KCET 2026 Chemistry answer key: full questions, step-by-step solutions, weightage and objection guide for students
KCET 2026 Chemistry was held on April 23, 2026 , from 2:30 PM to 3:50 PM . This article gives the KCET 2026 Chemistry answer key, the full question list with correct options, short explanations, solved numericals, sectional weightage and a step-by-step KEA objection guide.
The chemistry paper was reported by students as relatively easier than Physics and covered both 1st and 2nd PUC syllabi. Chemistry carries 60 marks out of 180 (one-third of total KCET marks), so a strong performance here can lift your overall rank quickly.
Quick exam recap
| Event | Detail |
|---|---|
| Exam | KCET 2026 Chemistry |
| Date | 2026-04-23 |
| Time | 2:30 PM – 3:50 PM |
| Marks | 60 marks (part of 180) |
| Difficulty (student feedback) | Relatively easier than Physics; balanced 1st & 2nd PUC coverage |
The paper mixed reaction-based and memory questions with a good number of numerical problems. If you revised NCERT and PUC thoroughly, you could score 45+ marks here.
Paper composition and syllabus coverage
KCET Chemistry drew from both 1st and 2nd PUC topics. The question types were:
- Reaction and mechanism-based questions from Organic Chemistry.
- Numerical and formula-based problems in Physical Chemistry.
- Memory-based factual questions in Inorganic Chemistry.
Most test-takers saw more weight from 2nd PUC topics in Organic Chemistry and p‑block elements.
Sectional weightage and topic checklist
| Section | Approx. questions | Nature of questions | Typical topics to check in NCERT/PUC |
|---|---|---|---|
| Organic Chemistry | 23–25 | Reaction-based, named reactions, mechanisms | Reaction mechanisms, functional group conversions, iodoform, nitration, reagents like PCC |
| Physical Chemistry | 18–20 | Numerical, mole concept, kinetics, solutions | Stoichiometry, kinetics, electrochemistry, thermochemistry, cell constant calculations |
| Inorganic Chemistry | 15–18 | Memory-based, p-block and d-block facts | Oxidation states, lanthanoid behaviour, trends, important reactions |
Use this checklist to verify which PUC chapters you must re-open quickly if you plan to file objections or review answers.
KCET 2026 Chemistry answer key: full question list with answers and concise explanations
Below is the question table published after the exam. Each row gives the question, the correct answer as noted in the published table and a short explanation or pointer to verify in NCERT/PUC textbooks.
| Q. No / Question | Correct answer | Short explanation / reference |
|---|---|---|
| Conductivity of a centimolar solution of KCl at 298 K (0.21 Ω⁻¹cm⁻¹). Resistance of cell = 60 Ω. Calculate cell constant. | 12.6 cm⁻¹ | Cell constant = κ × R. If κ = 0.21 Ω⁻¹cm⁻¹ and R = 60 Ω, cell constant = 0.21 × 60 = 12.6 cm⁻¹. Verify electrochemistry chapter (NCERT XII). |
| 2.69 V — the reaction will not occur (Non-Spontaneous) | Non-spontaneous | If required cell EMF differs from applied potential, reaction feasibility follows Ecell vs applied V. See electrochemistry notes. |
| During electrolysis of acidified water, if 16 g of O₂ gas is formed, what is the volume of H₂ gas liberated at the cathode under STP conditions? | 22.4 L | 16 g O₂ = 0.5 mol. Reaction 2H₂O → O₂ + 2H₂ gives 1 mol H₂ per mol O₂; 0.5 mol O₂ → 1 mol H₂ → 22.4 L at STP. |
| Which member of the Lanthanoid series is well known to exhibit a +4 oxidation state? | Cerium (Ce) | Ce commonly shows +4 (CeO₂). See inorganic chemistry (lanthanoids). |
| For a 1st order change R → P, concentration of R changes (details incomplete in table). The rate when [R] = 0.01 M is: | 3.47 × 10⁻⁴ M min⁻¹ | Answer reported; refer to first order kinetics formula and rate constant calculation in NCERT XII kinetics. |
| Activation energy for X → Y is 150 kJ mol⁻¹. ΔH = -135 kJ mol⁻¹. Activation energy for Y → X is: | 285 kJ mol⁻¹ | Ea(reverse) = Ea(forward) - ΔH = 150 - (-135) = 285 kJ mol⁻¹. |
