The replacement of hydrogen atom(s) in an aliphatic or aromatic hydrocarbon by halogen atom(s) results in the formation of alkyl halide (haloalkane) and aryl halide (haloarene) respectively. Haloalkanes contain halogen atom(s) attached to the sp3 hybridised carbon atom of an alkyl group, whereas haloarenes contain halogen atom(s) attached to sp2 hybridised carbon atom(s) of an aryl group.
-- NCERT Class 12 Chemistry, Ch. 6, p. 159C–X Bond
8 MCQs9-step worked example
Source: NCERT Organic Compounds Containing HalogensPYQ coverage: NEET 2021, 2023Official key: NTA-verifiedLast updated: 24 Sep 2026
C–X Bond, explained for NEET
The C–X bond gives you two trends running in opposite directions, and most lost marks here come from quoting the wrong one. Bond length increases from C–F to C–I. Bond enthalpy decreases over the same series. Iodine is the largest halogen, so C–I is the longest bond — and therefore the weakest, not the strongest. Size buys you length, not strength.
The reason is orbital overlap. Carbon bonds using a 2p orbital; the halogen contributes 2p (F), 3p (Cl), 4p (Br) or 5p (I). As the halogen's orbital grows more diffuse and sits further from its nucleus, overlap with carbon's compact 2p orbital gets poorer. Poorer overlap means a longer, weaker bond. NCERT Class 12 Chemistry Part 2, Chapter 6, page 163 tabulates this: C–F is both shortest and strongest.
The bond is also polar, because every halogen is more electronegative than carbon. The shared pair sits closer to X, leaving carbon with a partial positive charge — this is what makes the carbon electrophilic and sets up the nucleophilic substitution you meet in the next lesson.
Dipole moment does not follow electronegativity alone. Among the haloalkanes the order is CH₃Cl > CH₃F, even though fluorine is the more electronegative atom, because dipole moment is charge separation times bond length, and the longer C–Cl bond wins. Beyond chlorine the falling electronegativity dominates again, so bromine and iodine drop below fluorine: CH₃Cl (1.860 D) > CH₃F (1.847 D) > CH₃Br (1.830 D) > CH₃I (1.636 D) (page 164, Table 6.2).
Watch out for one more thing: haloalkanes have higher boiling points than alkanes of comparable mass, and this is a van der Waals and dipole–dipole effect in the bulk liquid — not a statement about C–X bond strength. A weak bond in a high-boiling liquid is not a contradiction.
Watch-out: when a stem says "strongest" or "most stable," reach for bond enthalpy. When it says "longest," reach for atomic size. They point opposite ways.
Can you answer these C–X Bond MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Which of the following correctly states the trend in C–X bond length in the haloalkanes CH₃F, CH₃Cl, CH₃Br and CH₃I?
Show answer and why every option is right or wrong
Answer: D. Bond length increases with the size of the halogen atom, so C–I is longest and C–F shortest. NCERT Class 12 Chemistry Part 2, Chapter 6, page 163 tabulates the values.
Why A is wrong: A is wrong because it reverses the trend — it is the bond enthalpy order, not the length order. Confusing the two directions is the highest-cost error on this topic.
Why B is wrong: B is wrong because the four bond lengths differ substantially, from about 139 pm for C–F to about 214 pm for C–I.
Why C is wrong: C is wrong because bond length varies monotonically with halogen size; there is no zig-zag in this series.
MCQ 2Easy RecallPractice
Why is the C–X bond in a haloalkane polar?
Show answer and why every option is right or wrong
Answer: C. Every halogen is more electronegative than carbon, so the shared pair lies closer to X, giving X a partial negative and carbon a partial positive charge. NCERT Class 12 Chemistry Part 2, Chapter 6, page 159.
Why A is wrong: A is wrong because lone-pair repulsion affects bond angles, not the direction of bond polarity. Polarity comes from the electronegativity difference.
Why B is wrong: B is wrong because hybridisation describes the geometry at carbon; an sp³ carbon bonded to four identical atoms would give non-polar bonds.
Why D is wrong: D is wrong because the C–X single bond in a haloalkane has no double-bond character; polarity does not require it.
MCQ 3Concept TrapPractice
A student claims that C–I is the strongest of the four C–X bonds because iodine is the largest halogen and therefore forms the most extensive overlap with carbon. Identify the flaw.
Show answer and why every option is right or wrong
Answer: A. Larger size means a more diffuse valence orbital sitting further from the nucleus, which overlaps carbon's 2p orbital poorly. The result is the longest and weakest bond in the series. NCERT Class 12 Chemistry Part 2, Chapter 6, page 163.
Why B is wrong: B is wrong because bond enthalpy falls from C–F to C–I; C–I is the weakest, not the strongest. This is the specific reversal the stem is testing.
