Redox reactions practice · Medium

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.

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.

Option D (8) is correct because the balanced redox reaction involves a loss of 8 electrons during the oxidation of iron.

Option A (+3) is correct because the oxidation number of N in NOCl is calculated assuming NO exists as $NO^+$ and balancing the charges.

Option A ($2 \mathrm{Pb}(\mathrm{NO}_3)_2 \rightarrow 2 \mathrm{PbO}+4 \mathrm{NO}_2+\mathrm{O}_2$) is correct because the oxidation state of nitrogen and oxygen changes within the same compound during decomposition.

Option C (2.5) is correct because the oxidation state of sulfur is calculated by considering the overall charge of the compound and the known oxidation states of other elements.

The correct answer is A ($\mathrm{K}_2 \mathrm{HPO}_3$) because it does not contain an O-H bond directly attached to a highly electronegative atom, which is necessary for hydrogen bonding.

Option B (2,0) is correct because the oxidation numbers of the carbon atoms in C3O2 are +2, 0, and +2, respectively.

Option D (Statement-I is incorrect but statement-II is correct) is correct because the equivalent weight of KMnO4 in acidic medium is M/5, but in strongly basic medium, it is M/1, and the electron gain is correctly stated.

Option C (${X_3}{\left( {Y{Z_4}} \right)_2}$) is correct because the sum of the oxidation numbers in the compound must be zero.

Option B (8/3) is correct because the average oxidation state of Fe in $F{e_3}{O_4}$ is calculated to be 8/3.

Option A is correct because both statements are true. The average oxidation number of sulfur in $\mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6$ is indeed 2.5, and two of the four sulfur atoms are not directly bonded to oxygen atoms.

Option C ($\mathrm{S}_2 \mathrm{O}_3^{-}$) is correct because thiosulfate ions react with iodine, which is responsible for the blue color with starch, causing the color to disappear.

Option A (0 to -1 and +5) is correct because in the reaction, chlorine undergoes both reduction (0 to -1) and oxidation (0 to +5).

Option C ($\mathrm{C}{\mathrm{u}_2}{\mathrm{I}_2} + {\mathrm{K}_2}\mathrm{S}{\mathrm{O}_4} + {\mathrm{I}_2}$) is correct because Cu2+ oxidizes I- to I2, and Cu2+ is reduced to Cu+ forming Cu2I2.

Option C ($+ 2,\mathrm{ } + 7$) is correct because +2 and +7 are the most stable oxidation states of manganese (Mn) due to stable electronic configurations.

Option D (+5) is correct because the oxidation number of phosphorus in all the given compounds is +5.

Option B (7.75) is correct because the equivalent mass is calculated by dividing the molar mass of nitrate by the number of electrons transferred in the reduction to ammonium.

Option B (Assertion is true, Reason is true and Reason is not correct explanation for Assertion) is correct because fluorine is a stronger oxidizing agent than iodine due to its higher reduction potential, and it also has greater electronegativity, but electronegativity is not the direct reason for its stronger oxidizing power.

Option B ($\mathrm{FeC}{\mathrm{l}_3}$) is correct because $\mathrm{FeC}{\mathrm{l}_3}$ cannot be further oxidized by $\mathrm{KMn}{\mathrm{O}_4}$ as iron is already in its highest oxidation state (+3).

Option B ($- 1$) is correct because hydrogen is more electronegative than calcium, so it has a -1 oxidation state.

Option C ($\mathrm{CrO}_4^{2-} \rightarrow \mathrm{Cr}_2 \mathrm{O}_7^{2-}$) is correct because the oxidation state of Cr remains +6 in both species.

Option C is correct because in the given reaction, the oxidation state of none of the elements changes.

Option C (If Assertion is True but the Reason is False) is correct because Cr+2 is indeed a reducing agent and Mn+3 is an oxidizing agent, but the reason given (Mn+3 having a d5 configuration) is not the correct explanation.

Option C (6) is correct because the value of 'n' is determined by balancing the redox reaction, specifically the number of electrons transferred during the oxidation and reduction processes.

Option A (5) is correct because the balanced half-reaction shows that 5 electrons are required to reduce MnO₄⁻ to Mn²⁺.