Question

Which of the following represents the correct trend for the mentioned property?
A. F $>$ P $>$ S $>$ B \quad -- First Ionization Energy
B. Cl $>$ F $>$ S $>$ P \quad -- Electron Affinity
C. K $>$ Al $>$ Mg $>$ B \quad -- Metallic character
D. K₂O $>$ Na₂O $>$ MgO $>$ Al₂O₃ \quad -- Basic character
Choose the correct answer from the options given below:

• A. A, B and D only
• B. A and B only
• C. B and C only
• D. A, B, C and D

Hint:

Mastering periodic trends is crucial. Pay special attention to exceptions like the electron affinity of F and Cl, and the ionization energy of elements with half-filled or fully-filled orbitals (e.g., Be vs B, N vs O, P vs S).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:

The question asks us to identify the correct statements related to periodic trends. Each statement must be evaluated using standard principles of periodicity such as ionization energy, electron affinity, metallic character, and basic nature of oxides.

Step 2: Detailed Explanation:

Statement A: First Ionization Energy trend: F > P > S > B

Ionization energy generally increases across a period due to increasing nuclear charge and decreases down a group due to increased atomic size and shielding.

Approximate ionization energy values (kJ/mol):
F ≈ 1681, P ≈ 1012, S ≈ 1000, B ≈ 801

Fluorine has the highest ionization energy among the given elements due to its small size and high effective nuclear charge. Phosphorus has a half-filled 3p³ configuration, which is relatively stable, so its ionization energy is slightly higher than that of sulphur (3p⁴). Boron has the lowest ionization energy among the four.

Conclusion: Statement A is correct.

Statement B: Electron Affinity trend: Cl > F > S > P

Electron affinity generally increases across a period and decreases down a group, with some important exceptions.

Chlorine has a higher electron affinity than fluorine because the very small size of fluorine causes strong electron-electron repulsion when an extra electron is added. Sulphur has higher electron affinity than phosphorus because phosphorus has a stable half-filled p-orbital configuration.

Approximate electron affinity values (kJ/mol):
Cl ≈ 349, F ≈ 328, S ≈ 200, P ≈ 72

Conclusion: Statement B is correct.

Statement C: Metallic character trend: K > Al > Mg > B

Metallic character increases down a group and decreases across a period.

Potassium is the most metallic among the given elements. However, in period 3, magnesium is more metallic than aluminium because metallic character decreases from left to right across a period.

The correct order within period 3 is: Mg > Al. Since the given order states Al > Mg, it is incorrect.

Conclusion: Statement C is incorrect.

Statement D: Basic character of oxides trend: K₂O > Na₂O > MgO > Al₂O₃

Basic character of metallic oxides increases down a group and decreases across a period.

K₂O is more basic than Na₂O because potassium is below sodium in group 1. Across period 3, basic character decreases: Na₂O (strongly basic) > MgO (basic) > Al₂O₃ (amphoteric).

Conclusion: Statement D is correct.

Step 3: Final Answer:

Statements A, B, and D are correct, while statement C is incorrect. Hence, the correct option is the one that includes A, B, and D only.