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Given below are the two statements: one is labeled as Assertion (A) and the other is labeled as Reason (R). Assertion (A): There is a considerable increase in covalent radius from N to P. However from As to Bi only a small increase in covalent radius is observed. Reason (R): covalent and ionic radii in a particular oxidation state increases down the group. In the light of the above statement, choose the most appropriate answer from the options given below:

Solution & Explanation

### 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 (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). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The p-Block Elements

Reference Study Guides

The P Block Elements

More The p-Block Elements Previous-Year Questions — Page 4

Q80 jee_main_2024_29_january_evening Anomalous Behaviour of Oxygen
Anomalous behaviour of oxygen is due to its
  • A. Large size and high electronegativity
  • B. Small size and low electronegativity
  • C. Small size and high electronegativity
  • D. Large size and low electronegativity

Solution

### Related Formula \text{Anomalous properties of second-period elements.} ### Core Logic The anomalous properties of oxygen compared to other chalcogens stem directly from its position in the second period of the periodic table. It is characterized by: 1. An exceptionally **small atomic radius**. 2. Highly pronounced **electronegativity**. 3. Complete absence of low-energy valence d-orbitals. ### Step 1: Selection Verification Therefore, the combination of small size and high electronegativity is the correct choice, matching option (3). ### Pattern Recognition All first members of periodic blocks (textN, textO, textF) deviate significantly from their heavier group members due to their high charge density, high electronegativity, and lack of d-orbitals. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: p-Block Elements
Q73 jee_main_2024_27_jan_morning Properties of Boron
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Melting point of Boron (2453text K) is unusually high in group 13 elements. Reason (R) : Solid Boron has very strong crystalline lattice. In the light of the above statements, choose the most appropriate answer from the options given below;
  • A. Both (A) and (R) are correct but (R) is not the correct explanation of (A)
  • B. Both (A) and (R) are correct and (R) is the correct explanation of (A)
  • C. (A) is true but (R) is false
  • D. (A) is false but (R) is true

Solution

### Core Logic Boron forms a highly compact, robust icosahedral covalent polymeric three-dimensional framework structure (B_12 units). This extremely solid, dense crystalline lattice organization requires immense thermal activation energy to rupture, explaining why its melting point (2453text K) is uniquely elevated among Group 13 elements. Both statements are true and (R) is the perfect explanation. ### Chapter Mix Class 11 Chemistry: p-Block Elements
Q90 jee_main_2024_27_jan_morning Oxidation States of Sulphur
From the given list, the number of compounds with +4 oxidation state of Sulphur: textSO_3, textH_2textSO_3, textSOCl_2, textSF_4, textBaSO_4, textH_2textS_2textO_7
Numerical Answer. Answer: 3 to 3

Solution

### Step 1: Audit oxidation numbers individually
CompoundOxidation State of Sulphur Calculation
textSO_3x + 3(-2) = 0 implies x = +6
textH_2textSO_32(+1) + x + 3(-2) = 0 implies x = +4
textSOCl_2x + (-2) + 2(-1) = 0 implies x = +4
textSF_4x + 4(-1) = 0 implies x = +4
textBaSO_4+2 + x + 4(-2) = 0 implies x = +6
textH_2textS_2textO_72(+1) + 2x + 7(-2) = 0 implies 2x = 12 implies x = +6
### Step 2: Sum the targets The compounds displaying an exact +4 assignment are textH_2textSO_3, textSOCl_2, and textSF_4. The total number is 3. ### Pattern Recognition Sulfurous derivatives, thionyl groupings, and tetrafluoride configurations typically feature the +4 oxidation level state. ### Chapter Mix Class 11 Chemistry: Redox Reactions Class 12 Chemistry: p-Block Elements
Q65 jee_main_2024_29_jan_morning Group 14 Elements Physical Properties
Given below are two statements : Statement I : The electronegativity of group 14 elements from Si to Pb gradually decreases. Statement II : Group 14 contains non-metallic, metallic, as well as metalloid elements. In the light of the above statements, choose the most appropriate from the options given below:
  • A. textStatement I is false but Statement II is true
  • B. textStatement I is true but Statement II is false
  • C. textBoth Statement I and Statement II are true
  • D. textBoth Statement I and Statement II are false

Solution

### Core Logic **Analyzing Statement I:** The electronegativity values for Group 14 elements according to the Pauling scale are approximately: - Carbon (C): 2.5 - Silicon (Si): 1.8 - Germanium (Ge): 1.8 - Tin (Sn): 1.8 - Lead (Pb): 1.9 The electronegativity values from Si to Pb are almost identical, and it slightly increases at Pb due to the poor shielding effect of d and f-orbitals (inert pair effect). It does not "gradually decrease." Therefore, Statement I is false. **Analyzing Statement II:** Group 14 consists of: - Carbon (C): Non-metal - Silicon (Si) & Germanium (Ge): Metalloids - Tin (Sn) & Lead (Pb): Metals Therefore, the group contains non-metals, metalloids, and metals. Statement II is true. ### Step 1: Final Conclusion Statement I is false, but Statement II is true. ### Pattern Recognition Electronegativity in Group 13 and 14 does not follow a strict linear decrease due to d-block and f-block contraction (poor shielding by d and f electrons). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: The p Block Elements
Q65 jee_main_2024_30_january_evening Group 16 Elements
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: H_2Te is more acidic than H_2S. Reason R: Bond dissociation enthalpy of H_2Te is lower than H_2S. In the light of the above statements, Choose the most appropriate from the options given below.
  • A. textBoth A and R are true but R is NOT the correct explanation of A.
  • B. textBoth A and R are true and R is the correct explanation of A.
  • C. textA is false but R is true.
  • D. textA is true but R is false.

Solution

### 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 H_2Te is lower than that of H_2S. Because the Te-H bond is weaker and more easily broken, it ionizes to release H^+ ions more readily than H_2S. Thus, H_2Te is more acidic than 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The p Block Elements

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