Classification of Elements and Periodicity in Properties Previous Year Questions — JEE Main Chemistry
36 past-year Classification of Elements and Periodicity in Properties questions from JEE Main (Chemistry) — page 3.
Q40 (2025)
Given below are two statements:
Statement (I) : The first ionization energy of Pb is greater than that of Sn
Statement (II) : The first ionization energy of Ge is greater than that of Si.
In the light of the above statements, choose the correct answer from the options given below :
- \text{Statement I is true but Statement II is false}
- \text{Both Statement I and Statement II are false}
- \text{Statement I is false but Statement II is true}
- \text{Both Statement I and Statement II are true}
### Core Logic
Let's analyze the first ionization energy ($\text{IE}_1$) values for Group 14 elements ($\mathrm{C, Si, Ge, Sn, Pb}$):
Generally, ionization energy decreases down a group as atomic size increases. However, heavy post-transition elements exhibit an anomaly:
* Analysis of Statement I:
Moving from $\mathrm{Sn}$ to $\mathrm{Pb}$, the $4f$ orbital subshell becomes fully filled. Because $4f$ electrons provide very poor shielding, the outer valence electrons experience a significantly higher effective nuclear charge ($Z_{\text{eff}}$). This inert pair effect contractive behavior tightly binds the outer electrons, making the first ionization energy of Lead higher than that of Tin:
$$\text{IE}_1(\mathrm{Pb}) = 715\text{ kJ/mol} > \text{IE}_1(\mathrm{Sn}) = 708\text{ kJ/mol}$$
Thus, Statement I is true.
* Analysis of Statement II:
Following normal periodic trends down the group, the ionization energy decreases from Silicon to Germanium due to the increasing atomic radius:
$$\text{IE}_1(\mathrm{Si}) = 786\text{ kJ/mol} > \text{IE}_1(\mathrm{Ge}) = 761\text{ kJ/mol}$$
Therefore, the claim that $\mathrm{Ge} > \mathrm{Si}$ is false.
### Pattern Recognition
The overall first ionization energy trend for Group 14 is: $\mathrm{C > Si > Ge > Pb > Sn}$. Notice that Lead breaks the downward trend and has a higher ionization energy than Tin due to poor shielding by $4f$ electrons (lanthanide contraction).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Class 11 Chemistry: The p-Block Elements
Q45 (2025)
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:
- \text{Group 2}
- \text{Group 13}
- \text{Group 4}
- \text{Group 14}
### Core Logic
Let's examine the differences between successive ionization energies to find where the largest jump occurs:
* $\text{IE}_1 = 800\text{ kJ/mol}$
* $\text{IE}_2 = 2427\text{ kJ/mol}$
* $\text{IE}_3 = 3658\text{ kJ/mol}$
* $\text{IE}_4 = 25024\text{ kJ/mol}$
* $\text{IE}_5 = 32824\text{ kJ/mol}$
Notice the massive, multi-fold jump between $\text{IE}_3$ and $\text{IE}_4$ ($3658 \rightarrow 25024\text{ kJ/mol}$).
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 $\rightarrow$ 3 valence electrons $
ightarrow$ Group 13.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q35 (2025)
Which of the following Statements are NOT true about the periodic table?
A. The properties of elements are function of atomic weights.
B. The properties of elements are function of atomic numbers.
C. Elements having similar outer electronic configuration are arranged in same period.
D. An element's location reflects the quantum numbers of the last filled orbital.
E. The number of elements in a period is same as the number of atomic orbitals available in energy level that is being filled.
Choose the correct answer from the options given below:
- A, C and E Only
- D and E Only
- A and E Only
- B, C and E Only
### Core Logic
Evaluating modern electronic system structures:
- Statement A is false: Modern periodic organization links traits to atomic *numbers*, not weights (which defined old Mendeleev frameworks).
- Statement C is false: Elements possessing identical outer configurations occupy the same chemical *group*, not the same horizontal period.
- Statement E is false: The total count of active elements across any discrete period is equal to double ($2 \times$) the count of functional available atomic orbitals in the relevant energy level.
### Pattern Recognition
Carefully identify double-negatives or questions asking for the 'NOT true' choice inside variable selection items.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q39 (2025)
Given below are two statements:
Statement (I): According to the Law of Octaves, the elements were arranged in the increasing order of their atomic number.
Statement (II): Meyer observed a periodically repeated pattern upon plotting physical properties of certain elements against their respective atomic numbers.
In the light of the above statements, choose the correct answer from the options given below:
- Statement I is false but Statement II is true
- Both Statement I and Statement II are true
- Statement I is true but Statement II is false
- Both Statement I and Statement II are false
### Related Formula
Early historical periodic classification principles depended on atomic weight attributes:
$$\text{Periodic Property} = f(\text{Atomic Weight})$$
### Core Logic
Evaluating historical accuracy:
- **Statement I**: Newlands' Law of Octaves arranged elements in the increasing order of their **atomic weights** (not atomic numbers). Hence, Statement I is false.
- **Statement II**: Lothar Meyer plotted physical properties like atomic volume, melting point, and boiling point against **atomic weight** (not atomic number). Hence, Statement II is false.
### Step 1: Conclusion
Since both statements incorrectly reference atomic number instead of atomic weight, both Statement I and Statement II are false.
### Pattern Recognition
Almost all classical periodic classifiers (Newlands, Döbereiner, de Chancourtois, Lothar Meyer, Mendeleev) relied strictly on *atomic weight*. The pivot to *atomic number* occurred later with Moseley's X-ray studies and the Modern Periodic Law.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q1317 (2025)
An element 'E' has the ionisation enthalpy value of $374 \, \mathrm{kJ \, mol^{-1}}$ . 'E' reacts with elements A, B, C and D with electron gain enthalpy values of $-328$ , $-349$ , $-325$ and $-295 \, \mathrm{kJ \, mol^{-1}}$ , respectively.
The correct order of the products EA, EB, EC and ED in terms of ionic character is :
- $\mathrm{EB} > \mathrm{EA} > \mathrm{EC} > \mathrm{ED}$
- $\mathrm{ED} > \mathrm{EC} > \mathrm{EA} > \mathrm{EB}$
- $\mathrm{EA} > \mathrm{EB} > \mathrm{EC} > \mathrm{ED}$
- $\mathrm{ED} > \mathrm{EC} > \mathrm{EB} > \mathrm{EA}$
### Related Formula
$$\text{Ionic Character} \propto \lvert \Delta H_{\text{IE}} - \Delta H_{\text{EGE}} \rvert$$
### Core Logic
The relative ionic quality of a standard binary bond rises as the gap scale between ionization enthalpy and negative electron gain enthalpy parameters widens. Comparing the values:
* For B: $\Delta H_{\text{EGE}} = -349 \mathrm{~kJ/mol}$ (largest energy release) $\rightarrow$ Highest ionic character.
* For A: $\Delta H_{\text{EGE}} = -328 \mathrm{~kJ/mol}$.
* For C: $\Delta H_{\text{EGE}} = -325 \mathrm{~kJ/mol}$.
* For D: $\Delta H_{\text{EGE}} = -295 \mathrm{~kJ/mol}$ (smallest energy release) $\rightarrow$ Lowest ionic character.
Arranging them in descending order of ionic character yields:
$$\mathrm{EB} \gt \mathrm{EA} \gt \mathrm{EC} \gt \mathrm{ED}$$
### Pattern Recognition
A highly exothermic electron gain enthalpy value favors easier anion production, widening electronegativity variations to enhance ionic bond properties.
### Chapter Mix
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
