Consider the following elements In, Tl, Al, Pb, Sn and Ge. The most stable oxidation states of elements with highest and lowest first ionisation enthalpies, respectively, are

### Core Logic **Analyzing Statement I:** The electronegativity values for Group 14 elements according to the Pauling scale are approximately: - Carbon (C): 2.5$2.5$ - Silicon (Si): 1.8$1.8$ - Germanium (Ge): 1.8$1.8$ - Tin (Sn): 1.8$1.8$ - Lead (Pb): 1.9$1.9$ The electronegativity values from Si to Pb are almost identical, and it slightly increases at Pb due to the poor shielding effect of d and f-orbitals (inert pair effect). It does not "gradually decrease." Therefore, Statement I is false. **Analyzing Statement II:** Group 14 consists of: - Carbon (C): Non-metal - Silicon (Si) & Germanium (Ge): Metalloids - Tin (Sn) & Lead (Pb): Metals Therefore, the group contains non-metals, metalloids, and metals. Statement II is true. ### Step 1: Final Conclusion Statement I is false, but Statement II is true. ### Pattern Recognition Electronegativity in Group 13 and 14 does not follow a strict linear decrease due to d-block and f-block contraction (poor shielding by d and f electrons). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: The p Block Elements

### Core Logic As we move down Group 16, the atomic size of the central atom increases. The increased size of Tellurium compared to Sulphur leads to a longer and weaker Element-Hydrogen bond. Consequently, the bond dissociation enthalpy of H_2Te$H_2Te$ is lower than that of H_2S$H_2S$. Because the Te-H$Te-H$ bond is weaker and more easily broken, it ionizes to release H^+$H^+$ ions more readily than H_2S$H_2S$. Thus, H_2Te$H_2Te$ is more acidic than H_2S$H_2S$, making both the assertion and reason true, with the reason correctly explaining the assertion. ### Pattern Recognition Down the group for p-block hydrides: Size increases → Bond length increases → Bond strength decreases → Acidity increases. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The p Block Elements

### 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