| The intermediates in heteropolar reactions are: options: free radicals, cations, anions, both anions and cations | Both anions and cations | Heteropolar reactions proceed via ionic intermediates (both signs). |
| Statement I: Nitrogen in pyridine cannot be estimated by Kjeldahl's. Statement II: Nitrogen in pyridine changes to ammonium on conc. H₂SO₄ in Kjeldahl's method. Choose correct | Statement I is true but Statement II is false | Pyridine N is basic but not converted to ammonium by conc. H₂SO₄ the same way; verify nitrogen estimation limitations. |
| C–Cl bond in methyl chloride compared to C–Cl bond in chlorobenzene is: Longer and stronger / Shorter and stronger / Shorter and weaker / Longer and weaker | Longer and stronger | Reported exam answer. For verification, see discussion of C(sp3)-Cl vs C(sp2)-Cl in organic chapters. |
| The compound with molecular formula C₃₀H₄₂ is: | Elcosane | Name provided in answer table. Verify hydrocarbon nomenclature references. |
| Compound from which chlorobenzene cannot be prepared easily: Aniline / Benzene / Phenol / Benzene diazonium chloride | Aniline | Aniline needs diazotisation then Sandmeyer to get chlorobenzene; alternatives exist but question asked 'cannot be prepared easily'—answer in paper is Aniline as per key. |
| In SN1 reaction, alkyl halide that on hydrolysis produces racemic mixture: t-BuBr / 2-bromobutane / isopropyl bromide / methyl bromide | 2-bromobutane | SN1 gives planar carbocation in some secondary cases leading to racemisation; reported answer: 2-bromobutane. |
| R–CH₂OH is converted into R–CHO by reacting with: Alkaline KMnO₄ / LiAlH₄ / Na/C₂H₅OH / PCC | PCC (Pyridinium chlorochromate) | PCC oxidises primary alcohols to aldehydes without further oxidation. See organic reagents list. |
| Glycerol is a trihydric alcohol. It contains: (one primary, one secondary, one tertiary) / (two primary and one secondary) / (two secondary and one primary) / (one primary and two tertiary) | Two primary and one secondary alcoholic groups | Glycerol structure CH₂OH–CHOH–CH₂OH has two primary and one secondary OH. |
| Carboxylic acids are more acidic than phenols because: Formation of dimers / Intermolecular H-bonding / More covalent nature / More resonance stabilisation of conjugate base | More resonance stabilisation of their conjugate base | Carboxylate ion is stabilized by resonance; phenoxide less so. |
| Compound that does not give iodoform test: Ethanal / Acetone / Ethanoic acid / Acetophenone | Ethanoic acid | Iodoform test requires methyl ketone or ethanol; acetic acid does not give positive iodoform. |
| Nitration of aniline in strong acidic medium gives significant amount of m-nitroaniline. Reason: (options given) | In strong acidic medium, aniline is present as anilinium ion. | Anilinium ion is meta-directing, so nitration gives m-nitroaniline. |
| Basic strength of alkylamines in aqueous phase is not decided by: Inductive effect / Solvation effect / Steric hindrance / Hyperconjugation effect | Hyperconjugation effect | Solvation and inductive effects are key in aqueous basicity; exam answer listed hyperconjugation as not decisive. |
| Statement I: Staggered conformation of ethane is more stable than eclipsed. Statement II: The torsional strain in staggered conformation is more. Choose | Statement I is true but Statement II is false | Staggered is more stable; torsional strain is lower (not higher) in staggered form. |
| Law of definite proportions — options mapping answers a-ii b-iv c-i d-iii (reported) | Answer as given in table | Mapping provided in answer table; consult stoichiometry section. |
Note: Each answer above is taken from the published question table after the exam. Use NCERT Class 11/12 and your PUC textbooks to verify references when preparing an objection.
