Q.Classify the following amines as primary, secondary or tertiary:
Concept understanding — Basicity Order Amines
Basicity of Amines: From Intuition to Precision
Imagine you have a nitrogen atom with a lone pair of electrons — that pair is like a key that can grab a proton (H+). The more willing the nitrogen is to share that pair, the stronger the base. That's the core idea.
The Intuition: What Makes a Nitrogen "Want" a Proton?
Amines are organic derivatives of ammonia (NH3), where one or more hydrogen atoms are replaced by alkyl groups (R). Alkyl groups are electron-donating — they push electron density toward the nitrogen. More alkyl groups = more electron density on nitrogen = stronger base.
So you'd expect: tertiary > secondary > primary > ammonia. That's the inductive effect argument.
But reality is more interesting. In water (the usual solvent for basicity measurements), the order is:
Secondary > Primary > Tertiary > Ammonia
Why the reversal for tertiary amines? Because basicity isn't just about the free amine — it's about the stability of the conjugate acid (the ammonium ion RNH3+) once the proton is grabbed.
The Precise Explanation: Three Factors at Play
1. Inductive Effect (pushes electron density)
Alkyl groups donate electrons through sigma bonds. More alkyl groups = more electron density on N = stronger base. This favours: tertiary > secondary > primary > ammonia.
2. Solvation Effect (stabilises the conjugate acid)
Once the amine grabs a proton, the resulting ammonium ion is positively charged. Water molecules surround it, stabilising the charge through hydrogen bonding. The more hydrogens on the nitrogen, the more H-bonds can form.
Primary ammonium (RNH3+) has 3 H's → best solvation.
Secondary (R2NH2+) has 2 H's → good solvation.
Tertiary (R3NH+) has 1 H → poor solvation.
This favours: primary > secondary > tertiary.
3. Steric Hindrance (blocks solvation)
Bulkier alkyl groups physically block water molecules from approaching the charged nitrogen. This further reduces solvation in tertiary amines.
The Net Result: The Actual Order in Water
| Amine Type | Inductive Effect | Solvation | Steric Hindrance | Net Basicity (pKb) |
|---|---|---|---|---|
| Ammonia (NH3) | Weakest | Best (3 H's) | None | 4.75 |
| Primary (RNH2) | Moderate | Good (2 H's) | Minimal | ~3.4 |
| Secondary (R2NH) | Strong | Moderate (1 H) | Some | ~3.2 |
| Tertiary (R3N) | Strongest | Poor (0 H's) | Significant | ~4.2 |
pKb is the negative log of the base dissociation constant. Lower pKb = stronger base. So secondary (pKb ≈ 3.2) is the strongest, then primary (≈3.4), then tertiary (≈4.2), then ammonia (4.75).
The Final Answer
The basicity order of amines in aqueous solution is:
Secondary > Primary > Tertiary > Ammonia
This is the standard order for simple alkyl amines (methyl, ethyl, etc.). The inductive effect dominates from ammonia to secondary, but the loss of solvation and steric hindrance in tertiary amines drops them below primary.
A Quick Check: Why Not Tertiary?
If you only considered electron donation, tertiary would win. But the ammonium ion R3NH+ has only one hydrogen to hydrogen-bond with water, and the three bulky alkyl groups physically block water molecules. The conjugate acid is poorly stabilised, so the equilibrium shifts back toward the free amine — making it a weaker base than expected.
This order applies to aliphatic amines in water. Aromatic amines (like aniline) are much weaker bases because the lone pair is delocalised into the benzene ring. Also, in the gas phase (no solvent), the order reverts to tertiary > secondary > primary > ammonia — confirming that solvation is the key reason for the reversal.
The Takeaway
Basicity is a tug-of-war between:
- Inductive effect (wants tertiary to win)
- Solvation + sterics (wants primary to win)
Secondary amines hit the sweet spot — strong inductive donation and decent solvation — making them the strongest bases in water.
The basicity order of amines in aqueous solution is one of the most frequently asked comparison-type questions in the NCERT Class 12 Chemistry chapter on amines, regularly appearing in CBSE boards, JEE Main and NEET. Anyone searching "basicity order of amines class 12 chemistry" or trying to understand why secondary amines outrank primary and tertiary amines will find the inductive-versus-solvation trade-off above is the standard NCERT-aligned explanation.
Why this formula?
Basicity Order of Amines: Why It Holds
Let's build this from first principles — understanding why amines have different basicities is essential for exams and for deeper chemistry.
1. What Does "Basicity" Mean Here?
Basicity of an amine is its ability to accept a proton (H+).
The stronger the base, the more readily it grabs H+.
The equilibrium is:
R3N+H2O⇌R3NH++OH−
The base dissociation constant Kb measures this:
Kb=[R3N][R3NH+][OH−]
A larger Kb means a stronger base.
2. The Key Factor: Electron Density on Nitrogen
The lone pair on nitrogen is what accepts H+.
More electron density on nitrogen → stronger base.
Why? Because a proton (H+) is attracted to negative charge. If the nitrogen's lone pair is more "available" (less tightly held), it binds H+ more easily.
3. The Two Competing Effects
✓ Inductive Effect (Electron-Donating)
Alkyl groups (−CH3, −C2H5, etc.) are electron-donating via the inductive effect.
They push electron density toward nitrogen.
- More alkyl groups → more electron density on N → stronger base.
This suggests:
3∘>2∘>1∘>NH3
✗ Solvation Effect (Hydration of the Conjugate Acid)
When the amine accepts H+, it forms R3NH+ (a positively charged ion).
This ion is stabilized by water molecules (hydration).
- More H atoms on the NH+ group → more hydrogen bonding with water → better stabilization of the conjugate acid.
- Better stabilization of R3NH+ → stronger base (because the equilibrium shifts toward protonation).
This suggests:
1∘>2∘>3∘ (since 1∘ has 3 H's, 2∘ has 2, 3∘ has 1)
4. The Actual Order (Aqueous Solution)
In water, the observed order for aliphatic amines is:
2° > 1° > 3° > NH3
Wait — that's not simply "more alkyl = stronger". Why?
The Reasoning (Step by Step)
-
From NH3 to 1∘: Adding one alkyl group increases electron density (inductive effect) → stronger base.
So 1∘>NH3.
-
From 1∘ to 2∘: Adding a second alkyl group further increases electron density → even stronger base.
So 2∘>1∘.
