Redox reactions practice · Mixed difficulty

Option D (${ClF}_3$) is correct because the oxidation number of Cl in ${ClF}_3$ is +3, calculated by setting the sum of oxidation numbers to zero.

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

Option B (4.8 g) is correct because it is calculated based on the stoichiometry of the decomposition reaction of KClO3.

Option C ($+ 1, + 1\,and\, +1$) is correct because alkali metals always have an oxidation state of +1 in their compounds.

Option A (Sn(s)) is correct because Sn(s) is oxidized from oxidation state 0 to +4, thus it is the reducing agent.

Option D (+2 and +1 respectively) is correct because fluorine is more electronegative than oxygen, leading to positive oxidation states for oxygen in these compounds.

Option B ($-1$ to $+7$) is correct because iodine, having 7 valence electrons, can exhibit oxidation states ranging from -1 to +7.

Option B ($\mathrm{M}{\mathrm{n}^{2 + }}$) is correct because in acidic medium, permanganate ion ($\mathrm{KMn}{\mathrm{O}_4}$) is reduced to manganese(II) ion ($\mathrm{M}{\mathrm{n}^{2 + }}$).

Option D ($\mathrm{MnO}_4^{-}$) is correct because manganese is in its highest oxidation state (+7) and can only be reduced, thus acting as an oxidizing agent.

Option D ($\mathrm{CaCO}_3$ (s) $\stackrel{\Delta}{\longrightarrow} \mathrm{CaO}$ (s) $+\mathrm{CO}_2(\mathrm{g})$) is correct because there is no change in oxidation state for any of the elements in this decomposition reaction.

Option B is correct because the reduction of $\mathrm{BrO}_3^ -$ to $\mathrm{Br}{\mathrm{O}^ - }$ involves a decrease in oxidation state, requiring a reducing agent to donate electrons.

Option A (electrodes) is correct because electrodes are conductors through which electricity enters or leaves an object, substance, or device.

Option D (All of these) is correct because $\mathrm{Cl}_2$, $\mathrm{MnO}_4^{2-}$, and $\mathrm{NO}_2$ all undergo disproportionation in an alkaline medium, where the same element is both oxidized and reduced.

Option D is correct because in the reaction SnCl2 + HgCl2 -> SnCl4 + Hg, Sn is oxidized and Hg is reduced, indicating a redox reaction.

Option B is correct because, without s-p mixing, ${O}_2$ has unpaired electrons in its $\pi^*$ antibonding orbitals, making it paramagnetic.

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

Option A (metal displacement reaction) is correct because zinc displaces copper from copper sulfate.

Option A ($\mathrm{CaC}{\mathrm{O}_3} \to \mathrm{CaO} + \mathrm{C}{\mathrm{O}_2}$) is correct because there is no change in the oxidation states of any element during the reaction.

Option C (Will change to dark green) is correct because $Cr^{6+}$ is reduced to $Cr^{3+}$ which is green in color.

Option D ($\underline{\mathrm{N}_2} \mathrm{O}, \mathrm{H}_3 \underline{\mathrm{PO}_2}$) is correct because the oxidation number of N in N2O and P in H3PO2 are both +1.

Option A is correct because it accurately defines the equivalent weight of a reductant or oxidant as the molecular weight divided by the number of electrons lost or gained per molecule.

Option D (All of the above) is correct because oxidants decrease in oxidation number, gain electrons, and oxidize others while reducing themselves.

Option B (+1) is correct because iron commonly exhibits oxidation states of 0, +2, and +3, but +1 is less frequently observed in its stable compounds.

Option A (+6) is correct because CrO5 has a butterfly structure with two peroxide linkages, leading to +6 oxidation state for Cr.

Option D (All of the above) is correct because a good indicator should exhibit a sharp and clear color change and be sensitive to the equivalence point.