The correct orders among the following are : - Atomic radius: mathrmB < mathrmAl < mathrmGa < mathrmIn < mathrmTl - Electronegativity: mathrmAl < mathrmGa < mathrmIn < mathrmTl < mathrmB - Density: mathrmTl < mathrmIn < mathrmGa < mathrmAl < mathrmB - 1^mathrmst Ionisation Energy: mathrmIn < mathrmAl < mathrmGa < mathrmTl < mathrmB Choose the correct answer from the options given below :

Solution & Explanation

### Related Formula Group 13 elements (mathrmB, mathrmAl, mathrmGa, mathrmIn, mathrmTl) show highly anomalous periodic trends due to the intervention of filled d-orbitals (d-block contraction in mathrmGa) and f-orbitals (lanthanoid contraction in mathrmTl). ### Core Logic Evaluate each specified trend against official physical constants: - **Atomic radius**: Due to d-block contraction, gallium (mathrmGa) is smaller than aluminum (mathrmAl): textRadius (pm): mathrmB(88) < mathrmGa(135) < mathrmAl(143) < mathrmIn(167) < mathrmTl(170) Hence, the given order is *Incorrect*. - **Electronegativity**: Electronegativity first decreases from mathrmB to mathrmAl, then increases down the group due to poor shielding of d and f electrons: textElectronegativity: mathrmAl(1.5) < mathrmGa(1.6) < mathrmIn(1.7) < mathrmTl(1.8) < mathrmB(2.0) Hence, this order is *Correct*. ### Step 1: Analyze density and ionization energy trends - **Density**: Increases down the group as atomic mass increases much faster than atomic volume: textDensity (g/cm^3text): mathrmB(2.35) < mathrmAl(2.70) < mathrmGa(5.90) < mathrmIn(7.31) < mathrmTl(11.85) Hence, the given order is *Incorrect* (it is completely reversed). - **1^mathrmst Ionisation Energy**: Shows an irregular trend due to ineffective shielding by d and f electrons: textIE_1mathrm~(kJ/mol): mathrmIn(558) < mathrmAl(577) < mathrmGa(579) < mathrmTl(589) < mathrmB(801) Hence, this order is *Correct*. ### Step 2: Conclusion Only the Electronegativity (B) and 1^mathrmst Ionisation Energy (D) orders are correct, matching Option (1). ### Pattern Recognition Group 13 elements do not follow monotonic trends. The poor shielding of 3d^10 and 4f^14 electrons increases the effective nuclear charge on valence electrons, causing anomalies in atomic radius (mathrmGa < mathrmAl) and pulling electronegativities and ionization energies upward as you go further down. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: The p-Block Elements Class 11 Chemistry: Classification of Elements and Periodicity in Properties

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Q49 jee_main_2025_28_jan_evening Group 15 Elements - Hydrides and Bonding
A group 15 element forms dpi-dpi bond with transition metals. It also forms hydride, which is a strongest base among the hydrides of other group members that form dpi-dpi bond. The atomic number of the element is ______.
Numerical Answer. Answer: 15 to 15

Solution

### 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: NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3 ### Core Logic Let's analyze the properties specified: 1. The element must be able to form dpi-dpi bonds with transition metals. Nitrogen cannot form these bonds because it lacks vacant d-orbitals in its valence shell. Therefore, nitrogen is excluded. 2. Among the remaining elements (Phosphorus, Arsenic, Antimony, Bismuth) that contain available d-orbitals, basic strength decreases down the group. Phosphorus forms phosphine (PH_3), which is the strongest base among the remaining members. ### Step 1: Identify the Atomic Number The identified element is Phosphorus (P). The atomic number of Phosphorus is 15. ### Pattern Recognition Pay attention to qualifying statements like *'among elements that form dpi-dpi bonds'*. This explicitly excludes second-period elements (like Nitrogen), making Phosphorus (Z=15) the top choice for basicity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: p-Block Elements
Q68 jee_main_2024_01_february_morning Group 15 Elements
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : PH_3 has lower boiling point than NH_3. Reason (R): In liquid state NH_3 molecules are associated through vander waal's forces, but PH_3 molecules are associated through hydrogen bonding. In the light of the above statements, choose the most appropriate answer from the options given below:
  • A. textBoth (A) and (R) are correct and (R) is not the correct explanation of (A)
  • B. text(A) is not correct but (R) is correct
  • C. textBoth (A) and (R) are correct but (R) is the correct explanation of (A)
  • D. text(A) is correct but (R) is not correct

Solution

### Core Logic NH_3 undergoes extensive intermolecular hydrogen bonding due to the high electronegativity and small size of Nitrogen. 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: PH_3 has a lower boiling point than NH_3 because breaking H-bonds in NH_3 requires more energy. Reason (R) is incorrect: It falsely claims NH_3 has van der Waals association and 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: NH_3 > PH_3 purely due to H-bonding in NH_3. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The p-Block Elements Class 11 Chemistry: Chemical Bonding and Molecular Structure
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

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