-
From 2∘ to 3∘: Here, the solvation effect becomes dominant.
The 3∘ amine has only one H on the NH+ group → poor hydration of the conjugate acid.
The inductive effect is still present, but the loss of solvation outweighs the gain in electron density.
So 3∘ is weaker than 2∘.
Thus, the net order is:
2° > 1° > 3° > NH3
5. The Key Formula(e) to Remember
For gas phase (no solvent):
Only inductive effect matters:
3° > 2° > 1° > NH3
For aqueous solution (common in exams):
Both effects matter — solvation dominates for 3∘:
2° > 1° > 3° > NH3
For aromatic amines (e.g., aniline):
The lone pair is delocalized into the benzene ring → much weaker base.
Aliphatic amines > Aromatic amines
6. Quick Exam Tip
If a question asks "basicity order of amines in water", always write:
2° > 1° > 3° > NH3
And explain:
- Inductive effect increases from NH3 to 3∘
- But solvation of the conjugate acid decreases from 1∘ to 3∘
- The balance gives the above order.
7. Summary Table
| Amine Type | Inductive Effect | Solvation of R3NH+ | Net Basicity (aq) |
|---|---|---|---|
| NH3 | Weakest | Best (3 H's) | Weakest |
| 1∘ | Moderate | Good (2 H's) | Moderate |
| 2∘ | Strong | Moderate (1 H) | Strongest |
| 3∘ | Strongest | Poor (0 H's) | Weaker than 2∘ |
Final takeaway:
Basicity is not just about "more alkyl = stronger". The solvation of the conjugate acid is the deciding factor in water. Always reason from both effects.
Count how many carbon atoms are attached to the nitrogen: one C = primary, two C = secondary, three C = tertiary.
- naphthalen-1-amine: N bonded to one aryl carbon (plus two H) → primary.
- N,N-dimethylnaphthalen-1-amine: N bonded to one aryl C and two CH3 carbons → tertiary.
- (C2H5)2CHNH2: N bonded to the single CH carbon (plus two H) → primary.
- (C2H5)2NH: N bonded to two ethyl carbons (plus one H) → secondary.
✓Final answer
- primary,
- tertiary,
- primary,
- secondary.
An amine is classed by how many carbon atoms are joined directly to its nitrogen: one → primary (1°), two → secondary (2°), three → tertiary (3°). The number of hydrogens on nitrogen (2, 1, 0) mirrors this. Applying the rule gives (i) 1°,
(ii) 3°,
(iii) 1°,
(iv) 2°.
Concept
Amines are viewed as derivatives of ammonia, NH3, in which one, two or three of the N–H bonds are replaced by N–C bonds. The classification depends only on how many carbon atoms are bonded to nitrogen, not on whether those carbons are part of a ring or a chain, and not on the size of the group:
- Primary (1°): R–NH2 (one C on N)
- Secondary (2°): R2NH (two C on N)
- Tertiary (3°): R3N (three C on N)
Applying the rule to each compound
- naphthalen-1-amine, C10H7–NH2. The nitrogen is joined to a single carbon — the C-1 of the naphthalene ring — and still carries two hydrogens (–NH2). One carbon on nitrogen ⇒ primary amine (an aromatic/aryl amine).
- N,N-dimethylnaphthalen-1-amine, C10H7–N(CH3)2. Here the nitrogen is bonded to the aryl carbon of the naphthalene ring and to the two carbons of the two methyl groups. Three carbons on nitrogen, no N–H left ⇒ tertiary amine.
- (C2H5)2CHNH2 (pentan-3-amine). The –NH2 is attached to the central methine carbon (the CH), which itself bears two ethyl groups. Nitrogen sees only that one carbon and keeps two hydrogens; the ethyl groups are further out on the carbon skeleton, not on nitrogen ⇒ primary amine. (This part tests whether you count carbons on N, not carbons in the molecule.)
- (C2H5)2NH (diethylamine). Nitrogen is bonded to the two carbons of two ethyl groups and retains one hydrogen ⇒ secondary amine.
✓Final answer
- naphthalen-1-amine — primary (1°);
- N,N-dimethylnaphthalen-1-amine — tertiary (3°);
- (C2H5)2CHNH2 (pentan-3-amine) — primary (1°);
- (C2H5)2NH (diethylamine) — secondary (2°).
Method: Classifying Amines (Primary/Secondary/Tertiary) by Counting C-N Bonds
Core Concept
An amine's class depends ONLY on how many carbon atoms are bonded directly to the nitrogen atom - one carbon on N gives a primary amine, two gives secondary, three gives tertiary - regardless of whether those carbons belong to a ring, a chain, or how large the attached groups are.
Steps
- Locate the nitrogen atom in the given structure.
- Count only the bonds going from N directly to a carbon atom (ignore N-H bonds and ignore carbons further away in the molecule that are not bonded to N itself).
- Map the count to a class: 1 carbon on N gives primary (1 degree); 2 carbons give secondary (2 degree); 3 carbons give tertiary (3 degree).
- Repeat independently for every compound in the set - classification never depends on comparing compounds to each other.
Applying the Method to Each Sub-Part
- naphthalen-1-amine, C10H7-NH2: N is bonded to exactly one carbon (the aryl C-1 of naphthalene) and keeps two H -> primary.
- N,N-dimethylnaphthalen-1-amine, C10H7-N(CH3)2: N is bonded to the aryl C-1 AND to the two methyl carbons - three C-N bonds, no N-H left -> tertiary.
- (C2H5)2CHNH2: the -NH2 is bonded only to the single central CH carbon; the two ethyl groups are attached to THAT carbon, not to nitrogen, so N still sees only one carbon -> primary (the trap here is counting the ethyl carbons as if they were on N).
- (C2H5)2NH: N is bonded to the two ethyl carbons and keeps one H -> secondary.
Key Exam Point
Sub-part (iii) is specifically designed to catch students who classify by "how many carbons are in the molecule" instead of "how many carbons are bonded to N" - always trace only the bonds directly touching the nitrogen atom.
Here are the common mistakes students make when classifying amines from drawn structures and condensed formulas, along with how to avoid each.
Mistake 1: Counting all the carbons in the molecule instead of the carbons bonded to nitrogen
Students see a big structure (like the naphthalene ring in part (i)) or a heavily branched formula and assume "many carbons = higher class."
- The Error: Calling naphthalen-1-amine "tertiary" because the ring has ten carbons.
