The Question is from neet-chemistry classification-of-elements-and-periodicity-in-prop , and it was created by the author A.RAMA NAGESWARARAO

Question

In which of the following arrangements the order is not correct according to property indicated against it?

A.RAMA NAGESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Periodic Trends
	**Related Topics**: Atomic Structure, Electronic Configuration
	**Prerequisites**: Understanding of periodic table trends for ionic size, ionization energy, electron affinity and atomic radius.
## SOLUTION PATHWAY
**Given Information**:
	* Four options with different properties and elements or ions.
	* Need to find the option where the order is NOT correct.
**Method & Steps**:
	1. **Analyze Option A (Ionic Size):**
		Reasoning:  Isoelectronic species (same number of electrons) have ionic size inversely proportional to the nuclear charge.  $Al^{3+}$ has the highest nuclear charge, followed by $Mg^{2+}$, then $Na^+$, and finally $F^-$ with the lowest nuclear charge (and additional electrons).  Therefore, the size order should be $Al^{3+} < Mg^{2+} < Na^+ < F^-$.
		Formula/Rule: Ionic size decreases with increasing nuclear charge for isoelectronic species.
	2. **Analyze Option B (Ionization Energy):**
		Reasoning: Generally, ionization energy increases across a period from left to right. However, there are exceptions. Nitrogen has a half-filled p-subshell which is more stable, hence it has a higher IE1 than oxygen. So the correct order is B < C < O < N.
		Key Point: Half-filled and fully-filled subshells have extra stability resulting in higher ionization energies.
	3. **Analyze Option C (Electron Affinity):**
		Reasoning:  Electron affinity generally increases across a period (becomes more negative) and decreases down a group. However, chlorine has a more negative EA than fluorine due to its smaller size and higher electron-electron repulsion.
		Key Point: Electron affinity is a complex property. Be mindful of exceptions.
	4. **Analyze Option D (Metallic Radius):**
		Reasoning:  Metallic radius increases down a group due to increasing number of electron shells.
		Formula/Rule: Metallic radius increases down a group.
	5. **Final Step:**
		Result: Option B is the only incorrect option.
		Verification: Review of periodic trends confirms the other options are correct.
 ## OPTION ANALYSIS
[A] :x: Correct order of ionic size.
[B] ✓: Incorrect order of ionization energy.  The correct order is B < C < O < N.
[C] :x: Correct order of electron affinity (with the Cl-F exception).
[D] :x: Correct order of metallic radius.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. **Incorrectly applying general trends**: Not considering exceptions like the half-filled and fully-filled stability or Cl vs F electron affinity.
		2. **Confusion between atomic radius and ionic radius**:  Remember that ionic radius can be different from atomic radius.
	**Quick Tips**:
		1.  Draw the periodic table and write down the trends for each property to visualize them.
		2.  Remember the exceptions and reasons for them.

A.RAMA NAGESWARARAO · Answer

Increasing size $:{\rm{A}}{{\rm{l}}^{3 + }} < M{{\rm{g}}^{2 + }} < N{{\rm{a}}^ + } < {{\rm{F}}^ - }$

A.RAMA NAGESWARARAO · Answer

The correct Option is "Increasing $I{E_1}{\rm{\;}}:{\rm{B}} < C < N < O$ "

A.RAMA NAGESWARARAO · Answer

Increasing $E{A_1}{\rm{\;}}:{\rm{I}} < Br < F < Cl$

A.RAMA NAGESWARARAO · Answer

Increasing metallic radius : ${\rm{Li}} < Na < K < Rb$

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