Previous year questions: Redox Reactions

Ten questions follow from the official JEE Main 2026 papers and the NCERT Exemplar. Each heading gives the paper and the marks. In a statement question, judge each statement separately before choosing the option.

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

Given below are two statements. Statement I: Oxidising power of halogens decreases in the order F2 > Cl2 > Br2 > I2, which is the basis of the layer test. Statement II: The layer test to identify Br2 and I2 in aqueous solution involves the oxidation of bromide or iodide into Br2 or I2 with Cl2, which is a displacement redox reaction. In the light of the statements, choose the correct answer: (a) both statements are true (b) both are false (c) Statement I is true but Statement II is false (d) Statement I is false but Statement II is true.

Answer: The oxidising power of halogens does decrease down the group, because the tendency to gain an electron falls as the atom gets larger, and this order is behind the displacement reactions used in the layer test. In the layer test chlorine is added to a solution containing bromide or iodide, and chlorine being the stronger oxidising agent displaces bromine or iodine, which is a redox displacement. Both statements are correct, so option (a).

Hinglish: Halogens ki oxidising power group mein neeche jaate hue ghatti hai, isliye F2 > Cl2 > Br2 > I2 sahi hai. Layer test mein Cl2 bromide ya iodide ko oxidise karke Br2 ya I2 banata hai, jo displacement redox hai. Dono statements sahi hain, answer (a).

Q2 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

Which of the following sets includes all the species that will change the orange colour of K2Cr2O7 in acidic medium? (a) Fe2+, Sn2+, I minus, S2 minus (b) S2 minus, Fe3+, I minus, C2O4 2 minus (c) Fe2+, NO2 minus, SO2, Sn4+ (d) Fe3+, SO4 2 minus, S2 minus, Sn4+

Answer: In acidic medium dichromate is a strong oxidising agent and is reduced to the green chromium three plus ion, so the orange colour changes. Species that can be oxidised by dichromate change it: iron two plus goes to iron three plus, tin two plus goes to tin four plus, iodide goes to iodine and sulphide goes to sulphur. Options (b), (c) and (d) each contain a species that is already in its higher oxidation state and cannot be oxidised further, so they cannot all change the colour. Option (a) contains only reducing species, so it is the answer.

Hinglish: Acidic medium mein dichromate oxidising agent hai aur green Cr^3+ banta hai, isliye orange colour badal jaata hai. Fe^2+, Sn^2+, I^- aur S^2- oxidise ho sakte hain, isliye inka set sahi hai. Baaki options mein Fe^3+ ya Sn^4+ jaise already oxidised species hain. Answer (a).

Q3 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

Given below are two statements. Statement I: The covalency of oxygen is generally two but it can exceed up to four. The oxidation state of oxygen in SO2 is minus two and in OF2 it is plus two. Statement II: The anomalous behaviour of oxygen when compared with the other elements of group 16 is due to its small size and high electronegativity. Choose the correct answer: (a) both statements are true (b) both are false (c) Statement I is true but Statement II is false (d) Statement I is false but Statement II is true.

Answer: Oxygen normally forms two covalent bonds, but in species such as the hydronium ion and in larger molecules its covalency can be higher, and the statement that it can exceed up to four is accepted. In SO2, oxygen is more electronegative than sulphur and takes minus two, while in OF2 fluorine is more electronegative and forces oxygen to plus two. Statement I is therefore correct. The anomalous behaviour of oxygen is indeed due to its small size and high electronegativity, so Statement II is also correct and the answer is (a).

Hinglish: SO2 mein S se kam electronegative hone ki wajah se oxygen -2 hai, OF2 mein F zyada electronegative hai isliye oxygen +2 hai; Statement I sahi. Oxygen ka anomalous behaviour small size aur high electronegativity se hota hai, isliye Statement II bhi sahi. Answer (a).

Q4 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

Correct statements from the following are: A. Potassium dichromate is an oxidising agent and it oxidises FeSO4 to Fe2(SO4)3 in acidic medium. B. Sodium dichromate can be used as a primary standard in volumetric estimation. C. CrO4 2 minus and Cr2O7 2 minus are interconvertible in aqueous solution by varying the pH. D. The Cr-O-Cr bond angle in Cr2O7 2 minus is 126 degrees.

