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 to identify the correct statements regarding periodic trends among the given options. We need to evaluate each statement based on the principles of periodicity in properties of elements.
Step 2: Detailed Explanation:
Let's analyze each statement one by one:
Statement A: First Ionization Energy (IE) trend: F $>$ P $>$ S $>$ B.
\begin{itemize} \item General Trend: Ionization energy generally increases across a period (due to increasing nuclear charge) and decreases down a group (due to increasing atomic size and shielding).
\item Comparison: \begin{itemize} \item The IE values (in kJ/mol) are approximately: F(1681), P(1012), S(1000), B(801).
\item F is in period 2 and has a very high effective nuclear charge, hence the highest IE.
\item P (3p³) has a stable half-filled p-orbital configuration, so its IE is higher than S (3p⁴), from which an electron can be removed more easily to attain a stable half-filled configuration. So, P $>$ S is correct.
\item B has a much lower IE than F, P, and S.
\item Thus, the order F $>$ P $>$ S $>$ B is correct based on the values.
\end{itemize} \item Conclusion: Statement A is correct.
\end{itemize} Statement B: Electron Affinity (EA) trend: Cl $>$ F $>$ S $>$ P.
\begin{itemize} \item General Trend: Electron affinity generally increases across a period and decreases down a group.
\item Comparison: \begin{itemize} \item Cl vs F: Chlorine has a higher electron affinity than Fluorine. This is an important exception. Due to the small size of the F atom, the incoming electron experiences significant repulsion from the already present electrons. The larger Cl atom can accommodate the incoming electron more easily. So, Cl $>$ F is correct.
\item S vs P: Sulphur (group 16) has a higher tendency to accept an electron to achieve a more stable configuration than Phosphorus (group 15), which has a stable half-filled p-orbital configuration. So, S $>$ P is correct.
\item The overall order based on values (in kJ/mol) Cl(349) $>$ F(328) $>$ S(200) $>$ P(72) is correct.
\end{itemize} \item Conclusion: Statement B is correct.
\end{itemize} Statement C: Metallic character trend: K $>$ Al $>$ Mg $>$ B.
\begin{itemize} \item General Trend: Metallic character decreases across a period and increases down a group.
\item Comparison: \begin{itemize} \item K is an alkali metal (group 1), and Mg is an alkaline earth metal (group 2). Al (group 13) and B (group 13) are further to the right. K is in period 4, while the others are in periods 3 and 2.
\item K is the most metallic among the given elements.
\item In period 3, metallic character decreases from left to right. Therefore, Mg (group 2) is more metallic than Al (group 13). The given order Al $>$ Mg is incorrect.
\end{itemize} \item Conclusion: Statement C is incorrect.
\end{itemize} Statement D: Basic character of oxides trend: K₂O $>$ Na₂O $>$ MgO $>$ Al₂O₃.
\begin{itemize} \item General Trend: The basic character of metallic oxides increases down a group and decreases across a period.
\item Comparison: \begin{itemize} \item K₂O vs Na₂O: K is below Na in group 1. Metallic character increases down the group, so the basicity of oxides also increases. K₂O is more basic than Na₂O. Correct.
\item Na₂O vs MgO vs Al₂O₃: These elements are in period 3. Across a period, metallic character decreases, and the basic character of oxides decreases. Na₂O (strongly basic) $>$ MgO (basic) $>$ Al₂O₃ (amphoteric). Correct.
\end{itemize} \item Conclusion: Statement D is correct.
\end{itemize} Step 3: Final Answer:
Statements A, B, and D are correct, while statement C is incorrect. Therefore, the correct option includes A, B, and D only.