### Related Formula textInert pair effect implies textStability of (+n-2) text oxidation state increases down the main p-block groups.$\text{Inert pair effect} \implies \text{Stability of } (+n-2) \text{ oxidation state increases down the main p-block groups.}$ ### Core Logic Let's analyze the group 13 stability dynamics: - **Inert Pair Effect**: Down Group 13, the reluctance of inner ns^2$ns^2$ electrons to participate in bonding increases. Thus, for Thallium (Tl), the +1$+1$ oxidation state is significantly more stable than the +3$+3$ oxidation state (textTl^+ > textTl^3+$\text{Tl}^+ > \text{Tl}^{3+}$). This validates statement (D). [cite: 1020, 1032] - Because textTl^3+$\text{Tl}^{3+}$ is highly unstable, it eagerly captures two electrons to reduce to textTl^+$\text{Tl}^+$, acting as a **powerful oxidizing agent**, verifying statement (A). [cite: 1020, 1023] - Aluminum is small and highly electropositive. Its standard reduction potential is heavily negative (E^0 = -1.66text V$E^0 = -1.66\text{ V}$), meaning textAl^3+$\text{Al}^{3+}$ resists reduction and remains highly stable in solution, validating statements (B) and (E). [cite: 1026, 1027, 1038] ### Step 1: Eliminating Flawed Entries Statement (C) states that both are highly stable in solution, which is false since textTl^3+$\text{Tl}^{3+}$ is highly unstable and readily oxidizes surrounding species. Thus, the valid statements are (A), (B), (D), and (E) only. ### Pattern Recognition Inert pair shortcuts: For heavy p-block blocks (like textTl, Pb, Bi$\text{Tl, Pb, Bi}$), the lowest oxidation state (+1, +2, +3$+1, +2, +3$ respectively) is always favored over the maximum group valence. Consequently, their high-valence ions act as excellent oxidizers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: The p-Block Elements

### Core Logic Oxygen exists naturally as a discrete diatomic element molecule system (O_2$O_2$), displaying an atomicity count equal to 2. In contrast, sulphur forms a puckered multi-atom ring structure configuration (S_8$S_8$), presenting a larger atomicity value equal to 8. This high structural molecular mass significantly amplifies the surface area available for London dispersion forces. This creates much stronger intermolecular van der Waals attractions within molecular configurations of sulphur, accounting for its significantly elevated thermal properties relative to gaseous oxygen molecules. ### Pattern Recognition O_2$O_2$ molecular setups form simple gaseous configurations, while S_8$S_8$ molecular configurations establish thick, heavy crown-packed chains. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: The p-Block Elements

### Related Formula Specific metal sulfide precipitates exhibit characteristic colors used in qualitative inorganic analysis schemes. ### Core Logic Evaluating the colors of the specified inorganic sulfides: - (NH_4)_2S$(NH_4)_2S$ (Ammonium sulfide solution / yellow ammonium sulfide compound matrix) rightarrow$\rightarrow$ **Yellow** - PbS$PbS$ (Lead sulfide) rightarrow$\rightarrow$ **Black** - CuS$CuS$ (Copper sulfide) rightarrow$\rightarrow$ **Black** - As_2S_3$As_2S_3$ (Arsenic(III) sulfide) rightarrow$\rightarrow$ **Yellow** - As_2S_5$As_2S_5$ (Arsenic(V) sulfide) rightarrow$\rightarrow$ **Yellow** ### Step 1: Identifying Valid Items The sulfides matching the yellow color description are (A), (D), and (E). ### Pattern Recognition In qualitative salt analysis, arsenic belongs to Group IIB and precipitates as a bright yellow sulfide (As_2S_3$As_2S_3$). Transition metal sulfides like PbS$PbS$ and CuS$CuS$ typically form dark black precipitates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: p-Block Elements

### Related Formula Basic strength order among Group 15 hydrides drops down the group due to an increase in size and a decrease in charge density: NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$ ### Core Logic Let's analyze the properties specified: 1. The element must be able to form dpi-dpi$d\pi-d\pi$ bonds with transition metals. Nitrogen cannot form these bonds because it lacks vacant d$d$-orbitals in its valence shell. Therefore, nitrogen is excluded. 2. Among the remaining elements (Phosphorus, Arsenic, Antimony, Bismuth) that contain available d$d$-orbitals, basic strength decreases down the group. Phosphorus forms phosphine (PH_3$PH_3$), which is the strongest base among the remaining members. ### Step 1: Identify the Atomic Number The identified element is Phosphorus (P$P$). The atomic number of Phosphorus is 15$15$. ### Pattern Recognition Pay attention to qualifying statements like *'among elements that form dpi-dpi$d\pi-d\pi$ bonds'*. This explicitly excludes second-period elements (like Nitrogen), making Phosphorus (Z=15$Z=15$) the top choice for basicity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: p-Block Elements