The correct order of C, N, O and F in terms of second ionisation potential is

Related Formula

Atomic radius decreases across a period due to increase in effective nuclear charge (Zeff$Z_{\text{eff}}$) and increases down a group due to addition of new electronic shells.

Core Logic

Let us analyze the elements in option (3):

• Be$\mathrm{Be}$ and Mg$\mathrm{Mg}$ belong to Group 2. Down the group, atomic radius increases: Mg > Be$\mathrm{Mg} > \mathrm{Be}$.
• Mg$\mathrm{Mg}$, Al$\mathrm{Al}$, and Si$\mathrm{Si}$ belong to Period 3. Across the period, atomic radius decreases: Mg > Al > Si$\mathrm{Mg} > \mathrm{Al} > \mathrm{Si}$.

Combining these trends, the correct decreasing order is:

Therefore, the order given in option (3) Be > Mg > Al > Si$\mathrm{Be} > \mathrm{Mg} > \mathrm{Al} > \mathrm{Si}$ is incorrect.

Pattern Recognition

Sees: Group 2 and Period 3 comparison. Trap: Assuming atomic radius always increases down a group regardless of cross-period shifts; Mg$\mathrm{Mg}$ is larger than Be$\mathrm{Be}$, making Be > Mg$\mathrm{Be} > \mathrm{Mg}$ fundamentally incorrect. Shortcut: Check adjacent group/period boundaries to immediately isolate inversions.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Related Formula

Core Logic

Statement A: Electron gain enthalpy can be endothermic (e.g., noble gases, noble metals). (False)

Statement B: Removal of an electron requires energy to overcome nuclear attraction; ionization is strictly endothermic. (True)

Statement C: Be$\mathrm{Be}$ (1s² 2s²$1s^2 2s^2$) has a fully filled stable 2s$2s$ subshell, requiring higher ionization energy than B$\mathrm{B}$ (1s² 2s² 2p¹$1s^2 2s^2 2p^1$). (True)

Statement D: Electronegativity of carbon varies with oxidation state / partial charge (Zeff$Z_{\text{eff}}$ in CCl₄ > CH₄$\mathrm{CCl}_4 > \mathrm{CH}_4$). (False)

Statement E: Cs$\mathrm{Cs}$ (or Fr$\mathrm{Fr}$) is the most electropositive element in group I, not Li$\mathrm{Li}$. (False)

Step 1: Final Conclusion

Statements B and C are correct.

Pattern Recognition

Ionization energy = ALWAYS endothermic. Be vs B IE1 = Be higher due to stable full 2s$2s$ subshell.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Core Logic

Let's evaluate the anomalies within Group 13 trends:

• Atomic Radius Order: Due to the poor shielding effect of the filled 3d$3d$ electron subshell in Gallium (transition contraction), its outer valence shell experiences a stronger nuclear pull. Consequently, the atomic radius sequence exhibits an inversion:

Therefore, the pair (Al < Ga)$(Al < Ga)$ provided in the problem statement is incorrect.

• Ionic Radius Order (M³⁺$M^{3+}$): In the fully ionized +3$+3$ configuration state, the transition contraction anomaly is overridden by standard shell count physics. The ionic radius follows the uniform down-the-group trend:

Hence, Gallium (Ga$Ga$) has the larger ionic radius between the two elements. Its atomic number is 31.

Pattern Recognition

Transition contraction heavily distorts the atomic radius profile of Gallium, but standard descending size progression rules are fully restored when evaluating the ionic M³⁺$M^{3+}$ radius profile.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 12 Chemistry: The p-Block Elements

Related Formula

Core Logic

Let's review the periodic table positioning and trends:

• Across a period from left to right, effective nuclear charge (Zeff$Z_{\text{eff}}$) increases, drawing electrons closer to the nucleus. Hence, atomic radius decreases.
• Mg$\text{Mg}$ and Al$\text{Al}$ sit in Period 3. Since Al$\text{Al}$ is further to the right (Z=13$Z=13$) than Mg$\text{Mg}$ (Z=12$Z=12$), the radius of Al$\text{Al}$ is smaller than that of Mg$\text{Mg}$:

Thus, the stated trend rMg < rAl$r_{\text{Mg}} < r_{\text{Al}}$ is incorrect.

Step 1: Cross-Checking Valid Trends

• rBr < rK$r_{\text{Br}} < r_{\text{K}}$: Correct, as potassium sits at the start of Period 4, bromine sits near the end.
• rRb < rCs$r_{\text{Rb}} < r_{\text{Cs}}$: Correct, atomic size increases down Group 1.
• rAt < rCs$r_{\text{At}} < r_{\text{Cs}}$: Correct, cesium sits far below and left compared to astatine.

Pattern Recognition

Period 3 size ranking baseline: Na > Mg > Al > Si > P > S > Cl$\text{Na} > \text{Mg} > \text{Al} > \text{Si} > \text{P} > \text{S} > \text{Cl}$. Moving rightward always drops the radius size unless noble gas van der Waals constraints interfere.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Core Logic

Let's analyze the first ionization energy (IE₁$\text{IE}_1$) values for Group 14 elements (C, Si, Ge, Sn, Pb$\mathrm{C, Si, Ge, Sn, Pb}$):

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 Sn$\mathrm{Sn}$ to Pb$\mathrm{Pb}$, the 4f$4f$ orbital subshell becomes fully filled. Because 4f$4f$ electrons provide very poor shielding, the outer valence electrons experience a significantly higher effective nuclear charge (Zeff$Z_{\text{eff}}$). This inert pair effect contractive behavior tightly binds the outer electrons, making the first ionization energy of Lead higher than that of Tin:

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:

Therefore, the claim that Ge > Si$\mathrm{Ge} > \mathrm{Si}$ is false.

Pattern Recognition

The overall first ionization energy trend for Group 14 is: C > Si > Ge > Pb > Sn$\mathrm{C > Si > Ge > Pb > Sn}$. Notice that Lead breaks the downward trend and has a higher ionization energy than Tin due to poor shielding by 4f$4f$ electrons (lanthanide contraction).

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 11 Chemistry: The p-Block Elements