- How to Avoid: The classification rule looks at one atom only — the nitrogen. Count the carbon atoms bonded directly to N:
- one C on N → primary (1°)
- two C on N → secondary (2°)
- three C on N → tertiary (3°)
- Example: In part (i), the nitrogen is bonded to a single ring carbon (C-1 of naphthalene) and two hydrogens → primary, no matter how large the ring system is.
Mistake 2: Classifying by the carbon skeleton next to the amine carbon (the "alcohol/haloalkane habit")
Students carry over the alkyl-halide/alcohol convention — where 1°/2°/3° describes the carbon bearing the functional group — and apply it to the amine.
- The Error: Calling (C2H5)2CHNH2 (part (iii)) a "secondary amine" because the CH carbon bearing the NH2 carries two ethyl groups (it is a secondary carbon).
- How to Avoid: For amines, the degree is a property of the nitrogen, not of the carbon it sits on. In (C2H5)2CHNH2, nitrogen sees only one carbon (the CH) and keeps two hydrogens → primary amine. The two ethyl groups are further out on the skeleton, not on nitrogen.
Mistake 3: Misreading −N(CH3)2 as "two substituents, so secondary"
- The Error: For part (ii), counting the two methyl groups of −N(CH3)2 and stopping there → "secondary."
- How to Avoid: Count every C–N bond, including the bond to the ring/parent chain. In N,N-dimethylnaphthalen-1-amine the nitrogen is bonded to the aryl carbon and to two methyl carbons — three C–N bonds, no N–H left → tertiary (3°). The "N,N-" prefix in a name is itself a signal that nitrogen carries two extra groups besides the parent.
Mistake 4: Thinking aryl amines classify differently from alkyl amines
- The Error: Treating a ring carbon on N as "not counting" (or counting it differently) because it is aromatic.
- How to Avoid: An aryl carbon bonded to nitrogen counts exactly like an alkyl carbon. Aniline (C6H5NH2) and naphthalen-1-amine are both primary amines — one C on N, two H on N — just aromatic ones.
Mistake 5: Not using the N–H count as a cross-check
- The Error: Deciding the class from the drawing alone and never verifying.
- How to Avoid: The hydrogens on nitrogen mirror the classification: 2 H → 1°, 1 H → 2°, 0 H → 3°. In part (iv), (C2H5)2NH has exactly one N–H → secondary, consistent with its two C–N bonds. If your carbon count and hydrogen count disagree, you have misread the structure — recount.
Quick Reference Table
| Part | Compound | C atoms on N | H atoms on N | Class |
|---|---|---|---|---|
| (i) | naphthalen-1-amine | 1 (aryl C) | 2 | Primary (1°) |
| (ii) | N,N-dimethylnaphthalen-1-amine | 3 (aryl C + 2 CH₃) | 0 | Tertiary (3°) |
| (iii) | (C2H5)2CHNH2 | 1 (the CH carbon) | 2 | Primary (1°) |
| (iv) | (C2H5)2NH | 2 (two ethyl C) | 1 | Secondary (2°) |
Final tip: Circle the nitrogen atom and draw only its four bonds before classifying. Everything outside those bonds — ring size, branching, chain length — is irrelevant to whether the amine is 1°, 2° or 3°.
Showing the 12 most recent of 27 on this concept.
- TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQQ.What is the amino acid which does not contain −NH2 group in its structure? (A) Tyrosine (B) Alanine (C) Proline (D) Methionine
›Reveal solutionSolution
The key idea is that proline is an imino acid, not a standard amino acid — its side chain loops back to form a secondary amine, so it lacks a free −NH2 group. The correct option is (C).
Amino acids are the building blocks of proteins. The standard structure of an α-amino acid has a central carbon (α-carbon) bonded to four groups: an amino group (−NH2), a carboxyl group (−COOH), a hydrogen atom, and a variable side chain (R). That −NH2 group is what makes it an "amino" acid.
But not every molecule classified as an amino acid follows this exact pattern. Proline is the classic exception. Its side chain is not a simple alkyl group — instead, it forms a five-membered ring that connects back to the nitrogen of the amino group. This turns the primary amine (−NH2) into a secondary amine (>NH), which is part of a ring. Because of this, proline is technically an imino acid, not a true amino acid. It still has a carboxyl group and an α-carbon, but the free −NH2 group is absent.
Let's check each option:
-
Tyrosine — has a standard −NH2 group attached to the α-carbon. Its side chain is a phenolic ring. No structural modification to the amino group. So it contains −NH2.
-
Alanine — the simplest of the four here, with a methyl group as the side chain. The α-carbon carries a normal −NH2 group. So it contains −NH2.
-
Proline — as described, the side chain is a three-carbon chain that loops back and bonds to the nitrogen atom, forming a pyrrolidine ring. The nitrogen now has only one hydrogen (instead of two), and is part of the ring. There is no free −NH2 group. This is the correct answer.
-
Methionine — has a standard −NH2 group. Its side chain contains a sulfur atom in a thioether linkage. No effect on the amino group. So it contains −NH2.
Watch outA common mistake is to think that all 20 standard amino acids have a free −NH2 group. Proline is the only one that does not — its nitrogen is part of a ring, making it a secondary amine. In exam questions, "amino acid" is often used loosely to include proline, but the defining feature here is the absence of −NH2.
TipIf you ever forget, remember the mnemonic: Proline is the "pro" at making rings — its side chain curls back to tie up the amino group. No free −NH2 means no primary amine.
✓Final answerThe correct option is (C) Proline.
-
- TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQQ.Which of the following compounds will be suitable for estimation of nitrogen by Kjeldahl’s method? (A) I & V only (B) I, II, III only (C) II & V only (D) III & IV only
›Reveal solutionSolution
Kjeldahl's method estimates nitrogen by converting it to ammonium sulfate, then ammonia. It is suitable for nitrogen in amines, amides, and amino acids, but not for nitrogen in nitro, azo, or heterocyclic ring compounds. Based on common examples, compounds like aniline (an amine) and urea (an amide) are suitable.
Concept and Intuition
Kjeldahl's method is a quantitative analytical technique used to determine the nitrogen content in organic compounds. The core idea is to convert all the nitrogen present in the sample into a measurable form, specifically ammonia, which can then be quantified.
The method relies on a crucial chemical transformation:
- Digestion: The organic compound is heated with concentrated sulfuric acid in the presence of a catalyst (like CuSO4) and potassium sulfate (K2SO4). This process oxidizes the organic matter and converts the nitrogen present into ammonium sulfate, (NH4)2SO4.
