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Given below are two statements: Statement I: The N-N single bond is weaker and longer than that of P-P single bond Statement II: Compounds of group 15 elements in +3 oxidation states readily undergo disproportionation reactions. In the light of above statements, choose the correct answer from the options given below:

Solution & Explanation

### Core Logic Let us analyze both statements systematically: * **Statement I:** The textN-textN single bond is indeed weaker than the textP-textP single bond due to high inter-electronic repulsion between non-bonding lone pairs on the small nitrogen atoms. However, nitrogen has a smaller atomic size than phosphorus, making the textN-textN single bond shorter (140text pm) compared to the textP-textP single bond (221text pm). Thus, Statement I is false because it incorrectly claims it is longer. * **Statement II:** In Group 15, only nitrogen and phosphorus compounds in the +3 oxidation state readily undergo disproportionation. As we go down the group (As, Sb, Bi), the +3 oxidation state becomes increasingly stable due to the inert pair effect, meaning they do not readily undergo disproportionation. Hence, Statement II is false as a general trend across all group 15 elements. ### Step 1: Verification of Conclusions Since the textN-textN bond is shorter and heavier elements in the +3 state do not disproportionate readily, both statements are evaluated to be false. ### Pattern Recognition Sees: "textN-textN single bond longer" ightarrow Absolute error. Small atoms form short bonds, always. Inert pair effect stabilizes +3 lower down the group, preventing disproportionation reactions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: p-Block Elements

Reference Study Guides

More The p-Block Elements Previous-Year Questions — Page 2

Q37 2025 Group 15 Elements
Given below are two statements: Statement I: Nitrogen forms oxides with +1 to +5 oxidation states due to the formation of ppi - ppi bond with oxygen. Statement II: Nitrogen does not form halides with +5 oxidation state due to the absence of d-orbital in it. In the light of given statements, choose the correct answer from the options given below:
  • A. textStatement I is true but Statement II is false
  • B. textBoth Statement I and Statement II are false.
  • C. textStatement I is false but Statement II is true.
  • D. textBoth Statement I and Statement II are true.

Solution

### Core Logic * **Statement I is true:** Nitrogen has a small atomic size and high electronegativity, allowing it to form strong multiple ppi - ppi bonds with oxygen atoms. This enables stable oxide forms across a wide range of oxidation numbers spanning +1 to +5 (e.g., N_2O_5). * **Statement II is true:** Nitrogen is a second-period element with a valence shell configuration of 2s^2 2p^3. It completely lacks vacant 2d orbitals. As a result, it cannot expand its octet beyond a covalency of 4, preventing the synthesis of pentahalides like NX_5. ### Pattern Recognition Second-period elements are structurally bounded by a maximum covalency of 4 due to the complete absence of d-orbitals. This explains why NF_5 is unstable/non-existent while PF_5 is easily synthesized. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The p-Block Elements
Q44 2025 Inert Pair Effect and Oxidation States
The correct statements from the following are: (A) textTl^3+ is a powerful oxidising agent (B) textAl^3+ does not get reduced easily (C) Both textAl^3+ and textTl^3+ are very stable in solution (D) textTl^+ is more stable than textTl^3+ (E) textAl^3+ and textTl^+ are highly stable Choose the correct answer from the options given below:
  • A. text(A), (B), (C), (D) and (E)
  • B. text(A), (B), (D) and (E) only
  • C. text(B), (D) and (E) only
  • D. text(A), (C) and (D) only

Solution

### Related Formula textInert pair effect implies textStability 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 electrons to participate in bonding increases. Thus, for Thallium (Tl), the +1 oxidation state is significantly more stable than the +3 oxidation state (textTl^+ > textTl^3+). This validates statement (D). [cite: 1020, 1032] - Because textTl^3+ is highly unstable, it eagerly captures two electrons to reduce to textTl^+, 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), meaning textAl^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+ 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), the lowest oxidation state (+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
Q38 2025 Group 16 Elements Physical Properties
The large difference between the melting and boiling points of oxygen and sulphur may be explained on the basis of
  • A. Atomic size
  • B. Atomicity
  • C. Electronegativity
  • D. Electron gain enthalpy

Solution

### Core Logic Oxygen exists naturally as a discrete diatomic element molecule system (O_2), displaying an atomicity count equal to 2. In contrast, sulphur forms a puckered multi-atom ring structure configuration (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 molecular setups form simple gaseous configurations, while S_8 molecular configurations establish thick, heavy crown-packed chains. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: The p-Block Elements

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