Question:medium

Which of the following is an INCORRECT match?

Show Hint

Alkali metals (Group 1) like Potassium ($\text{K}$) are always the most metallic elements in their respective periods. Seeing Potassium placed at the very bottom of the metallic character chain in Option (A) immediately flags it as the incorrect match.
Updated On: May 20, 2026
  • Metallic character: $\text{Al} > \text{Mg} > \text{B} > \text{K}$
  • First Ionization Enthalpies: $\text{Na} < \text{Mg} > \text{Al} < \text{Si}$
  • Electron Gain enthalpy: $\text{F} < \text{Cl} > \text{Br} > \text{I}$
  • Ionic size: $\text{Na}^+ > \text{Mg}^{+2} > \text{Al}^{+3} > \text{Si}^{+4}$
Show Solution

The Correct Option is A

Solution and Explanation

Understanding the Concept: Periodic properties follow predictable trends across periods and down groups:
Metallic character (electropositive nature) increases down a column as ionization energy falls and decreases left-to-right across a period as effective nuclear charge increases.
Ionization Enthalpy exhibits a stable anomaly between Group 2 ($\text{ns}^2$ configuration stability) and Group 13 ($\text{ns}^2\text{np}^1$).
Ionic size among isoelectronic species decreases as the nuclear atomic charge increases.

Step 1: Evaluate Option (A).
Potassium ($\text{K}$) is an alkali metal belonging to Group 1, making it highly electropositive with the lowest ionization energy in its row. Hence, it possesses a significantly higher metallic character than alkaline earth metals ($\text{Mg}$) or post-transition metals ($\text{Al}$). The actual correct decreasing order of metallic character is $\text{K}>\text{Mg}>\text{Al}>\text{B}$. Thus, match (A) is completely incorrect, and represents the targeted option choice.
Step 2: Verify why the other options are correct periodic trends.

Option (B) is a correct match: $\text{Mg}$ ($1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2$) has a fully filled subshell, which requires more energy to disrupt than the lone $3\text{p}$ electron of $\text{Al}$, causing the observed dip ($\text{Mg}>\text{Al}$).
Option (C) is a correct match: Chlorine has a higher electron affinity than fluorine because fluorine's tiny $2\text{p}$ orbital experiences significant electron-electron repulsion.
Option (D) is a correct match: These ions are isoelectronic ($10$ electrons). As nuclear charge increases from $\text{Na}$ ($+11$) to $\text{Si}$ ($+14$), the nucleus pulls the electron cloud tighter, reducing ionic size.
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