Answer: Statement A is correct, because acidified dichromate oxidises iron two plus to iron three plus and itself goes to chromium three plus. Statement B is false, because sodium dichromate is hygroscopic and is not a primary standard. Statement C is correct: in acidic solution chromate converts to dichromate and in basic solution dichromate converts back to chromate. Statement D is correct, the Cr-O-Cr linkage in dichromate is bent at about 126 degrees. The correct statements are A, C and D.

Hinglish: A sahi hai kyunki acidic dichromate Fe^2+ ko Fe^3+ oxidise karta hai. B galat hai kyunki sodium dichromate hygroscopic hai, primary standard nahi. C sahi hai: acidic mein chromate se dichromate, basic mein wapas chromate. D sahi hai, Cr-O-Cr angle lagbhag 126° hai. Sahi statements A, C aur D.

Q5 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

The correct statements among the following are: A. Mo(VI) and W(VI) are less stable than Cr(VI). B. Ce4+ and Tb4+ are oxidants while Eu2+ and Yb2+ are reductants. C. Cm and Am have seven unpaired electrons. D. Actinoid contraction is greater from element to element than lanthanoid contraction. Choose the correct option.

Answer: Statement A is false: in group 6 the plus six state becomes more stable down the group, so Mo(VI) and W(VI) are more stable than Cr(VI), which is a strong oxidising agent. Statement B is true: cerium four plus and terbium four plus are good oxidants, while europium two plus and ytterbium two plus are reducing. Statement C is false as written: curium has eight unpaired electrons in its ground state configuration and americium has seven, so both do not have seven. Statement D is true, since the actinoid contraction is greater because the 5f electrons shield poorly. The correct statements are B and D.

Hinglish: A galat hai kyunki group mein neeche jaate hue +6 state zyada stable hoti hai, isliye Mo(VI) aur W(VI) zyada stable hain. B sahi hai: Ce^4+, Tb^4+ oxidants aur Eu^2+, Yb^2+ reductants. C galat hai kyunki unpaired electron counts match nahi karte. D sahi hai kyunki actinoid contraction lanthanoid se zyada hai. Answer B aur D.

Q6 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

Correct statements from the following are: A. Nitrogen in oxidation states from plus one to plus four disproportionates in acid medium. B. Nitrogen has the ability to form d pi - p pi multiple bonds with itself and other elements with small size and high electronegativity. C. N-N single bond is stronger than P-P single bond. D. Nitrogen has the highest density in its group due to small size. E. The maximum covalency of nitrogen is four since it has only four valence orbitals for bonding.

Answer: Statement A is correct, because nitrogen in the intermediate plus one to plus four states disproportionates in acid to species such as nitric acid and nitric oxide. Statement B is false, because nitrogen has no d orbitals and cannot form d pi - p pi bonds. Statement C is false: the N-N single bond is weaker than the P-P single bond because of lone pair repulsion, which is why nitrogen prefers triple bonds. Statement D is false: nitrogen is a gas and has a low density. Statement E is correct, since nitrogen has only four valence orbitals and cannot exceed a covalency of four. The correct statements are A and E.

Hinglish: A sahi hai, nitrogen ke +1 se +4 states acid mein disproportionate karte hain. B galat hai kyunki nitrogen ke paas d orbitals nahi hain. C galat hai, N-N single bond lone pair repulsion ki wajah se P-P se kamzor hai. D galat hai, nitrogen gas hai aur density kam hai. E sahi hai, chaar valence orbitals ki wajah se covalency maximum four hai. Answer A aur E.

Q7 (NCERT Exemplar, 1 mark)

Which of the following is not an example of a redox reaction? (i) CuO + H2 giving Cu + H2O (ii) Fe2O3 + 3CO giving 2Fe + 3CO2 (iii) 2K + F2 giving 2KF (iv) BaCl2 + H2SO4 giving BaSO4 + 2HCl

Answer: In the first three reactions oxidation numbers change: copper goes from plus two to zero in (i), iron goes from plus three to zero in (ii), and potassium and fluorine go from zero to plus one and minus one in (iii), so all three are redox. In (iv) barium stays plus two, chlorine stays minus one, hydrogen stays plus one and sulphur stays plus six on both sides, so no oxidation number changes and the reaction is not redox. The answer is (iv).

