Redox reactions practice · Mixed difficulty

Option B ($O{F_2} > {O_3} > K{O_2} > Ba{O_2}$) is correct because it represents the correct decreasing order of oxidation states of oxygen in the given compounds.

The correct answer is A because the oxidation state of phosphorus changes from 0 to +1 in KH2PO2 (oxidation) and from 0 to -3 in PH3 (reduction).

Option B (– 2) is correct because the overall charge of the complex ion is calculated by considering the oxidation state of Nickel and the charges of the cyanide ligands.

Option B is correct because the strongest oxidizing agent has the highest reduction potential, and the weakest has the lowest. Ag+/Ag has the highest, and Mg2+/Mg has the lowest.

Option C (+3, +4, +5) is correct because it accurately calculates the oxidation number of P in each compound based on the standard rules for assigning oxidation numbers.

The correct answer is D (If both Assertion & Reason are False) because MnO4- can be reduced to different oxidation states, and decrease in oxidation number is reduction, not oxidation.

The correct answer is A (${Mn}^{2+}<{Cl}^{-}<{Cr}^{3+}<{Cr}$) because the reducing power increases as the reduction potential decreases.

The correct answer is 4 because H_2O_2, Cu⁺, Cl_2, and ClO_3^- can undergo disproportionation reactions.

Option B ($\mathrm{Mg}^{2+}$) is correct because the strongest oxidizing agent is the one with the highest reduction potential.

Option A (0) is correct because when K2Cr2O7 is converted to K2CrO4, the oxidation state of chromium remains +6 in both compounds.

Option C (6) is correct because an element with a common oxidation state of -2 typically has 6 valence electrons in its outermost shell, as it needs to gain 2 electrons to achieve a stable octet.

Option A ((a) and (b) only) is correct because reactions (a) and (b) are disproportionation reactions where the same element is both oxidized and reduced.

The correct answer is D (All of these) because higher reduction potential corresponds to a stronger oxidizing agent, lower oxidation potential means a stronger reducing agent, and the oxidation state of oxygen in O3 is 0.

Option D (Both 1 and 2) is correct because zinc reduces copper ions to copper metal, which deposits on the zinc rod, and the zinc rod dissolves as it is oxidized.

Option A (3.55) is correct because the percentage of available chlorine is calculated based on the stoichiometry of the reaction and the amount of hypo used.

Option C ($Steam + \mathrm{ }iron\mathrm{ } \to \mathrm{ }hydrogen\mathrm{ } + \mathrm{ }iron\mathrm{ }oxide$) is correct because the oxidation state of oxygen in steam changes, indicating reduction.

Option A is correct because both the assertion and reason are true, and the reason correctly explains why oxygen has a +2 oxidation state in ${F}_2 {O}$.

Option A ($\mathrm{F}_2$) is correct because fluorine ($F_2$) has the highest standard reduction potential ($E^{\circ}=+2.5 \mathrm{~V}$), making it the strongest oxidizing agent.

Option C (3) is correct because NH3, NaN3 and Mg3N2 have nitrogen in negative oxidation state.

Option D (3) is correct because the +3 oxidation state is the most stable and common oxidation state among the lanthanoids due to the stability of the resulting electronic configurations.

Option B (Oxidation number method) is correct because the oxidation number method assumes complete electron transfer from less to more electronegative atoms.

Option A (5) is correct because the following species can reduce $\mathrm{Fe}^{+3}(aq)$ to $\mathrm{Fe}^{+2}(aq)$: $(\mathrm{NH}_4)_2 \mathrm{~S}, \mathrm{HI}, \mathrm{Sn}^{+2}(aq), \mathrm{SO}_2, \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$.

The correct answer is A (F) because fluorine is the strongest oxidizing agent and cannot be oxidized by MnO2.

Option B (2) is correct because only F and Na exhibit only one non-zero oxidation state.