It is noticed that Pb²⁺$Pb^{2+}$ is more stable than Pb⁴⁺$Pb^{4+}$ but Sn²⁺$Sn^{2+}$ is less stable than Sn⁴⁺$Sn^{4+}$. Observe the following reactions PbO₂ + Pb arrow 2PbO; ΔᵣG°(1)$PbO_{2} + Pb \rightarrow 2PbO; \Delta_{r}G^{\circ}(1)$ SnO₂ + Sn arrow 2SnO; ΔᵣG°(2)$SnO_{2} + Sn \rightarrow 2SnO; \Delta_{r}G^{\circ}(2)$ Identify the correct set from the following.

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 H₂Te$H_2Te$ is lower than that of H₂S$H_2S$. Because the Te-H$Te-H$ bond is weaker and more easily broken, it ionizes to release H^+$H^+$ ions more readily than H₂S$H_2S$.

Thus, H₂Te$H_2Te$ is more acidic than H₂S$H_2S$, 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.

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: NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃$NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$. 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: NH₃ < PH₃ < AsH₃ < SbH₃ < BiH₃$NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$. Statement B is correct.

Statement C: Following the logic from B, NH₃$NH_3$ is a very weak reducing agent (mild), while BiH₃$BiH_3$ 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 NH₃$NH_3$, the lone pair is concentrated in a small sp³$sp^3$ 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.

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 (Zeff$Z_{eff}$).

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).

Chapter Mix

Class 12 Chemistry: The p-Block Elements

Core Logic

(A) All group 16 elements form oxides of the EO₂$EO_2$ and EO₃$EO_3$ type where E = S$E = S$, Se, Te or Po. These are generally acidic in nature. (Correct) (B) SO₂$SO_2$ is reducing while TeO₂$TeO_2$ 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 H₂S$H_2S$ to H₂Te$H_2Te$ down the group because the E-H$E-H$ bond dissociation enthalpy decreases as atomic size increases. (Incorrect) (D) The ozone molecule (O₃$O_3$) 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.

Chapter Mix

Class 12 Chemistry: The p-Block Elements

Core Logic

Statement I: In the trivalent state, most compounds of Group 13 elements (like BCl₃$BCl_3$, AlCl₃$AlCl_3$) 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(H₂O)₆]³⁺$[Al(H_2O)_6]^{3+}$ ion where the hybridization state of Al is sp³d²$sp^3d^2$. Thus, Statement II is also true.

Step 1: Final Conclusion

Both Statement I and Statement II are correct, pointing to option (4).

Chapter Mix

Class 11 Chemistry: The p-Block Elements