### Core Logic Statement A: As we move down Group 15, the size of the central atom increases, causing the E-H bond length to increase and its bond strength to decrease. Thus, thermal stability decreases in the order: NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$. Statement A is correct. Statement B: Because the E-H bond strength decreases down the group, it becomes easier to release hydrogen. Therefore, the reducing ability increases down the group: NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$. Statement B is correct. Statement C: Following the logic from B, NH_3$NH_3$ is a very weak reducing agent (mild), while BiH_3$BiH_3$ is the strongest reducing agent among them. Statement C is false. Statement D: Basicity depends on the availability of the lone pair for donation. In NH_3$NH_3$, the lone pair is concentrated in a small sp^3$sp^3$ orbital (high electron density). Moving down the group, the lone pair occupies larger orbitals with more s-character (Drago's rule), decreasing electron density and making the lone pair less available. Thus, basicity decreases down the group. Statement D is false. ### Step 1: Final Conclusion Only statements A and B are correct. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The p Block Elements

### Core Logic Assertion (A) is a factual statement directly from NCERT: In Group 15, there is a large increase in covalent radius from Nitrogen to Phosphorus, but from Arsenic to Bismuth, the increase is very small. Reason (R) states that covalent and ionic radii increase down the group. This is a generally true statement. ### Step 1: Explaining the discrepancy While (R) is a true general trend, it does not correctly explain the anomaly highlighted in (A). The correct explanation for the small increase in radius from As to Bi is the poor shielding effect of completely filled d and f-orbitals in the heavier members, leading to a higher effective nuclear charge (Z_eff$Z_{eff}$). ### Final Conclusion Therefore, both (A) and (R) are true, but (R) is not the correct explanation for (A). ### Pattern Recognition Whenever radius anomalies occur in heavier p-block elements (like small increases), the underlying cause is almost always d-block or f-block contraction (poor shielding). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The p-Block Elements

### Core Logic Statement I: S_8$S_8$ disproportionates in alkaline medium to give sulphide (S^2-$S^{2-}$) and thiosulphate (S_2O_3^2-$S_2O_3^{2-}$). The reaction is: S_8 + 12OH^- rightarrow 4S^2- + 2S_2O_3^2- + 6H_2O$S_8 + 12OH^- \rightarrow 4S^{2-} + 2S_2O_3^{2-} + 6H_2O$ This is a correct statement. Statement II: ClO_4^-$ClO_4^-$ contains chlorine in its highest possible oxidation state of +7$+7$. A species can undergo disproportionation only if the element can exist in both higher and lower oxidation states than the present one. Since chlorine cannot exceed +7$+7$, ClO_4^-$ClO_4^-$ cannot undergo disproportionation. This is an incorrect statement. ### Step 1: Final Conclusion Statement I is correct, and Statement II is incorrect. Therefore, option (1) is the correct answer. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Redox Reactions Class 12 Chemistry: The p-Block Elements

### Core Logic (A) All group 16 elements form oxides of the EO_2$EO_2$ and EO_3$EO_3$ type where E = S$E = S$, Se, Te or Po. These are generally acidic in nature. (Correct) (B) SO_2$SO_2$ is reducing while TeO_2$TeO_2$ is an oxidising agent due to the inert pair effect increasing the stability of lower oxidation states down the group. (Correct) (C) The reducing property increases from H_2S$H_2S$ to H_2Te$H_2Te$ down the group because the E-H$E-H$ bond dissociation enthalpy decreases as atomic size increases. (Incorrect) (D) The ozone molecule (O_3$O_3$) has the structure O=O^+-O^-$O=O^+-O^-$. Counting the lone pairs: the central oxygen has 1, the double-bonded oxygen has 2, and the single-bonded negatively charged oxygen has 3. Total lone pairs = 1 + 2 + 3 = 6$1 + 2 + 3 = 6$. (Incorrect) ### Step 1: Final Selection Since A and B are the only correct statements, option 4 is the right choice. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The p-Block Elements

### Core Logic Statement I: In the trivalent state, most compounds of Group 13 elements (like BCl_3$BCl_3$, AlCl_3$AlCl_3$) are covalent due to small cation size and high charge, and they are readily hydrolysed in water. Thus, Statement I is true. Statement II: Trichlorides on hydrolysis in normal water can form tetrahedral [M(OH)_4]^-$[M(OH)_4]^-$ species depending on pH, but in an acidified aqueous solution, aluminium forms the octahedral [Al(H_2O)_6]^3+$[Al(H_2O)_6]^{3+}$ ion where the hybridization state of Al is sp^3d^2$sp^3d^2$. Thus, Statement II is also true. ### Step 1: Final Conclusion Both Statement I and Statement II are correct, pointing to option (4). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: The p-Block Elements