Given below are two statements : Statement I: In the oxalic acid vs mathrmKMnO_4$\mathrm{KMnO}_4$ (in the presence of dil mathrmH_2mathrmSO_4$\mathrm{H}_2\mathrm{SO}_4$ ) titration the solution needs to be heated initially to 60^circmathrmC$60^{\circ}\mathrm{C}$ , but no heating is required in Ferrous ammonium sulphate (FAS) vs mathrmKMnO_4$\mathrm{KMnO}_4$ titration (in the presence of dil mathrmH_2mathrmSO_4$\mathrm{H}_2\mathrm{SO}_4$ ) Statement II : In oxalic acid vs mathrmKMnO_4$\mathrm{KMnO}_4$ titration, the initial formation of mathrmMnSO_4$\mathrm{MnSO}_4$ takes place at high temperature, which then acts as catalyst for further reaction. In the case of FAS vs mathrmKMnO_4$\mathrm{KMnO}_4$ , heating oxidizes mathrmFe^2+$\mathrm{Fe}^{2+}$ into mathrmFe^3+$\mathrm{Fe}^{3+}$ by oxygen of air and error may be introduced in the experiment. In the light of the above statements, choose the correct answer from the options given below:

### Core Logic In the binary compound A_2B$A_2B$, the atom 'B' exhibits a -2 oxidation state, which is its lowest possible oxidation state (meaning it has achieved a stable noble gas configuration by accepting 2 electrons). A_2B rightarrow 2A^+ + B^2-$$A_2B \rightarrow 2A^+ + B^{2-}$$ If the element B gains 2 electrons to complete its octet (8 electrons), it must originally have 8 - 2 = 6$8 - 2 = 6$ electrons in its valence shell in the neutral atomic state. ### Step 1: Determine Valence Electrons B^2-$B^{2-}$ has a complete octet. Neutral state B = 8 - 2 = 6$8 - 2 = 6$ valence electrons. (This corresponds to Group 16 elements like Oxygen or Sulfur). ### Pattern Recognition Lowest negative oxidation state = (Group Number - 18) for p-block. If lowest ox state is -2, then Group number is 18 - 2 = 16$18 - 2 = 16$. Group 16 elements have 6 valence electrons. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Redox Reactions Class 11 Chemistry: Classification of Elements and Periodicity in Properties

### Core Logic The disproportionation of Cl_2$Cl_2$ in cold, dilute alkaline medium yields chloride (Cl^-$Cl^-$) and hypochlorite (ClO^-$ClO^-$) ions. Oxidation half-reaction: Cl_2 rightarrow 2ClO^- + 2e^-$Cl_2 \rightarrow 2ClO^- + 2e^-$ Balancing O and H in basic medium: Cl_2 + 4OH^- rightarrow 2ClO^- + 2H_2O + 2e^-$Cl_2 + 4OH^- \rightarrow 2ClO^- + 2H_2O + 2e^-$ Reduction half-reaction: Cl_2 + 2e^- rightarrow 2Cl^-$Cl_2 + 2e^- \rightarrow 2Cl^-$ Adding both half-reactions (electrons are already equal): 2Cl_2 + 4OH^- rightarrow 2ClO^- + 2Cl^- + 2H_2O$2Cl_2 + 4OH^- \rightarrow 2ClO^- + 2Cl^- + 2H_2O$ Dividing the entire equation by 2 to get the simplest integer coefficients: 1Cl_2 + 2OH^- rightarrow 1ClO^- + 1Cl^- + 1H_2O$1Cl_2 + 2OH^- \rightarrow 1ClO^- + 1Cl^- + 1H_2O$ ### Step 1: Coefficient Matching

By comparing with the given equation: a Cl_2 + b OH^- rightarrow c ClO^- + d Cl^- + e H_2O$a Cl_2 + b OH^- \rightarrow c ClO^- + d Cl^- + e H_2O$ We get: a = 1$a = 1$ b = 2$b = 2$ c = 1$c = 1$ d = 1$d = 1$ Thus, the values are 1, 2, 1, and 1. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Redox Reactions Class 12 Chemistry: The p Block Elements

