Given below are two statements: Statement I: H₂Se$\text{H}_2\text{Se}$ is more acidic than H₂Te$\text{H}_2\text{Te}$. Statement II: H₂Se$\text{H}_2\text{Se}$ has higher bond enthalpy for dissociation than H₂Te$\text{H}_2\text{Te}$. In the light of the above statements, choose the correct answer from the options given below:

Related Formula

Core Logic

Let's analyze the group 13 stability dynamics:

• Inert Pair Effect: Down Group 13, the reluctance of inner ns²$ns^2$ 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 (Tl^+ > Tl³⁺$\text{Tl}^+ > \text{Tl}^{3+}$). This validates statement (D). [cite: 1020, 1032]
• Because Tl³⁺$\text{Tl}^{3+}$ is highly unstable, it eagerly captures two electrons to reduce to Tl^+$\text{Tl}^+$, 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 (E⁰ = -1.66 V$E^0 = -1.66\text{ V}$), meaning Al³⁺$\text{Al}^{3+}$ 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 Tl³⁺$\text{Tl}^{3+}$ 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 Tl, 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.

Chapter Mix

Class 11 Chemistry: The p-Block Elements

Core Logic

Oxygen exists naturally as a discrete diatomic element molecule system (O₂$O_2$), displaying an atomicity count equal to 2. In contrast, sulphur forms a puckered multi-atom ring structure configuration (S₈$S_8$), 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

O₂$O_2$ molecular setups form simple gaseous configurations, while S₈$S_8$ molecular configurations establish thick, heavy crown-packed chains.

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:

• (NH₄)₂S$(NH_4)_2S$ (Ammonium sulfide solution / yellow ammonium sulfide compound matrix) arrow$\rightarrow$ Yellow
• PbS$PbS$ (Lead sulfide) arrow$\rightarrow$ Black
• CuS$CuS$ (Copper sulfide) arrow$\rightarrow$ Black
• As₂S₃$As_2S_3$ (Arsenic(III) sulfide) arrow$\rightarrow$ Yellow
• As₂S₅$As_2S_5$ (Arsenic(V) sulfide) arrow$\rightarrow$ 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 (As₂S₃$As_2S_3$). Transition metal sulfides like PbS$PbS$ and CuS$CuS$ typically form dark black precipitates.

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:

Core Logic

Let's analyze the properties specified:

• The element must be able to form dπ-dπ$d\pi-d\pi$ bonds with transition metals. Nitrogen cannot form these bonds because it lacks vacant d$d$-orbitals in its valence shell. Therefore, nitrogen is excluded.
• Among the remaining elements (Phosphorus, Arsenic, Antimony, Bismuth) that contain available d$d$-orbitals, basic strength decreases down the group. Phosphorus forms phosphine (PH₃$PH_3$), 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 dπ-dπ$d\pi-d\pi$ bonds'. This explicitly excludes second-period elements (like Nitrogen), making Phosphorus (Z=15$Z=15$) the top choice for basicity.

Chapter Mix

Class 12 Chemistry: p-Block Elements

Core Logic

NH₃$NH_3$ undergoes extensive intermolecular hydrogen bonding due to the high electronegativity and small size of Nitrogen. PH₃$PH_3$ (Phosphine) molecules are only held together by weak van der Waals (dispersion) forces because Phosphorus is less electronegative and larger, unable to form strong hydrogen bonds.

Step 1: Evaluate Statements

Assertion (A) is correct: PH₃$PH_3$ has a lower boiling point than NH₃$NH_3$ because breaking H-bonds in NH₃$NH_3$ requires more energy. Reason (R) is incorrect: It falsely claims NH₃$NH_3$ has van der Waals association and PH₃$PH_3$ has hydrogen bonding. It is exactly the opposite.

Pattern Recognition

N, O, and F are the only atoms electronegative enough to form stable hydrogen bonds in simple hydrides. Boiling point anomaly: NH₃ > PH₃$NH_3 > PH_3$ purely due to H-bonding in NH₃$NH_3$.

Chapter Mix

Class 12 Chemistry: The p-Block Elements Class 11 Chemistry: Chemical Bonding and Molecular Structure