Organic N-compound+conc. H2SO4catalyst, heat(NH4)2SO4+CO2+H2O
- Distillation: The ammonium sulfate solution is then treated with an excess of a strong base, typically sodium hydroxide (NaOH). This liberates ammonia gas (NH3).
(NH4)2SO4+2NaOH⟶Na2SO4+2NH3+2H2O
- Titration: The liberated ammonia gas is absorbed in a known excess volume of a standard acid (e.g., H2SO4 or HCl). The unreacted acid is then back-titrated with a standard base to determine the amount of ammonia absorbed, and consequently, the amount of nitrogen in the original sample.
The critical limitation of Kjeldahl's method is that not all forms of nitrogen can be quantitatively converted to ammonium sulfate under the digestion conditions. For the method to be effective, the nitrogen must be in a reduced state or a state that can be easily reduced to ammonia.
ImportantCompounds suitable for Kjeldahl's method:
- Amines (primary, secondary, tertiary)
- Amides
- Amino acids and proteins
- Ammonium salts
Compounds NOT suitable for Kjeldahl's method:
- Nitro compounds (e.g., nitrobenzene, −NO2)
- Azo compounds (e.g., azobenzene, −N=N−)
- Nitrogen in heterocyclic rings (e.g., pyridine, quinoline)
- Diazo compounds
- Nitriles (some sources list them as unsuitable, or requiring modified conditions)
The reason these unsuitable compounds cannot be estimated is that their nitrogen atoms are either in a highly oxidized state (like nitro compounds) or are part of very stable structures (like heterocyclic rings or azo groups) that do not readily convert to ammonium sulfate under the standard Kjeldahl digestion conditions.
Let's analyze the given compounds based on these principles. Since the specific structures for I, II, III, IV, V are not provided, we will consider common examples that represent the types of compounds typically tested in such questions.
Step-by-step Analysis
We will assume the compounds are:
- I: Aniline (C6H5NH2)
- II: Nitrobenzene (C6H5NO2)
- III: Pyridine (C5H5N)
- IV: Azobenzene (C6H5−N=N−C6H5)
- V: Urea (H2N−CO−NH2)
-
Compound I: Aniline
- Aniline is a primary aromatic amine. The nitrogen atom is directly bonded to a carbon atom and two hydrogen atoms.
- This type of nitrogen is readily converted to ammonium sulfate upon digestion with concentrated sulfuric acid.
- Therefore, aniline is suitable for Kjeldahl's method.
-
Compound II: Nitrobenzene
- Nitrobenzene contains a nitro group (−NO2). The nitrogen in a nitro group is in a highly oxidized state.
- Under the standard Kjeldahl digestion conditions, this nitrogen is not quantitatively reduced to ammonia.
- Therefore, nitrobenzene is not suitable for Kjeldahl's method.
-
Compound III: Pyridine
- Pyridine is a heterocyclic compound where the nitrogen atom is part of an aromatic ring system.
- The nitrogen in such a stable ring structure is not quantitatively converted to ammonium sulfate during the digestion process.
- Therefore, pyridine is not suitable for Kjeldahl's method.
-
Compound IV: Azobenzene
- Azobenzene contains an azo group (−N=N−). In azo compounds, the nitrogen atoms are directly bonded to each other.
- These nitrogen-nitrogen bonds are not easily broken, and the nitrogen is not quantitatively converted to ammonium sulfate under Kjeldahl conditions.
- Therefore, azobenzene is not suitable for Kjeldahl's method.
-
Compound V: Urea
- Urea is an amide. It contains nitrogen atoms in amide groups (−CONH2).
- Nitrogen in amide groups is readily converted to ammonium sulfate upon digestion with concentrated sulfuric acid.
- Therefore, urea is suitable for Kjeldahl's method.
Based on this analysis, compounds I (Aniline) and V (Urea) are suitable for estimation of nitrogen by Kjeldahl's method.
✓Final answerBased on the common examples representing the types of compounds, compounds I (Aniline) and V (Urea) are suitable for estimation of nitrogen by Kjeldahl’s method. The correct option is (A).
- TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQQ.The atomic numbers of elements A, B, D, E respectively are 7, 17, 13, 11. The element which forms basic oxide among them is (A) E (B) B (C) D (D) A
›Reveal solutionSolution
The basic character of an oxide depends on the metallic character of the element. Among the given elements, sodium (atomic number 11) is the most metallic and forms a strongly basic oxide. The correct option is (A) E.
The key idea here is that oxides of metals are generally basic, while oxides of non-metals are acidic. So to find which element forms a basic oxide, we need to identify which one is a metal — and more specifically, which one is the most metallic.
Let’s decode the atomic numbers given:
- A: atomic number 7 → Nitrogen (N) — a non-metal
- B: atomic number 17 → Chlorine (Cl) — a non-metal
- D: atomic number 13 → Aluminium (Al) — a metal, but with some amphoteric character
- E: atomic number 11 → Sodium (Na) — a highly reactive metal
Now, let’s work through the reasoning step by step.
-
Metallic character and basicity of oxides
Metallic elements tend to lose electrons and form positive ions. Their oxides react with water to give hydroxides, which release OH⁻ ions — that’s what makes them basic. Non-metals, on the other hand, form covalent oxides that are typically acidic (they release H⁺ in water).
-
Identify the metals among the given elements
From the atomic numbers:
- 11 (Na) and 13 (Al) are metals.
- 7 (N) and 17 (Cl) are non-metals. So only D and E are candidates for forming basic oxides.
-
Compare the basic nature of Na₂O and Al₂O₃
Sodium oxide (Na₂O) dissolves in water to form sodium hydroxide (NaOH), a strong base. Aluminium oxide (Al₂O₃) is amphoteric — it can react with both acids and bases, but its basic character is much weaker than that of a typical metal oxide. In fact, Al₂O₃ behaves more like an acidic oxide when reacting with strong bases.
-
Which one is clearly basic?
Sodium (E) forms a strongly basic oxide. Aluminium (D) forms an amphoteric oxide, not purely basic. So among the options, E is the correct choice.
Watch outA common mistake is to think that all metal oxides are strongly basic. Aluminium oxide is a metal oxide but is amphoteric — it can act as both an acid and a base. So don’t pick D just because it’s a metal.