Solved numericals: step-by-step worked solutions
Below are clear, quick solutions for representative numerical questions you will commonly face in KCET-type papers.
1) Electrolysis gas volume (from table)
Question: During electrolysis of acidified water, if 16 g of O₂ is formed, what is the volume of H₂ liberated at STP?
Solution: - Molar mass of O₂ = 32 g mol⁻¹. - Moles of O₂ = 16 / 32 = 0.5 mol. - Reaction: 2 H₂O → O₂ + 2 H₂. For every 1 mol O₂, 2 mol H₂ form. - Moles H₂ from 0.5 mol O₂ = 0.5 × 2 = 1.0 mol. - At STP, 1 mol gas = 22.4 L. So volume = 22.4 L.
Tip: Always balance the redox equation first and convert mass → moles → gas volume.
2) Activation energy reverse reaction
Question: Ea (X → Y) = 150 kJ mol⁻¹ , ΔH = -135 kJ mol⁻¹ . Find Ea (Y → X).
Solution: - ΔH = Ea(forward) - Ea(reverse) ⇒ Ea(reverse) = Ea(forward) - ΔH. - Ea(reverse) = 150 - (−135) = 285 kJ mol⁻¹.
Tip: Watch the sign of ΔH. If reaction is exothermic (negative ΔH), Ea for reverse is larger.
3) Cell constant calculation (worked)
Question: Conductivity κ = 0.21 Ω⁻¹ cm⁻¹ , resistance R = 60 Ω . Cell constant = ?
Solution: - Conductance G = 1/R = 1/60 Ω⁻¹ = 0.01667 Ω⁻¹. - Relationship: κ = G × cell constant ⇒ cell constant = κ / G = κ × R. - cell constant = 0.21 × 60 = 12.6 cm⁻¹.
Tip: Use consistent units. Conductivity units Ω⁻¹ cm⁻¹ × resistance (Ω) gives cm⁻¹.
4) Typical kinetics quick method (example idea)
When given first-order decay data, use ln([A]₀/[A]) = kt. Plug values directly and keep time units consistent (minutes or seconds as given).
General tip: Carry at least three significant digits during calculation to avoid round-off error.
Memory-based inorganic Q&A and mnemonics
High-frequency inorganic facts in this paper:
- Lanthanoid with +4 state: Cerium (Ce) .
- p‑block and d‑block oxidation state patterns are common question sources.
- Keep quick mnemonics ready for group trends and common oxidation states.
Quick mnemonic for remembering oxidation states of Ce: "Ce can be 3 or 4" — tie to CeO₂ examples in lab.
Verify all factual answers against NCERT Class 11/12 inorganic chapters and your PUC textbooks before filing an objection.
KCET 2026 Chemistry answer key: Answer release, objections and how-to guide
KEA promised to update the provisional answer key after the exam on April 23, 2026 . Once the provisional key is live you can file objections.
How to file an objection (step-by-step):
- Visit kea.kar.nic.in and log in using your CET number and date of birth .
- Choose subject: Chemistry and enter your version code and the question number you want to challenge.
- Provide a clear, valid reference. Prefer NCERT or the PUC textbook chapter and quote exact page/line if possible.
- Attach supporting document or screenshot (where portal allows). Follow the format prompted on the portal.
- Submit before the objection deadline shown on KEA. Objection filing is free .
What to include in your objection:
- Subject, version code, question number.
- Short reason why key is wrong.
- Valid reference (NCERT/PUC) with chapter/section. If the portal allows uploaded files, attach the page image.
After submission:
- KEA reviews all objections.
- If an objection is approved, the final answer key will be revised and scores recalculated.
- KEA will publish the final key and updated response to objections; follow announcements on kea.kar.nic.in.
Note: The official objection deadline date was not specified in the published table we used. Check kea.kar.nic.in immediately after the provisional key appears and file within the window shown.
