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Give below are two statements: Statement-I: Noble gases have very high boiling points. Statement-II: Noble gases are monoatomic gases. They are held together by strong dispersion forces. Because of this they are liquefied at very low temperature. Hence, they have very high boiling points. In the light of the above statements, choose the correct answer from the options given below:

Solution & Explanation

### Core Logic Statement I and II are False. Noble gases have low boiling points. Noble gases are held together by weak dispersion forces. ### Pattern Recognition Noble gases are characterized by extremely weak intermolecular forces (London dispersion forces) because they are monoatomic and non-polar, which directly results in very low boiling and melting points. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The p-Block Elements

Reference Study Guides

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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