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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The P Block Elements

More The p-Block Elements Previous-Year Questions — Page 6

Q72 jee_main_2024_31_jan_evening Oxygen Family (Group 16 Elements)
Choose the correct statements from the following A. All group 16 elements form oxides of general formula EO_2 and EO_3 where E = S, Se, Te and Po. Both the types of oxides are acidic in nature. B. TeO_2 is an oxidising agent while SO_2 is reducing in nature. C. The reducing property decreases from H_2S to H_2Te down the group. D. The ozone molecule contains five lone pairs of electrons. Choose the correct answer from the options given below:
  • A. text1. A and D only
  • B. text2. B and C only
  • C. text3. C and D only
  • D. text4. A and B only

Solution

### Core Logic (A) All group 16 elements form oxides of the EO_2 and EO_3 type where E = S, Se, Te or Po. These are generally acidic in nature. (Correct) (B) SO_2 is reducing while TeO_2 is an oxidising agent due to the inert pair effect increasing the stability of lower oxidation states down the group. (Correct) (C) The reducing property increases from H_2S to H_2Te down the group because the E-H bond dissociation enthalpy decreases as atomic size increases. (Incorrect) (D) The ozone molecule (O_3) has the structure O=O^+-O^-. Counting the lone pairs: the central oxygen has 1, the double-bonded oxygen has 2, and the single-bonded negatively charged oxygen has 3. Total lone pairs = 1 + 2 + 3 = 6. (Incorrect) ### Step 1: Final Selection Since A and B are the only correct statements, option 4 is the right choice. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The p-Block Elements
Q76 jee_main_2024_31_jan_evening Group 13 Elements (Boron Family)
Given below are two statements: Statement I: Group 13 trivalent halides get easily hydrolyzed by water due to their covalent nature. Statement II: AlCl_3 upon hydrolysis in acidified aqueous solution forms octahedral [Al(H_2O)_6]^3+ ion. In the light of the above statements, choose the correct answer from the options given below:
  • A. text(1) Statement I is true but statement II is false
  • B. text(2) Statement I is false but statement II is true.
  • C. text(3) Both statement I and statement II are false.
  • D. text(4) Both statement I and statement II are true.

Solution

### Core Logic Statement I: In the trivalent state, most compounds of Group 13 elements (like BCl_3, AlCl_3) are covalent due to small cation size and high charge, and they are readily hydrolysed in water. Thus, Statement I is true. Statement II: Trichlorides on hydrolysis in normal water can form tetrahedral [M(OH)_4]^- species depending on pH, but in an acidified aqueous solution, aluminium forms the octahedral [Al(H_2O)_6]^3+ ion where the hybridization state of Al is sp^3d^2. Thus, Statement II is also true. ### Step 1: Final Conclusion Both Statement I and Statement II are correct, pointing to option (4). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: The p-Block Elements
Q61 jee_main_2024_31_jan_morning Group 18 Elements
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:
  • A. textStatement I is false but Statement II is true.
  • B. textBoth Statement I and Statement II are true.
  • C. textStatement I is true but Statement II is false.
  • D. textBoth Statement I and Statement II are false.

Solution

### 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
Q79 jee_main_2024_31_jan_morning Group 14 Elements
Consider the oxides of group 14 elements SiO_2, GeO_2, SnO_2, PbO_2, CO and GeO. The amphoteric oxides are
  • A. GeO, GeO_2
  • B. SiO_2, GeO_2
  • C. SnO_2, PbO_2
  • D. SnO_2, CO

Solution

### Core Logic In Group 14 elements, the nature of oxides varies down the group: CO is neutral. CO_2 and SiO_2 are acidic. GeO_2 is distinctly acidic. SnO_2 and PbO_2 are amphoteric. ### Pattern Recognition Tin (Sn) and Lead (Pb) mostly form amphoteric oxides (e.g., SnO, SnO_2, PbO, PbO_2), distinguishing them from the acidic oxides of Carbon and Silicon. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: The p-Block Elements

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