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The successive 5 ionisation energies of an element are 800, 2427, 3658, 25024 and 32824 kJ/mol, respectively. By using the above values predict the group in which the above element is present:

Solution & Explanation

### Core Logic Let's examine the differences between successive ionization energies to find where the largest jump occurs: * textIE_1 = 800text kJ/mol * textIE_2 = 2427text kJ/mol * textIE_3 = 3658text kJ/mol * textIE_4 = 25024text kJ/mol * textIE_5 = 32824text kJ/mol Notice the massive, multi-fold jump between textIE_3 and textIE_4 (3658 ightarrow 25024text kJ/mol). This huge increase indicates that removing the 4th electron requires breaking into a stable, filled core noble gas shell configuration. This means the atom has exactly 3 valence electrons in its outermost shell (ns^2 np^1), which identifies it as a member of **Group 13** (the Boron family). ### Pattern Recognition To find the number of valence electrons, look for the ionization step where the value jumps drastically. The number of relatively low ionization steps before that jump equals the number of valence electrons. Here, 3 low steps ightarrow 3 valence electrons ightarrow Group 13. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Reference Study Guides

Classification Of Elements And Periodicity In Properties

More Classification of Elements and Periodicity in Properties Previous-Year Questions — Page 4

Q45 2025 Periodic Trends in Physical and Chemical Properties
Given below are two statements : Statement (I) : The radii of isoelectronic species increases in the order: mathrm M g ^ 2 + < mathrm N a ^ + < mathrm F ^ - < mathrm O ^ 2 - Statement (II) : The magnitude of electron gain enthalpy of halogen decreases in the order: mathrm C l > mathrm F > mathrm B r > mathrm I
  • A. Statement I is incorrect but Statement II is correct
  • B. Both Statement I and Statement II are incorrect.
  • C. Statement I is correct but Statement II is incorrect
  • D. Both Statement I and Statement II are correct

Solution

### Related Formula textIonic Radius propto frac1textNuclear Charge (Z) quad text(for Isoelectronic series) ### Core Logic Evaluating each statement systematically : * Statement (I) is correct: mathrmMg^2+, mathrmNa^+, mathrmF^-, mathrmO^2- all possess exactly 10 electrons (isoelectronic). As the positive nuclear charge decreases (Z = 12 for mathrmMg down to Z = 8 for mathrmO), the nucleus exerts less pull on the electron cloud, causing the ionic radius to increase : mathrmMg^2+ < mathrmNa^+ < mathrmF^- < mathrmO^2- * Statement (II) is correct: Chlorine has a higher electron gain enthalpy magnitude than fluorine due to lower electron-electron repulsion in its larger 3p orbital. The standard halogen trend follows: mathrmCl > mathrmF > mathrmBr > mathrmI Thus, both statements are correct. ### Pattern Recognition For species with the same number of electrons, a higher negative charge always leads to a larger electron cloud radius due to reduced nuclear traction. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

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