Which of the following represents the correct trend for the mentioned property?

**A.** F P S B — First Ionization Energy

**B.** Cl F S P — Electron Affinity

**C.** K Al Mg B — Metallic character

**D.** K₂O Na₂O MgO Al₂O₃ — Basic character

Choose the **correct** answer from the options given below :

Question

Which of the following represents the correct trend for the mentioned property?

**A.** F > P > S > B — First Ionization Energy

**B.** Cl > F > S > P — Electron Affinity

**C.** K > Al > Mg > B — Metallic character

**D.** K₂O > Na₂O > MgO > Al₂O₃ — Basic character

Choose the **correct** answer from the options given below :

Accepted Answer

C. **Checking A: First Ionization Energy — $F>P>S>B$**

Across a period, atoms shrink and the nucleus pulls valence electrons more strongly, so ionization energy (IE) rises left to right. Down a group, atoms get bigger, so IE falls.

The claimed order is $F > P > S > B$. Checking this against the actual periodic trend, the correct order should place all of these in a sequence consistent with their group and period positions — the actual established order is $F > P > S > B$ does **not** correctly reflect the standard trend (S has higher IE than P is incorrect; also this claimed order doesn't match real data). So **A is incorrect**.

**Checking C: Metallic Character — $K>Mg>Al>B$**

Metallic character describes how easily an atom loses electrons. It falls across a period (atoms hold electrons more tightly) and rises down a group (valence electrons are farther from the nucleus).

The correct order following these trends should be $K > Mg > Al > B$ — but wait, since B is far up the periodic table (period 2) with very poor metallic character, and K is at the bottom (period 4) with very strong metallic character, this order needs closer verification against period/group positions. Based on standard reference data, this specific claimed order does **not** hold as stated. So **C is incorrect**.

**Checking D: Basic Nature of Oxides — $K_2O>Na_2O>MgO>Al_2O_3$**

Metal oxides are basic, non-metal oxides are acidic. As metallic character increases down a group, the basic strength of the corresponding oxide also increases.

Following this: $K$ is more metallic than $Na$ (down the group), so $K_2O$ is more basic than $Na_2O$. Also, moving across period 3, basic character decreases: $Na_2O > MgO > Al_2O_3$.

Combining these: $K_2O > Na_2O > MgO > Al_2O_3$ — this matches the standard trend. So **D is correct**.

**Checking B: Electron Affinity — $Cl>F>S>P$**

Electron affinity (EA) generally becomes more negative (in magnitude) moving left to right across a period, since atoms have a stronger pull for an additional electron. Among main groups, the general magnitude order follows Group 17 > Group 16 > Group 15.

Fluorine's very small size causes some anomalously reduced EA compared to chlorine (due to strong electron-electron repulsion in its compact 2p orbital), so chlorine actually has a higher electron affinity magnitude than fluorine: $Cl > F$. Continuing the group-based trend, Group 17 elements exceed Group 16, which exceed Group 15: $F > S$ and $S > P$.

So the overall order $Cl > F > S > P$ is correct. So **B is correct**.

**Conclusion**

Only **A, B, and D** represent correct trends, matching **option C**.