TipIn periodic table trends, basic character of oxides increases down a group and decreases across a period. Sodium (Group 1) is far more metallic than aluminium (Group 13), so its oxide is more basic.
✓Final answerThe element that forms a basic oxide is E (sodium), so the correct option is (A) E.
- TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQQ.Identify the correct statements from the following I. Conjugate base of chloric acid is ClO− II. AlCl3 is a Lewis acid III. Conjugate base of NH3 is NH2− (A) I, II, III (B) I, II only (C) II, III only (D) I, III only
›Reveal solutionSolution
Conjugate acid-base pairs differ by one proton; Lewis acids accept electron pairs. Chloric acid is HClOX3, so its conjugate base is ClOX3X−, not ClOX−. AlClX3 is electron-deficient and accepts a lone pair, making it a Lewis acid. NHX3 loses a proton to give NHX2X−, so that pair is correct. Only statements II and III are true.
The core of this question is understanding two distinct definitions: the Brønsted-Lowry concept of conjugate acid-base pairs (where a conjugate base is what remains after the acid donates a proton), and the Lewis concept of acids (electron-pair acceptors). Each statement must be tested against these definitions.
-
Statement I: Conjugate base of chloric acid is ClOX−
Chloric acid is HClOX3. When it donates a proton (HX+), it loses one hydrogen atom and the charge adjusts. The species left is ClOX3X− (the chlorate ion). The ion ClOX− is the conjugate base of hypochlorous acid (HClO), not chloric acid. So statement I is false.
-
Statement II: AlClX3 is a Lewis acid
A Lewis acid is any species that can accept an electron pair. AlClX3 has an aluminum atom with only six electrons in its valence shell (three bonds to chlorine, no lone pairs). This electron deficiency means it readily accepts a lone pair from a donor (like ClX− in the classic dimerization to AlX2ClX6). Therefore, AlClX3 is indeed a Lewis acid. Statement II is true.
-
Statement III: Conjugate base of NHX3 is NHX2X−
Ammonia (NHX3) can act as a Brønsted acid, though weakly. If it donates a proton, it loses one HX+, leaving NHX2X− (the amide ion). This fits the definition: conjugate base = acid minus HX+. Statement III is true.
Watch outA common mistake is confusing the names of oxyacids. Chloric acid (HClOX3) is different from hypochlorous acid (HClO), chlorous acid (HClOX2), and perchloric acid (HClOX4). Each has a different conjugate base. Always check the formula of the acid first.
Since only statements II and III are correct, the answer is the option listing those two.
✓Final answerThe correct option is (C) II, III only.
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- TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQQ.Arrange the following in the correct order of their boiling points (C2H5)2O \hspace{1cm} CH3(CH2)3OH \hspace{1cm} CH3CH2CH(OH)CH3 \hspace{1cm} CH3–(CH2)3–CH3 I \hspace{2.5cm} II \hspace{2.5cm} III \hspace{2.5cm} IV (A) I > III > II > IV (B) II > I > III > IV (C) III > II > I > IV (D) II > III > IV > I
›Reveal solutionSolution
Hydrogen bonding (alcohols II, III) dominates; among molecules of similar mass the longer-chain alkane IV slightly outboils the weakly polar ether I, and straight-chain 1-butanol (II) beats branched 2-butanol (III). Order: II > III > IV > I — option (D).
Concept & Intuition
Boiling point tracks the strength of intermolecular forces: hydrogen bonding > dipole-dipole > London dispersion. Identify each species:
- I: (C2H5)2O — diethyl ether (polar, no O–H).
- II: CH3(CH2)3OH — 1-butanol (straight-chain primary alcohol).
- III: CH3CH2CH(OH)CH3 — 2-butanol (branched secondary alcohol).
- IV: CH3(CH2)3CH3 — n-pentane (nonpolar alkane).
All have similar molar mass (~72–74 g/mol), so hydrogen bonding is the deciding factor.
Step-by-step reasoning.
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Alcohols come first. II and III both have –OH and hydrogen-bond, so they boil highest (1-butanol ~117 °C, 2-butanol ~99 °C).
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Straight vs branched alcohol. Branching lowers surface area and weakens dispersion, so straight-chain II boils higher than III.
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Ether vs alkane. Diethyl ether (I, ~35 °C) is only weakly polar; n-pentane (IV, ~36 °C) is nonpolar but has a slightly longer carbon chain and marginally stronger dispersion, so IV boils just above I.
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Full order:
II(∼117∘C)>III(∼99∘C)>IV(∼36∘C)>I(∼35∘C).
Watch outDo not assume a polar molecule always outboils a nonpolar one of similar mass. Diethyl ether's dipole is weak, and n-pentane's slightly larger dispersion gives it a marginally higher boiling point.
TipFor nearly equal molar masses, the molecule with the longer, less-branched carbon chain (more surface contact) tends to boil higher — here n-pentane edges out diethyl ether.
✓Final answerThe correct option is (D).
ANSWER: D
- TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQQ.Identify the set of molecules which are not in the correct order of their dipole moments (A) HF>HCl>HBr (B) H2O>H2S>CO2 (C) H2S>HCl>HF (D) NH3>NF3>BF3
›Reveal solutionSolution
Dipole moment depends on both bond polarity and molecular geometry. Checking each option against known values shows that (A), (B), and (D) all list the correct order, but (C) reverses the true order — the actual order is HF > HCl > H₂S, not H₂S > HCl > HF. So the set that is not in the correct order is (C).
Concept & Intuition
A common trap is to assume greater electronegativity difference always means a larger dipole moment — but bond length and molecular geometry matter too. Even though HF has the largest electronegativity difference among the hydrogen halides, its dipole moment still comes out largest once you compare actual measured values, and H₂S (despite having two polar S–H bonds) has a smaller net dipole than either HCl or HF.
Step-by-step reasoning
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Option (A): HF > HCl > HBr
Measured dipole moments: HF ≈ 1.91 D, HCl ≈ 1.08 D, HBr ≈ 0.82 D.
→ Order HF > HCl > HBr is correct.
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Option (B): H₂O > H₂S > CO₂
H₂O is bent (~104.5°) with a large dipole (≈1.85 D) because oxygen is highly electronegative. H₂S is also bent but sulfur is less electronegative, giving a smaller dipole (≈0.97 D). CO₂ is linear and symmetric, so its two C=O bond dipoles cancel exactly, giving zero net dipole.
→ Order H₂O > H₂S > CO₂ is correct.
