonds & versatile carbon in Chapter 4 — Carbon and its Compounds. Without a clear idea of covalent bond, students cannot answer higher-order CBSE board questions involving tetravalency, catenation, allotropes. Board papers regularly include 2-mark and 3-mark questions on this concept, and competency-based questions often link covalent bond to real-life situations. Building clarity here pays off directly in board marks.

How is covalent bonds & versatile carbon tested in the Class 10 Science CBSE board exam?

The CBSE Class 10 Science board exam tests covalent bonds & versatile carbon through a mix of 1-mark MCQs, 2-mark short answers, 3-mark explanations with examples, 5-mark descriptive questions (often with diagrams or balanced equations) and 4-mark competency-based questions. Expect direct questions on covalent bond, tetravalency, catenation and application-based questions drawn from NCERT activities. Students who follow NCERT thoroughly and practice this chapter's questions consistently score in the 90%+ range.

What are the key terms to remember for covalent bonds & versatile carbon in Class 10 Science?

The key terms to remember for covalent bonds & versatile carbon in NCERT Class 10 Science Chapter 4 are: covalent bond, tetravalency, catenation, allotropes, diamond, graphite. Each of these concepts carries exam weightage and regularly appears in the CBSE board paper. Write clear one-line definitions of every term in your revision notes and revisit them before the exam. Linking these terms visually through a flowchart or concept map makes recall easier during the Class 10 Science board exam.

Is Covalent Bonds & Versatile Carbon included in the Class 10 Science syllabus for 2025–26 CBSE board exam?

Yes, Covalent Bonds & Versatile Carbon is a part of the NCERT Class 10 Science syllabus (2025–26) prescribed by CBSE. It falls under Chapter 4 — Carbon and its Compounds — and is examined in the annual board paper. The current syllabus retains the full treatment of covalent bond, tetravalency, catenation as per the NCERT textbook. Because CBSE bases every board question on NCERT, studying this part thoroughly ensures complete syllabus coverage and guarantees marks from this chapter.

How should I prepare covalent bonds & versatile carbon for the CBSE Class 10 Science board exam?

Prepare covalent bonds & versatile carbon for the CBSE Class 10 Science board exam in three steps. First, read this NCERT part carefully, highlighting definitions and diagrams of covalent bond, tetravalency, catenation. Second, solve every in-text question and end-of-chapter exercise — CBSE questions often come directly from NCERT. Third, practice competency-based and assertion-reason questions to sharpen reasoning. Write answers in the exam-style format (point-wise with diagrams) and time yourself. This method delivers confidence and full marks in the board exam.

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