Why C is wrong: C is wrong because iodine is the largest of the four halogens named — the premise about size is correct; only the inference about strength is wrong.
Why D is wrong: D is wrong because it discards the real cause. Orbital size and overlap quality drive the enthalpy trend here, and dismissing size leaves the trend unexplained.
MCQ 4Direct ApplicationPractice
Arrange CH₃F, CH₃Cl, CH₃Br and CH₃I in decreasing order of C–X bond enthalpy.
Show answer and why every option is right or wrong
Answer: C. Bond enthalpy falls as the halogen gets larger and overlap with carbon worsens, so C–F is strongest and C–I weakest. NCERT Class 12 Chemistry Part 2, Chapter 6, page 163.
Why A is wrong: A is wrong because it is the bond-length order applied to enthalpy. The two trends run in opposite directions.
Why B is wrong: B is wrong because it borrows the dipole-moment ordering, where CH₃Cl does lead CH₃F. Dipole moment and bond enthalpy are different quantities.
Why D is wrong: D is wrong because the enthalpy trend is monotonic in halogen size; no arrangement that interleaves F and Cl below Br can be correct.
MCQ 5Direct ApplicationPractice
Which haloalkane has the largest dipole moment?
Show answer and why every option is right or wrong
Answer: A. Dipole moment is the product of charge separation and bond length, so the longer C–Cl bond outweighs fluorine's higher electronegativity: CH₃Cl (1.860 D) > CH₃F (1.847 D) > CH₃Br (1.830 D) > CH₃I (1.636 D). NCERT Class 12 Chemistry Part 2, Chapter 6, page 164, Table 6.2.
Why B is wrong: B is wrong because reasoning from electronegativity alone puts CH₃F first, but the short C–F bond gives a smaller charge-separation distance, leaving it second (1.847 D), just behind CH₃Cl (1.860 D).
Why C is wrong: C is wrong because CH₃Br sits third (1.830 D), behind CH₃Cl and CH₃F; its longer bond does not compensate for bromine's lower electronegativity relative to chlorine.
Why D is wrong: D is wrong because CH₃I has both the lowest electronegativity difference in the series and, despite the longest bond, the smallest dipole moment.
MCQ 6Concept TrapPractice
In a haloalkane, which atom of the C–X bond bears the partial positive charge, and what does this imply about the site of attack in a substitution reaction?
Show answer and why every option is right or wrong
Answer: B. The more electronegative halogen pulls electron density away, leaving carbon electron-deficient (δ+) and hence open to nucleophilic attack. NCERT Class 12 Chemistry Part 2, Chapter 6, page 159; the mechanisms themselves are covered in the substitution lesson.
Why A is wrong: A is wrong because the halogen is the more electronegative partner and so carries δ−, not δ+.
Why C is wrong: C is wrong because it inverts the polarity. Carbon is the less electronegative atom in every C–X bond and therefore carries δ+.
Why D is wrong: D is wrong because the electronegativity difference guarantees a permanent dipole; the C–X bond is polar in every haloalkane.
MCQ 7CalculationPractice
Two statements are made about CH₃I. Statement I: the C–I bond is the longest of the four C–X bonds. Statement II: the C–I bond is therefore the most difficult to break. Evaluate both.
Show answer and why every option is right or wrong
Answer: B. Statement I is correct — iodine's size makes C–I longest. Statement II is incorrect: a longer bond arising from poorer overlap is easier to break, so C–I has the lowest bond enthalpy in the series. NCERT Class 12 Chemistry Part 2, Chapter 6, page 163.
Why A is wrong: A is wrong because the inference in Statement II reverses the enthalpy trend. Length and strength move in opposite directions across this series.
Why C is wrong: C is wrong on both counts: Statement I is in fact correct, and Statement II is in fact incorrect.
Why D is wrong: D is wrong because Statement I is a correct reading of the bond-length trend; only Statement II fails.
MCQ 8CalculationPractice
A haloalkane CH₃X has both a lower C–X bond enthalpy than CH₃Br and a smaller dipole moment than CH₃F. Which halogen is X, and which single property of X accounts for both observations?
Show answer and why every option is right or wrong
Answer: C. Only iodine lies below bromine in bond enthalpy and below fluorine in dipole moment. Its large, diffuse 5p orbital explains the weak bond, and its low electronegativity explains the small dipole. NCERT Class 12 Chemistry Part 2, Chapter 6, page 163.
Why A is wrong: A is wrong because C–F has the highest bond enthalpy in the series, so it cannot fall below C–Br.
Why B is wrong: B is wrong because CH₃Cl has the largest dipole moment of the four and so cannot be smaller than CH₃F's.
Why D is wrong: D is wrong because iodine satisfies both conditions; it is last in bond enthalpy and last in dipole moment.