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Option (C): H₂S > HCl > HF
Using the same measured values as above: H₂S ≈ 0.97 D, HCl ≈ 1.08 D, HF ≈ 1.91 D. The actual order is HF > HCl > H₂S — the complete reverse of what this option claims.
→ This order is incorrect.
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Option (D): NH₃ > NF₃ > BF₃
NH₃ is pyramidal with a lone pair reinforcing the bond dipoles, giving a net dipole of ≈1.47 D. NF₃ is also pyramidal, but because fluorine is more electronegative than nitrogen, the N–F bond dipoles point toward F, opposing the lone-pair contribution — this largely cancels out and leaves a much smaller net dipole (≈0.24 D). BF₃ is trigonal planar and symmetric, giving zero net dipole.
→ Order NH₃ > NF₃ > BF₃ is correct.
Watch outDon't assume more electronegative atoms always mean a larger dipole moment. In NF₃, the bond dipoles point away from the lone pair and partially cancel it, making its net dipole much smaller than NH₃'s — the opposite of what a naive electronegativity comparison with NH₃ would suggest.
TipFor molecules with lone pairs, always take the vector sum: the lone pair contributes its own dipole, and the bond dipoles may add to or subtract from it depending on their direction.
✓Final answerThe correct option is (C).
ANSWER: C
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- TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQQ.Observe the following species(i) NH3(ii) AlCl3(iii) SnCl4(iv) CO2(v) Ag+(vi) HSO4− How many of the above species act as Lewis acids? (A) 5 (B) 3 (C) 4 (D) 2
›Reveal solutionSolution
A Lewis acid is an electron-pair acceptor. Checking each species: AlCl₃, SnCl₄, CO₂, and Ag⁺ can all accept an electron pair, while NH₃ (a lone-pair donor) and HSO₄⁻ (a Brønsted acid/base with no accessible empty orbital) cannot. That gives 4 Lewis acids, so the correct option is (C).
Concept & Intuition
A Lewis acid is any species that can accept a pair of electrons — typically because it has an incomplete octet, a vacant low-energy orbital, or an electron-deficient centre that can coordinate a lone pair. Go through each species and ask: can it accept an electron pair?
Step-by-step reasoning
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NH₃ — Nitrogen has a lone pair and a complete octet; it donates electrons rather than accepting them.
→ Not a Lewis acid (it is a Lewis base).
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AlCl₃ — Aluminium has only six electrons around it (an incomplete octet) and readily accepts a lone pair (e.g., from Cl⁻ to form AlCl₄⁻).
→ Lewis acid.
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SnCl₄ — Tin can expand its octet using empty d-orbitals and accepts a lone pair to form species like SnCl₆²⁻.
→ Lewis acid.
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CO₂ — The carbon in CO₂ is strongly electron-deficient (δ⁺) and readily accepts a lone pair from a nucleophile, as in its reaction with OH⁻/H₂O to form HCO₃⁻/H₂CO₃. This is a standard textbook example of a Lewis acid, alongside BF₃, SO₂, and SO₃.
→ Lewis acid.
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Ag⁺ — A silver cation has vacant orbitals and readily accepts lone pairs from ligands (e.g., NH₃, forming [Ag(NH₃)₂]⁺).
→ Lewis acid.
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HSO₄⁻ — Sulfur here is already tetrahedrally coordinated with no accessible empty orbital, and the species behaves as a Brønsted acid/base (donating or accepting a proton), not as an electron-pair acceptor.
→ Not a Lewis acid.
Counting the Lewis acids: AlCl₃, SnCl₄, CO₂, Ag⁺ — that's 4 species.
Watch outDon't overlook CO₂ as a Lewis acid just because it has no formal incomplete octet — its electron-deficient carbon still readily accepts a lone pair from a nucleophile, which is exactly the Lewis-acid criterion.
TipElectron-deficient molecules such as BF₃, AlCl₃, CO₂, SO₂, and SO₃ are classic textbook examples of Lewis acids — keep this list in mind for quick recall.
✓Final answerThe correct option is (C).
ANSWER: C
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- TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQQ.In the given reaction sequence, Z is
[!FORMULA] C6H5CO2H(1) NH3(2) ΔXBr2 /NaOHYCHCl3 /KOHZ
(A) C6H5−Cl (B) C6H5−OH (C) C6H5−NC (D) C6H5−CN
›Reveal solutionSolution
Benzoic acid → benzamide → (Hofmann) aniline → (carbylamine reaction) phenyl isocyanide CX6HX5NC — option (C).
The concept first. Three named reactions, each with a give-away signature:
- Acid + ammonia + heat: forms the ammonium salt, which on heating dehydrates to the amide. (Acid → amide is the standard way in.)
- Hofmann bromamide degradation (BrX2/NaOH): converts an amide RCONHX2 into a primary amine RNHX2 with the loss of one carbon. The mechanism goes through an N-bromoamide, then a nitrene-like migration of R from carbon to nitrogen, giving an isocyanate that is hydrolysed.
- Carbylamine (isocyanide) test (CHClX3+KOH, heat): the diagnostic test for primary amines. KOH deprotonates chloroform to give dichlorocarbene :CClX2, which the amine attacks; loss of 2HCl gives the isocyanide R−N≡C, recognisable by its very unpleasant smell.
Step 1 — X.
CX6HX5COOH+NHX3CX6HX5COOX− NHX4X+Δ,−HX2OCX6HX5CONHX2
X = benzamide.
Step 2 — Y.
CX6HX5CONHX2BrX2/NaOHCX6HX5NHX2+NaX2COX3+NaBr
Seven carbons become six, and −NHX2 is now attached directly to the ring. Y = aniline.
Step 3 — Z.
CX6HX5NHX2+CHClX3+3KOHΔCX6HX5NC+3KCl+3HX2O
Z = phenyl isocyanide, CX6HX5−N≡C.
Step 4 — Beware the look-alike. Option (D), CX6HX5CN (benzonitrile), is bonded through carbon and would come from a diazonium salt with CuCN (Sandmeyer), not from the carbylamine reaction. Options (A) and (B) would need diazotisation followed by CuCl or by warm water.
✓Final answerZ is phenyl isocyanide, CX6HX5NC, so the correct option is (C).