### Core Logic For a species to undergo disproportionation, it must contain an element that is present in an intermediate oxidation state. This allows it to act both as an oxidizing agent (by being reduced to a lower state) and a reducing agent (by being oxidized to a higher state). Let's evaluate each species: 1. H_2O_2$H_2O_2$: Oxygen is in -1$-1$. Can go to 0$0$ (O_2$O_2$) and -2$-2$ (H_2O$H_2O$). (Yes) 2. ClO_3^-$ClO_3^-$: Chlorine is in +5$+5$. Can go to +7$+7$ (ClO_4^-$ClO_4^-$) and -1$-1$ (Cl^-$Cl^-$). (Yes) 3. P_4$P_4$: Phosphorus is in 0$0$. Can go to -3$-3$ (PH_3$PH_3$) and +1/+3/+5$+1/+3/+5$. (Yes) 4. Cl_2$Cl_2$: Chlorine is in 0$0$. Can go to -1$-1$ (Cl^-$Cl^-$) and +1$+1$ (ClO^-$ClO^-$). (Yes) 5. Ag$Ag$: Metal in 0$0$ state. Cannot show negative oxidation state. (No) 6. Cu^+1$Cu^{+1}$: Copper is in +1$+1$. Can go to 0$0$ (Cu$Cu$) and +2$+2$ (Cu^2+$Cu^{2+}$). (Yes) 7. F_2$F_2$: Fluorine is the most electronegative, only shows 0$0$ and -1$-1$. Cannot be oxidized to a positive state. (No) 8. NO_2$NO_2$: Nitrogen is in +4$+4$. Can go to +5$+5$ (HNO_3$HNO_3$) and +3$+3$ (HNO_2$HNO_2$). (Yes) 9. K^+$K^+$: Potassium is in its highest oxidation state +1$+1$. Cannot be oxidized further. (No) ### Step 1: Final Count The species that can undergo disproportionation are: H_2O_2$H_2O_2$, ClO_3^-$ClO_3^-$, P_4$P_4$, Cl_2$Cl_2$, Cu^+1$Cu^{+1}$, and NO_2$NO_2$. Total count is 6. ### Pattern Recognition Rule out maximum oxidation states (K^+$K^+$), minimum oxidation states, and elements like F_2$F_2$ which never show positive oxidation states. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Redox Reactions

### Core Logic This is a redox reaction in a slightly basic/neutral medium (as indicated by MnO_2$MnO_2$ product). Separate the reaction into two half-cells and balance using the ion-electron method. ### Step 1: Reduction Half-reaction MnO_4^- rightarrow MnO_2$$MnO_4^- \rightarrow MnO_2$$ Balance O by adding H_2O$H_2O$ and H by adding OH^-$OH^-$ (or balance with H^+$H^+$ then convert): MnO_4^- + 2H_2O rightarrow MnO_2 + 4OH^-$$MnO_4^- + 2H_2O \rightarrow MnO_2 + 4OH^-$$ Balance charge by adding e^-$e^-$: MnO_4^- + 2H_2O + 3e^- rightarrow MnO_2 + 4OH^- dots (1)$$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^- \dots (1)$$ ### Step 2: Oxidation Half-reaction I^- rightarrow I_2$$I^- \rightarrow I_2$$ Balance I: 2I^- rightarrow I_2$$2I^- \rightarrow I_2$$ Balance charge by adding e^-$e^-$: 2I^- rightarrow I_2 + 2e^- dots (2)$$2I^- \rightarrow I_2 + 2e^- \dots (2)$$ ### Step 3: Combining halves To cancel electrons, multiply equation (1) by 2 and equation (2) by 3: 2[MnO_4^- + 2H_2O + 3e^- rightarrow MnO_2 + 4OH^-]$$2[MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-]$$ 3[2I^- rightarrow I_2 + 2e^-]$$3[2I^- \rightarrow I_2 + 2e^-]$$ Add them together: 2MnO_4^- + 6I^- + 4H_2O rightarrow 2MnO_2 + 3I_2 + 8OH^-$$2MnO_4^- + 6I^- + 4H_2O \rightarrow 2MnO_2 + 3I_2 + 8OH^-$$ ### Step 4: Conclusion Comparing this with the given equation: 2MnO_4^- + bI^- + cH_2O rightarrow xI_2 + yMnO_2 + zOH^-$$2MnO_4^- + bI^- + cH_2O \rightarrow xI_2 + yMnO_2 + zOH^-$$ We see b = 6, c = 4, x = 3, y = 2, z = 8$b = 6, c = 4, x = 3, y = 2, z = 8$. Therefore, the value of z is 8. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Redox Reactions

### Related Formula 2MnO_4^- + 5C_2O_4^2- + 16H^+ rightarrow 2Mn^2+ + 10CO_2 + 8H_2O$$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$$ ### Core Logic From the balanced redox equation in an acidic medium, we can see the exact stoichiometry between permanganate, oxalate, and hydrogen ions. For 2$2$ moles of MnO_4^-$MnO_4^-$, 16$16$ moles of H^+$H^+$ are required. ### Step 1: Calculating for 1 mole For 1$1$ mole of MnO_4^-$MnO_4^-$, the number of moles of H^+$H^+$ ions required is: frac162 = 8$$\frac{16}{2} = 8$$ Note: The official NTA answer was given as 4 initially, but our experts confirm the standard balanced stoichiometry requires 8 moles of H^+$H^+$. We are outputting the chemically correct value of 8. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Redox Reactions Class 12 Chemistry: The d- and f-Block Elements