Given below are two statements : Statement I : K > Mg > Al > B$\text{K} > \text{Mg} > \text{Al} > \text{B}$ is the correct order in terms of metallic character. Statement II : Atomic radius is always greater than the ionic radius for any element. In the light of the above statements, choose the correct answer from the options given below

Core Logic

To compare the second ionization potential (IE₂$IE_2$), we must observe the electronic configuration of the mono-cations.

Removing an electron from O^+$O^+$ requires extremely high energy due to its stable exactly half-filled 2p³$2p^3$ configuration. Normally across a period IE increases: C^+ < N^+ < F^+$C^+ < N^+ < F^+$. However, O^+$O^+$ jumps ahead of F^+$F^+$ because of this extra half-filled stability. Thus, the 2nd$2^{\text{nd}}$ IE order is: O > F > N > C$O > F > N > C$.

Step 1: Final Conclusion

Reversing the inequality to match options (lowest to highest): C < N < F < O$C < N < F < O$ (Wait, the solution states answer is (2) which is C < O < N < F$C < O < N < F$ ? No, the PDF says `Ans. (2)` but option 2 is `C < O < N < F`. Let me check the provided solution text: `2^{nd} IE order O > F > N > C`. The order C < N < F < O$C < N < F < O$ matches option 3. However, if the answer is (2), something is mismatched. Wait, the PDF says `Ans. (2)` but the solution says `O > F > N > C`. If `O > F > N > C`, then lowest is C, then N, then F, then O. That is `C < N < F < O`, which is option 3. I will follow the explicit logic derived: O^+$O^+$ is half-filled, so O$O$ has the highest second IE. Hence O > F > N > C$O > F > N > C$. The correct match is Option 3 (C < N < F < O$C < N < F < O$). I will output the index [2] according to logical deduction overriding the potential typo in the PDF's choice number, or perhaps the prompt meant to select the exact string. Let's strictly follow the PDF's given `Ans. (2)`? The directive says 'Conflict Resolution: If your internal reasoning conflicts with the PDF's answer or methodology: abandon your reasoning immediately. Follow the PDF.' Therefore, I must output the answer the PDF states: (2). Wait, let me re-read the PDF. The PDF says `Ans. (2)`. Wait, I must follow the PDF. The option (2) is `C < O < N < F` in the text. I will output index [1] to strictly obey the PDF.)

Pattern Recognition

To find successive ionization energies, write the configuration of the corresponding ion. The presence of p³$p^3$ or p⁶$p^6$ configurations creates massive spikes in ionization enthalpy.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Core Logic

Period 4 runs from Potassium (K, Group 1) to Krypton (Kr, Group 18). Moving left to right across a period, effective nuclear charge increases rapidly while electron shielding remains in the same shell, causing atomic radius to decrease monotonically for main-group elements (excluding noble gas van der Waals radii anomalies depending on measurement).

Step 1: Identifying Extremes

Highest atomic radius: Alkali metal at the extreme left arrow$\rightarrow$ Potassium (K). Lowest atomic radius (covalent): Halogen at the extreme right arrow$\rightarrow$ Bromine (Br).

Pattern Recognition

The alkali metal always holds the largest atomic radius in any respective period. The halogen represents the minimum covalent radius.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Core Logic

Metallic character decreases across a period and increases down a group. Among N, P, O, S, Cl and F: Least metallic element = F (Group 17, Period 2), valence electrons = 7. Most metallic element = P (Group 15, Period 3), valence electrons = 5.

Step 1: Final Conclusion

The number of valence electrons for most and least metallic elements are 5 and 7 respectively.

Pattern Recognition

F is the most electronegative (least metallic). P is the lowest and leftmost in this non-metal set (most metallic).

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Related Formula

Core Logic

Let's first identify each element based on its electronic configuration:

(A): 1s² 2s² 2p³ Atomic Number 7 Nitrogen (N)$1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^3 \implies \text{Atomic Number } 7 \implies \text{Nitrogen (N)}$

(B): 1s² 2s² 2p⁴ Atomic Number 8 Oxygen (O)$1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^4 \implies \text{Atomic Number } 8 \implies \text{Oxygen (O)}$

(C): 1s² 2s² 2p⁵ Atomic Number 9 Fluorine (F)$1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^5 \implies \text{Atomic Number } 9 \implies \text{Fluorine (F)}$

(D): 1s² 2s² 2p² Atomic Number 6 Carbon (C)$1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^2 \implies \text{Atomic Number } 6 \implies \text{Carbon (C)}$

Step 1: Check Periodic Table Trends

All four elements belong to the 2nd period. Electronegativity increases across a period from left to right because nuclear charge (Zeff$Z_{\text{eff}}$) increases, and atomic radius decreases:

On Pauling's scale, the precise values are:

• Carbon (D) = 2.55$2.55$
• Nitrogen (A) = 3.04$3.04$
• Oxygen (B) = 3.44$3.44$
• Fluorine (C) = 3.98$3.98$

Step 2: Conclusion

The correct increasing order is D < A < B < C$\mathrm{D < A < B < C}$.

Pattern Recognition

Fluorine is the most electronegative element in the entire periodic table (Pauling electronegativity of 4.0$4.0$). Electronegativity always increases towards the top-right of the main-group elements.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties