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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More The p-Block Elements Previous-Year Questions — Page 5

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
Q79 jee_main_2024_30_january_evening Group 15 Hydrides
Choose the correct statements about the hydrides of group 15 elements. A. The stability of the hydrides decreases in the order mathrmNH_3 > mathrmPH_3 > mathrmAsH_3 > mathrmSbH_3 > mathrmBiH_3 B. The reducing ability of the hydrides increases in the order mathrmNH_3 < mathrmPH_3 < mathrmAsH_3 < mathrmSbH_3 < mathrmBiH_3 C. Among the hydrides, mathrmNH_3 is strong reducing agent while mathrmBiH_3 is mild reducing agent. D. The basicity of the hydrides increases in the order mathrmNH_3 < mathrmPH_3 < mathrmAsH_3 < mathrmSbH_3 < mathrmBiH_3 Choose the most appropriate from the option given below:
  • A. textB and C only
  • B. textC and D only
  • C. textA and B only
  • D. textA and D only

Solution

### Core Logic Statement A: As we move down Group 15, the size of the central atom increases, causing the E-H bond length to increase and its bond strength to decrease. Thus, thermal stability decreases in the order: NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3. Statement A is correct. Statement B: Because the E-H bond strength decreases down the group, it becomes easier to release hydrogen. Therefore, the reducing ability increases down the group: NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3. Statement B is correct. Statement C: Following the logic from B, NH_3 is a very weak reducing agent (mild), while BiH_3 is the strongest reducing agent among them. Statement C is false. Statement D: Basicity depends on the availability of the lone pair for donation. In NH_3, the lone pair is concentrated in a small sp^3 orbital (high electron density). Moving down the group, the lone pair occupies larger orbitals with more s-character (Drago's rule), decreasing electron density and making the lone pair less available. Thus, basicity decreases down the group. Statement D is false. ### Step 1: Final Conclusion Only statements A and B are correct. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The p Block Elements
Q64 jee_main_2024_30_jan_morning Group 15 Elements
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:
  • A. text(A) is false but (R) is true
  • B. textBoth (A) and (R) are true but (R) is not the correct explanation of (A)
  • C. text(A) is true but (R) is false
  • D. textBoth (A) and (R) are true and (R) is the correct explanation of (A)

Solution

### 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
Q64 jee_main_2024_31_jan_evening Properties of Sulphur and Halogens
Given below are two statements: Statement I: S_8 solid undergoes disproportionation reaction under alkaline conditions to form S^2- and S_2O_3^2- Statement II: ClO_4^- can undergo disproportionation reaction under acidic condition. In the light of the above statements, choose the most appropriate answer from the options given below:
  • A. textStatement I is correct but statement II is incorrect.
  • B. textStatement I is incorrect but statement II is correct
  • C. textBoth statement I and statement II are incorrect.
  • D. textBoth statement I and statement II are correct.

Solution

### Core Logic Statement I: S_8 disproportionates in alkaline medium to give sulphide (S^2-) and thiosulphate (S_2O_3^2-). The reaction is: S_8 + 12OH^- rightarrow 4S^2- + 2S_2O_3^2- + 6H_2O This is a correct statement. Statement II: ClO_4^- contains chlorine in its highest possible oxidation state of +7. A species can undergo disproportionation only if the element can exist in both higher and lower oxidation states than the present one. Since chlorine cannot exceed +7, ClO_4^- cannot undergo disproportionation. This is an incorrect statement. ### Step 1: Final Conclusion Statement I is correct, and Statement II is incorrect. Therefore, option (1) is the correct answer. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Redox Reactions Class 12 Chemistry: The p-Block Elements

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