The Question is from neet-chemistry the-s-block-elements , and it was created by the author A.RAMA NAGESWARARAO

Question

Which alkaline earth metal forms a peroxide MO$_2$ most readily, and why?

A.RAMA NAGESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: D. Barium (Ba)
## CONCEPT FRAMEWORK
	**Primary Concept**: Stability of Peroxides and Superoxides
	**Related Topics**: Group 2 elements (Alkaline Earth Metals), Oxidation States, Lattice Energy, Size of atoms and ions.
	**Prerequisites**: Periodic trends, Ionic bonding.
## SOLUTION PATHWAY
**Given Information**:
	We are looking for the alkaline earth metal that most readily forms a peroxide with the formula MO$_2$.
**Method & Steps**:
	1. **Trend in peroxide formation**: The stability of peroxides and superoxides increases down the group (from Be to Ba).
		Reasoning: As the size of the cation increases, the larger cation can better stabilize the larger peroxide anion (O$_2^{2-}$). This is due to lower charge density and reduced polarization of the anion.
	2. **Barium's position in the group**: Barium (Ba) is the largest alkaline earth metal.
		Working:  Ba is at the bottom of Group 2, below Mg, Ca, and Sr.
		Key Point: Larger cation size leads to greater stability of the peroxide.
	3. **Barium peroxide formation**: Barium forms the peroxide (BaO$_2$) most readily.
		Result: BaO$_2$ is the most stable peroxide among the alkaline earth metals.
		Verification: This aligns with the trend of increasing stability of peroxides down the group.
 ## OPTION ANALYSIS
[A] :x: Magnesium (Mg) forms MgO (oxide) more readily than MgO$_2$ (peroxide). Mg$^{2+}$ is relatively small and has high charge density, making it less suitable for stabilizing the large O$_2^{2-}$ anion.
[B] :x: Calcium (Ca) can form CaO$_2$, but it is less stable than BaO$_2$. Ca$^{2+}$ is larger than Mg$^{2+}$ but smaller than Ba$^{2+}$, so its peroxide is less stable.
[C] :x: Strontium (Sr) can also form SrO$_2$, but it is less stable than BaO$_2$. Sr$^{2+}$ is larger than Ca$^{2+}$ but smaller than Ba$^{2+}$, so its peroxide is less stable.
[D] ✓: Barium (Ba) forms BaO$_2$ most readily due to the large size and low charge density of Ba$^{2+}$, which effectively stabilizes the O$_2^{2-}$ anion.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. Assuming stability is solely based on electronegativity: While electronegativity plays a role, the size of the cation is the dominant factor in peroxide stability in this case.
		2. Ignoring the trend down the group: Remember that the stability of peroxides and superoxides generally increases as you move down Group 2 (and Group 1).
	**Quick Tips**:
		1. Relate size to stability: Larger cations stabilize larger anions better due to lower charge density.
		2. Remember the trend: Peroxide and superoxide stability increases down the group for alkali and alkaline earth metals.

A.RAMA NAGESWARARAO · Answer

Magnesium (Mg)

A.RAMA NAGESWARARAO · Answer

Calcium (Ca)

A.RAMA NAGESWARARAO · Answer

Strontium (Sr)

A.RAMA NAGESWARARAO · Answer

The correct Option is "Barium (Ba)"

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