Redox reactions practice · Mixed difficulty

Option B is correct because N2O5, upon hydrolysis, only forms HNO3, where nitrogen remains in the +5 oxidation state, thus not undergoing disproportionation.

Option C is correct because phosphorus undergoes both oxidation and reduction in this reaction, which is the definition of a disproportionation reaction.

Option D (SO2 and SO3) is correct because the difference in oxidation numbers of S in SO2 and SO3 is 2, which is the greatest among the given pairs.

Option D ($5/4$) is correct because it represents the number of moles of the potassium salt oxidized by one mole of permanganate ion based on the balanced redox reaction.

Option B (1) is correct because in $CrO_5$, there is one oxygen atom with a -2 oxidation state and four oxygen atoms involved in peroxide linkages.

Option B (1) is correct because in $CrO_5$, there is one oxygen atom with a -2 oxidation state and four oxygen atoms involved in peroxide linkages.

Option B (6) is correct because when a metal ion M³⁺ loses 3 electrons, its oxidation state increases by 3, resulting in +6.

The correct answer is D because bromine undergoes both oxidation and reduction in the reaction, acting as both an oxidizing and reducing agent (disproportionation).

Option D (III in haematite and II, III in magnetite) is correct because in haematite (Fe2O3), iron is in the +3 oxidation state, and in magnetite (Fe3O4), iron is present in both +2 and +3 oxidation states.

Option A (1/5 mole) is correct because the balanced redox reaction shows that 2 moles of $MnO_4^ -$ are required to oxidize 10 moles of iodide ions, thus 1/5 mole is needed for one mole of iodide.

Option B ($\mathrm{I}_2 < \mathrm{Br}_2 < \mathrm{Cl}_2 < \mathrm{F}_2$) is correct because the oxidizing power of halogens decreases down the group due to decreasing electronegativity and increasing atomic size.

Option C is correct because in the reaction, water gains oxygen, indicating it is being oxidized.

Option A (5) is correct because the manganese oxidation state changes from +7 to +2, requiring 5 electrons to be added to the left side of the equation.

Option A is correct because in the given reaction, MnO4^2- undergoes both oxidation and reduction, thus it is a disproportionation reaction.

Option A ($\mathrm{CrO}_4^{2 - } \to \mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }$) is correct because the oxidation state of Cr remains +6 on both sides of the reaction.

Option D ($\mathrm{Fe}$) is correct because iron (Fe) is a transition metal that commonly exhibits multiple oxidation states, such as +2 and +3.

The correct answer is A (1) because, based on the stoichiometry and equilibrium calculations, the Kp for the reaction is found to be 1.

The correct answer is D (Statement-I is incorrect but statement-II is correct) because the oxidation number of Ni in [Ni(CO)4] is 0, not +2, and CO is a neutral ligand.

Option A ($\mathrm{MnO}_4^ - \to \mathrm{M}{\mathrm{n}^{2 + }}$) is correct because the oxidation state of Mn changes from +7 to +2, involving a transfer of 5 electrons.

Option C is correct. In the reaction Na₂O₂ + H₂SO₄ → Na₂SO₄ + H₂O₂, the oxidation states of all elements remain unchanged.

Option C ($+ 5$) is correct because the oxidation number of P is calculated by setting up an equation based on the overall charge of the ion.

Option B is correct because both the assertion and reason are true, but the reason is not the correct explanation for the assertion. The oxidation number of carbon in HCHO is zero, and HCHO is a covalent compound, but the latter doesn't explain the former.

Option B (–3 to +5) is correct because nitrogen can exhibit oxidation states ranging from -3 (e.g., in NH3) to +5 (e.g., in N2O5).

Option B ($\mathrm{Gd}^{2+}$) is the best reducing agent because it can easily lose an electron to achieve the stable half-filled $f^7$ configuration of $\mathrm{Gd}^{3+}$.

Option A (increases) is correct because, across a period, the number of valence electrons increases, leading to a higher maximum oxidation number.

Option C ($\mathrm{HN_3}$) is correct because the oxidation number of nitrogen is fractional in hydrazoic acid.

Option D ($KIO_4$) is correct because iodine has the highest oxidation number (+7) in this compound.

The correct answer is D because the balanced equation shows a loss of 8 electrons during the oxidation of iron to form Fe3O4.

Option B (0) is correct because fluorine is the most electronegative element and can only have a negative oxidation state or zero when it is in its elemental form.

Option D (Both (A) and (R) are false) is correct because redox reactions involve electron transfer, while neutralization reactions involve the reaction of an acid and a base.