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The elements of Group 13 with highest and lowest first ionisation enthalpies are respectively:

Solution & Explanation

### Related Formula textGroup 13 Ionisation Enthalpy Order: B > Tl > Ga > Al > In ### Core Logic The first ionisation enthalpy trend for Group 13 elements is irregular due to poor shielding by d and f electrons: - **Boron (B)** is the smallest atom in the group with no d-orbital shielding issues, so it has the **highest** ionisation enthalpy. - As we move down, poor shielding by 3d electrons causes a slight increase at Gallium (Ga). Poor shielding by 4f electrons causes a sharp increase at Thallium (Tl). - **Indium (In)** ends up with the weakest effective attraction for its outermost valence electron, giving it the **lowest** first ionisation enthalpy. Therefore, the highest and lowest elements are **B & In** respectively. ### Pattern Recognition Group 13 does not follow a linear downward trend. Remember the characteristic 'W' shape or zig-zag pattern of its ionisation energies. Indium sits at the absolute minimum point of this curve. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: The 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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