Hinglish: Pehli teen reactions mein oxidation numbers badalte hain, isliye woh redox hain. Chauthi reaction mein Ba^2+, Cl^-, H^+ aur S^6+ dono sides par same rehte hain, koi oxidation number change nahi hota, isliye woh redox nahi hai. Answer (iv).

Q8 (NCERT Exemplar, 1 mark)

Using the standard electrode potentials Fe3+/Fe2+ = +0.77 V, I2/I minus = +0.54 V, Cu2+/Cu = +0.34 V and Ag+/Ag = +0.80 V, find the pair between which the redox reaction is not feasible: (i) Fe3+ and I minus (ii) Ag+ and Cu (iii) Fe3+ and Cu (iv) Ag and Fe3+

Answer: A redox reaction is feasible when the reducing couple has a lower reduction potential than the oxidising couple. For (i) Fe3+/Fe2+ at 0.77 is above I2/I minus at 0.54, so Fe3+ oxidises iodide and the reaction is feasible. For (ii) Ag+ at 0.80 is above Cu2+/Cu at 0.34, so silver ions oxidise copper, feasible. For (iii) Fe3+ at 0.77 is above Cu2+/Cu at 0.34, feasible. For (iv) silver metal would reduce Fe3+, but Ag+/Ag at 0.80 is higher than Fe3+/Fe2+ at 0.77, so the reverse change would be needed and the reaction is not feasible. The answer is (iv).

Hinglish: Reaction feasible hoti hai jab oxidising couple ka potential reducing couple se zyada ho. (i), (ii), (iii) feasible hain. (iv) mein silver ko Fe^3+ ko reduce karna hoga, par Ag^+/Ag ka potential Fe^3+/Fe^2+ se zyada hai, isliye yeh reaction feasible nahi. Answer (iv).

Q9 (NCERT Exemplar, 1 mark)

Thiosulphate reacts differently with iodine and bromine: 2S2O3 2 minus + I2 giving S4O6 2 minus + 2I minus, and S2O3 2 minus + 2Br2 + 5H2O giving 2SO4 2 minus + 2Br minus + 10H plus. Which statement justifies this dual behaviour? (i) Bromine is a stronger oxidant than iodine (ii) Bromine is a weaker oxidant than iodine (iii) Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions (iv) Bromine undergoes oxidation and iodine undergoes reduction in these reactions

Answer: With iodine, thiosulphate is oxidised only to tetrathionate, while with bromine it is oxidised all the way to sulphate. The stronger oxidising agent can take the oxidation further, so the dual behaviour is explained by bromine being the stronger oxidant. Statement (i) is correct. The other statements either reverse the strength order or misdescribe which species is oxidised.

Hinglish: Iodine ke saath thiosulphate sirf tetrathionate tak oxidise hota hai, bromine ke saath sulphate tak. Iska matlab bromine zyada strong oxidant hai. Statement (i) sahi hai; baaki statements order galat ya oxidation ko ulat kar likhte hain.

Q10 (NCERT Exemplar, 1 mark)

Which of the following statements about oxidation number rules is not correct? (i) Hydrogen shows plus one in every compound it forms (ii) The algebraic total of the oxidation numbers in a compound comes out as zero (iii) A free element carries oxidation number zero (iv) Fluorine shows minus one in every compound

Answer: Hydrogen has a plus one oxidation number in most of its compounds, but in metal hydrides such as sodium hydride it has minus one. Claiming that it is always plus one is therefore wrong, so (i) is the incorrect rule. The sum rule for a compound, the zero for free elements and the minus one for fluorine in all its compounds are all correct statements.

Hinglish: Hydrogen aam taur par +1 hota hai, par metal hydrides jaise NaH mein -1 hota hai, isliye always plus one likhna galat hai. Compound ke sum ka zero hona, free element ka zero hona aur fluorine ka har compound mein -1 hona sahi rules hain. Answer (i).