ANSWER: C
- TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQQ.Consider the following carbocations
[!FORMULA] CX6HX5CHX2X+ICHX2=CHX+IICHX3−CX+−H∣CHX3IIICHX3−CHX2X+IVHC ≡CX+V
Arrange the above carbocations in the order of decreasing stability (A) I > III > IV > II > V (B) V > II > IV > III > I (C) V > II > III > I > IV (D) I > III > IV > V > II›Reveal solutionSolution
Carbocation stability increases with greater delocalization/donation of electron density onto the electron-deficient carbon. Ranking these five: I (benzylic, resonance-stabilized) > III (secondary, two alkyl groups) > IV (primary, one alkyl group) > II (vinyl, no effective stabilization) > V (alkynyl, least stable of all). This is option (A).
Concept and intuition:
Carbocations are electron-deficient species; anything that spreads out (delocalizes) or donates into the positive charge makes them more stable. The most powerful stabilizer here is resonance (as in the benzyl cation), followed by hyperconjugation and the inductive effect from alkyl groups (more alkyl substitution = more stability). Vinyl and alkynyl carbocations are exceptionally unstable because the positive charge sits on an sp²/sp carbon, which holds electrons more tightly (higher s-character) and offers essentially no resonance stabilization from the adjacent π system, since the empty orbital does not align with it.
Step-by-step reasoning:
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Identify each carbocation's structure and stabilization mode.
- I: CX6HX5CHX2X+ — the benzyl carbocation. The positive charge is delocalized into the aromatic ring through resonance (several resonance structures place the charge on the ring carbons). This is the most stable of the five.
- II: CHX2=CHX+ — the vinyl carbocation. The cationic carbon is essentially sp-hybridized with the empty orbital lying in the plane of the molecule, orthogonal to the π system, so the adjacent double bond cannot donate into it by resonance. Very unstable.
- III: CHX3−CHX+−CHX3 — the isopropyl carbocation. This carbon bears two methyl groups and one hydrogen, so it is a secondary carbocation, stabilized by hyperconjugation and induction from its two alkyl groups.
- IV: CHX3−CHX2X+ — the ethyl carbocation. This carbon bears one methyl group and two hydrogens, so it is a primary carbocation, stabilized by only one alkyl group.
- V: HC≡CX+ — the ethynyl (alkynyl) carbocation. The cationic carbon is sp-hybridized (50% s-character), holding the bonding electrons closest to the nucleus and offering no resonance or hyperconjugative stabilization. This is the least stable of the five.
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Rank by known stability order.
General order: benzylic/allylic > tertiary > secondary > primary > vinyl > alkynyl. With only benzylic, secondary, primary, vinyl and alkynyl represented here:
I (benzylic) > III (secondary) > IV (primary) > II (vinyl) > V (alkynyl).
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Match with the options.
- (A) I > III > IV > II > V — matches exactly.
- (B) V > II > IV > III > I — reverses the order; wrong, since V is the least stable, not the most.
- (C) V > II > III > I > IV — also has V first; wrong.
- (D) I > III > IV > V > II — places V ahead of II, but V (alkynyl) is less stable than II (vinyl); wrong.
Watch outA common mistake is to think a triple bond's π electrons can stabilize an adjacent positive charge the way a benzene ring does. In fact, an sp carbon is more electron-withdrawing (more s-character) and destabilizes a cation further rather than stabilizing it. Likewise, the vinyl cation's empty orbital lies in the plane of the molecule, perpendicular to the π bond, so no resonance donation from the double bond is possible.
TipRemember the order: benzylic/allylic > tertiary > secondary > primary > vinyl > alkynyl — the more s-character at the cationic carbon (and the less alkyl/resonance support it has), the less stable the carbocation.
✓Final answerThe correct option is (A).
ANSWER: A
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- TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQQ.In which of the following, ions are correctly arranged with respect to their bond orders? (A) N22−→O2−→O22−→C22− (B) O22−→O2−→N22−→C22− (C) C22−>O2>O22>N22 (D) C22−→N22−→O2−→O22−
›Reveal solutionSolution
Bond order decreases as antibonding orbitals are progressively filled. Using molecular orbital theory, C22− has bond order 3, N22− has 2, O2− has 1.5, and O22− has 1. The correct arrangement is (D).
Bond order measures the net number of bonding electron pairs holding two atoms together. In molecular orbital theory, we calculate it as:
Bond order=21(electrons in bonding MOs−electrons in antibonding MOs)
Higher bond order means a stronger, shorter bond. When electrons are added to a molecule (forming anions), they occupy the next available orbital according to the aufbau principle. If that orbital is antibonding, the bond order drops.
Let me work through each ion systematically, counting electrons and applying the MO filling sequence.
Determining each bond order
1. C22−: 12 + 2 = 14 electrons
The neutral C2 molecule has 12 electrons. Adding two more gives 14 total.
For homonuclear diatomics of C and N (where 2s-2p mixing is significant), the order is:
σ2s<σ2s∗<π2px=π2py<σ2pz<π2px∗=π2py∗<σ2pz∗
Filling 14 electrons:
- σ2s2,σ2s∗2,π2px2,π2py2,σ2pz2
Bonding: 2 + 4 + 2 = 8
Antibonding: 2
Bond order = 21(8−2)=3
2. N22−: 14 + 2 = 16 electrons
Adding two electrons to the 14 in N2, they enter the degenerate π2p∗ orbitals:
- σ2s2,σ2s∗2,π2px2,π2py2,σ2pz2,π2px∗1,π2py∗1
Bonding: 8
Antibonding: 2 + 2 = 4
Bond order = 21(8−4)=2
3. O2−: 16 + 1 = 17 electrons
For oxygen (no s-p mixing), the σ2pz comes before the π2p∗ orbitals. Neutral O2 has 16 electrons with configuration ending in π2px∗1,π2py∗1 (two unpaired). Adding one electron:
- Configuration ends: π2px∗2,π2py∗1
Bonding: 8
Antibonding: 2 + 3 = 5
Bond order = 21(8−5)=1.5
4. O22−: 16 + 2 = 18 electrons
Adding two electrons to O2:
- Configuration ends: π2px∗2,π2py∗2
Bonding: 8
Antibonding: 2 + 4 = 6
Bond order = 21(8−6)=1
TipFor oxygen-family species, each electron added goes into antibonding π∗ orbitals, decreasing bond order by 0.5 per electron.
Arranging by bond order
From highest to lowest:
C22−(3)>N22−(2)>O2−(1.5)>O22−(1)
This matches option (D) exactly.
✓Final answerThe correct option is (D): C22−→N22−→O2−→O22− with bond orders 3, 2, 1.5, and 1 respectively.
- TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQQ.Match the following List-I (Molecule/ion) A) SnCl2 B) NH3 C) I3− D) SO3 List-II (Shape) I. Trigonal planar II. Linear III. Angular IV. Trigonal pyramidal The correct answer is (A) A - III, B - I, C - II, D - IV (B) A - IV, B - III, C - I, D - II (C) A - III, B - I, C - IV, D - II (D) A - III, B - IV, C - II, D - I
›Reveal solutionSolution
The key idea is to determine the molecular shape of each species using VSEPR theory (steric number and lone pairs). The correct matches are: SnCl₂ → Angular, NH₃ → Trigonal pyramidal, I₃⁻ → Linear, SO₃ → Trigonal planar, so the answer is option (D).
Concept & Intuition
Molecular geometry is predicted by the Valence Shell Electron Pair Repulsion (VSEPR) theory. The central atom’s steric number (number of bonded atoms + lone pairs) determines the electron-pair geometry, while the number of lone pairs modifies the final molecular shape. For each species, we count valence electrons, assign the central atom, and then apply VSEPR.
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SnCl₂ (A)
- Tin (Sn) has 4 valence electrons; each Cl contributes 1 for bonding, total = 4 + 2×1 = 6 electrons (3 pairs).
- Sn is central, forms two single bonds with Cl, leaving one lone pair on Sn.
- Steric number = 2 bonds + 1 lone pair = 3 → electron-pair geometry is trigonal planar.
- With one lone pair, the molecular shape is angular (bent, ~120°).
- So A matches III.
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NH₃ (B)
- Nitrogen has 5 valence electrons; each H contributes 1, total = 5 + 3×1 = 8 electrons (4 pairs).
- N forms three N–H bonds, leaving one lone pair.
- Steric number = 3 bonds + 1 lone pair = 4 → electron-pair geometry is tetrahedral.
- With one lone pair, the molecular shape is trigonal pyramidal.
- So B matches IV.
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I₃⁻ (C)
- Iodine has 7 valence electrons; three I atoms give 21, plus 1 for the negative charge = 22 electrons (11 pairs).
- The central I is bonded to two terminal I atoms. The central I has 7 valence electrons, uses 2 for bonds, leaving 5 electrons (2 lone pairs + 1 electron in a 3-center-4-electron bond). Actually, the standard VSEPR treatment: central I has 2 bonds and 3 lone pairs (steric number = 5), but the molecular shape is determined by the arrangement of atoms only.
- With 2 bonded atoms and 3 lone pairs, the electron-pair geometry is trigonal bipyramidal, but the lone pairs occupy equatorial positions, leaving the two I atoms axial → linear shape.
- So C matches II.
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SO₃ (D)
- Sulfur has 6 valence electrons; each O contributes 6, but for bonding we consider S as central with three double bonds (or resonance). Total valence = 6 + 3×6 = 24 electrons.
- S forms three bonds (each double bond counts as one region of electron density) and has no lone pairs.
- Steric number = 3 bonds + 0 lone pairs = 3 → electron-pair geometry is trigonal planar.
- With no lone pairs, the molecular shape is trigonal planar.
- So D matches I.
Watch outA common mistake is to think NH₃ is “trigonal planar” because it has three bonds, but the lone pair pushes the hydrogens down, making it pyramidal. Also, I₃⁻ is often misidentified as bent — but the three lone pairs on the central iodine force the two iodines into a straight line.
TipFor I₃⁻, remember that the central iodine has 3 lone pairs; the VSEPR “AX₂E₃” designation always gives a linear shape (the two atoms are axial, lone pairs equatorial).
Now match the lists:
A → III, B → IV, C → II, D → I.
This corresponds to option (D).
✓Final answerThe correct option is (D).
ANSWER: D
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- TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQQ.What is the correct order of melting temperature of C, Si, Ge? (A) C > Ge > Si (B) Si > C > Ge (C) C > Si > Ge (D) Si > Ge > C
›Reveal solutionSolution
Melting points of Group 14 elements decrease down the group due to decreasing bond strength, but carbon is anomalously high because of its small size and strong covalent bonding. The correct order is C > Si > Ge, which corresponds to option (C).
The key concept here is trends in melting points for Group 14 elements (carbon, silicon, germanium, tin, lead). All these elements form giant covalent (diamond-like) structures in their solid state, except for tin and lead which are metallic. The melting point depends on the strength of the covalent bonds holding the atoms together.
Why this approach works:
Melting a solid means breaking the bonds between atoms. For diamond (carbon), the bonds are extremely strong because carbon atoms are very small, allowing close overlap of orbitals and very short, strong bonds. As we go down the group, atoms get larger, bond lengths increase, and bond strength decreases. So melting points should generally decrease. But carbon is an outlier — its melting point is far higher than the others. Silicon and germanium follow the expected decreasing trend.
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Recognize the structure type
Carbon (as diamond), silicon, and germanium all crystallize in the diamond cubic structure — a giant covalent network. Melting requires breaking many strong covalent bonds simultaneously, so melting points are very high.
-
Understand the bond strength trend
Bond strength in a covalent network depends on orbital overlap. Smaller atoms have shorter bonds and better overlap. Carbon’s 2p orbitals are much smaller than silicon’s 3p or germanium’s 4p. Thus, the C–C bond is the strongest, Si–Si is weaker, and Ge–Ge is weaker still.
-
Apply the trend to the given elements
- Carbon (diamond): melting point ~3550°C (actually sublimes, but effectively the highest).
- Silicon: melting point ~1414°C.
- Germanium: melting point ~938°C. So the order is clearly C > Si > Ge.
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Check the options
(A) C > Ge > Si — incorrect, because Si melts higher than Ge.
(B) Si > C > Ge — incorrect, because C is far higher than Si.
(C) C > Si > Ge — correct.
(D) Si > Ge > C — incorrect, because C is highest.
Watch outA common mistake is to think that since silicon is a semiconductor and carbon is a nonmetal, silicon might have a higher melting point. But the small size of carbon gives it uniquely strong bonds — diamond is the hardest natural material and has the highest melting point of any element.
TipA quick memory aid: In Group 14, melting points decrease down the group, but carbon is in a league of its own. So the order is always C > Si > Ge > Sn > Pb (though Sn and Pb are metallic and have much lower melting points).
✓Final answerThe correct option is (C).
ANSWER: C
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