Score estimation: convert answers into marks
How to calculate raw marks from the answer key:
- Chemistry total = 60 marks .
- For each correct answer, add the mark shown in the KCET instruction (usually 1 mark per question in Chemistry unless stated otherwise).
- There is no negative marking in KCET Chemistry (confirm on exam instruction sheet). Use the official key to tally correct responses.
Scoring tips:
- If you are unsure about a question, keep it unmarked in your internal estimate rather than guessing. Conservative estimates reduce surprise in counselling.
- After provisional key, calculate two scores: (a) using the provisional key as-is, (b) using alternate answers where you have strong references and might file objections.
KCET marks vs estimated rank and college prospects
| Chemistry Score (out of 60) | Estimated KCET Rank (Engineering) | Likely college prospects |
|---|---|---|
| 55 – 60 | Top 500 | Top NITs and top government engineering colleges |
| 45 – 54 | 500 – 3,000 | UVCE, BIT and top state government colleges |
| 35 – 44 | 3,000 – 10,000 | Good private engineering colleges |
| 25 – 34 | 10,000 – 25,000 | Mid-level private colleges, management quota |
| Below 25 | 25,000+ | Explore other state-level options |
Remember: These are estimated ranges based on typical KCET patterns. Final rank depends on your total score (all three subjects) and overall exam performance.
Gap checklist and next steps after checking your answers
- Official timelines: KEA will publish provisional and final answer keys on kea.kar.nic.in . We do not have the exact objection window date in the table used here; check KEA and act quickly when the provisional key is live.
- If you plan to object: prepare NCERT/PUC references, screenshots and the exact version code and question number.
- Counselling readiness: keep scanned copies of all documents (10th, PUC marksheets, CET admit card, photo ID). Start tracking closing ranks for colleges you want.
- Alternate plan: list 3–4 fallback colleges by rank band in the table above so you can switch during counselling.
Downloadable checklist and printable summary
| Quick checklist (printable) | What to do |
|---|---|
| Score calculator | Tally correct answers from provisional key → convert to /60 |
| Objection pack | CET number, DOB, version code, question no., NCERT/PUC page screenshot |
| Documents for counselling | 10th, PUC marksheets, CET admit card, photo ID, passport photo scans |
| Revision targets | Revisit Organic named reactions, electrochemistry formulas, p-block facts |
Print this table and keep it with your notes while you wait for the final key.
What we will update
We will update this article when KEA publishes the provisional and final answer keys, and when KEA posts the exact objection window and responses. Keep an eye on kea.kar.nic.in for official notices.
FAQs (quick answers)
When was KCET 2026 Chemistry held?
KCET 2026 Chemistry was held on April 23, 2026 , from 2:30 PM to 3:50 PM .
How many marks is KCET Chemistry worth?
Chemistry is worth 60 marks , which is one-third of the KCET total of 180 marks.
Which section had the highest weightage?
Organic Chemistry had the highest weightage with approximately 23–25 questions.
How do I raise an objection to the provisional answer key?
Log in to kea.kar.nic.in with your CET number and DOB , select subject Chemistry , enter version code and question number, add valid reference (NCERT/PUC), and submit before the objection deadline shown on KEA.
Is there a fee to file objections?
No. Filing an objection is free of cost .
What references are accepted for objections?
Use NCERT textbooks and PUC textbooks as primary references. Quote chapter and paragraph or attach a screenshot if the portal supports uploads.
What score in Chemistry is considered high for KCET 2026?
A well-prepared student could score 45+ in Chemistry. Scores 55–60 typically place you in the Top 500 range for KCET engineering ranks (estimate).
Where will the final answer key appear?
KEA will publish both provisional and final answer keys on kea.kar.nic.in . Check the portal regularly for official updates.
If you want, gather your answers now and use the checklist above to calculate your provisional Chemistry score out of 60. Keep NCERT and your PUC textbooks handy if you plan to file an objection — you will need exact references.