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How do you solve a C–X Bond question? A worked example
- 1
Given
Four haloalkanes: CH₃F, CH₃Cl, CH₃Br, CH₃I. C–X bond lengths from NCERT Class 12 Chemistry Part 2, Chapter 6, page 163: C–F = 1.39 × 10² pm, C–Cl = 1.78 × 10² pm, C–Br = 1.93 × 10² pm, C–I = 2.14 × 10² pm.
- 2
Required
The decreasing order of C–X bond enthalpy, and the identity of the haloalkane whose C–X bond breaks most readily.
- 3
Concept
Bond enthalpy measures the energy needed to homolytically cleave one mole of the bond. It tracks the quality of orbital overlap, not the size of the atoms. As the halogen's valence p orbital grows larger and more diffuse (2p → 3p → 4p → 5p), its overlap with carbon's compact 2p orbital worsens. Poorer overlap gives a longer bond and a smaller bond enthalpy.
- 4
Formula
No algebraic formula applies. The governing relationship is qualitative and inverse:
increasing halogen size → increasing C–X bond length → decreasing C–X bond enthalpy - 5
Substitution
Rank the halogens by atomic radius: I > Br > Cl > F. This reproduces the bond-length order already given in the data: C–I (2.14 × 10² pm) > C–Br (1.93 × 10² pm) > C–Cl (1.78 × 10² pm) > C–F (1.39 × 10² pm).
- 6
Calculation
Invert the length order to obtain the enthalpy order:
C–F > C–Cl > C–Br > C–I
Note on exactness: nothing here is computed arithmetically, so no rounding arises. The bond lengths are written in scientific notation because "139 pm" and "214 pm" carry ambiguous trailing digits when quoted as bare integers; expressing them as 1.39 × 10² pm and 2.14 × 10² pm fixes them at three significant figures, matching the precision NCERT tabulates. The halogen count of four in this series is a counting integer and is exact. - 7
Final answer
Decreasing C–X bond enthalpy: C–F > C–Cl > C–Br > C–I. The bond that breaks most readily is C–I in CH₃I, the longest and weakest of the four.
- 8
Common trap
The dossier flags two negative-marking traps for exactly this question shape. First, quoting the bond-length trend when the stem asks for bond enthalpy — the two orders are exact reverses, so the error costs the full mark rather than producing a near-miss. Second, assuming C–I is strongest because iodine is the largest atom. Size increases length; it does not increase strength. A one-second habit fixes both: underline the word "length" or "strength" in the stem before you look at the options.
- 9
Similar NEET-style question
Among CH₃F, CH₃Cl, CH₃Br and CH₃I, identify the compound with the shortest C–X bond and state whether it also has the largest dipole moment. Justify in one line each. *(Answer: CH₃F has the shortest C–X bond, but CH₃Cl — not CH₃F — has the largest dipole moment, because dipole moment is charge separation times bond length and the longer C–Cl bond outweighs fluorine's greater electronegativity.)*
What to remember before solving C–X Bond questions
Halogen atoms are more electronegative than carbon, therefore, carbon-halogen bond of alkyl halide is polarised; the carbon atom bears a partial positive charge whereas the halogen atom bears a partial negative charge.
-- NCERT Class 12 Chemistry, Ch. 6, p. 163As we go down the group in the periodic table, the size of halogen atom increases. Fluorine atom is the smallest and iodine atom is the largest. Consequently the carbon-halogen bond length also increases from C-F to C-I.
-- NCERT Class 12 Chemistry, Ch. 6, p. 163Alkyl halides are colourless when pure. However, bromides and iodides develop colour when exposed to light. Many volatile halogen compounds have sweet smell.
-- NCERT Class 12 Chemistry, Ch. 6, p. 167The boiling points of organohalogen compounds are comparatively higher than the corresponding hydrocarbons because of strong dipole-dipole and van der Waals forces of attraction. These are slightly soluble in water but completely soluble in organic solvents.
-- NCERT Class 12 Chemistry, Ch. 6, p. 189C–X Bond questions from past NEET papers
2 questions from NEET 2021, 2023. Answers verified against NTA official keys.
NEET 2023
The given compound is an example of ______.
1Aryl halide
2Allylic halide
3Vinylic halide
4Benzylic halide
NTA Answer: Option 2(final)
NEET 2021
The correct sequence of bond enthalpy of 'C—X' bond is :
1CH₃–Cl > CH₃–F > CH₃–Br > CH₃–I
2CH₃–F < CH₃–Cl < CH₃–Br < CH₃–I
3CH₃–F > CH₃–Cl > CH₃–Br > CH₃–I
4CH₃–F < CH₃–Cl > CH₃–Br > CH₃–I
NTA Answer: Option 3(final)
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Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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