Relative stability of the contributing structures is :
Relative stability of the contributing structures is :
\((I)>(III)>(II)\)
\((I)>(II)>(III)\)
\((II)>(I)>(III)\)
\((III)>(II)>(I)\)
The Correct Option is B
Approach Solution - 1
To determine the relative stability of the given contributing structures, we need to consider several factors, including the distribution of charges, the number of covalent bonds, and the octet rule. Let’s analyze each structure in detail:
- Structure (I):
- This structure has no charges. All atoms are neutral, and each atom satisfies the octet rule, making it highly stable.
- There are multiple covalent bonds, contributing to greater stability due to shared electrons.
- Structure (II):
- This structure has a negative charge on the carbon atom, which is relatively stable because carbon can accommodate a negative charge.
- The octet rule is still satisfied for the atoms present.
- Structure (III):
- This structure has a positive charge on an oxygen atom, which is less stable than a negative charge on carbon because oxygen is more electronegative and prefers to be negatively charged.
- The octet rule is satisfied, but the charge distribution contributes to reduced stability.
Comparing these factors, we can deduce the relative stability as follows:
- Structure (I) is the most stable because it has no charges and satisfies the octet rule fully.
- Structure (II) is more stable than structure (III), as carbon with a negative charge is more stable than oxygen with a positive charge.
- Structure (III) is the least stable for the reasons mentioned above.
Hence, the correct order of stability is \((I) > (II) > (III)\).
Approach Solution -2
Explanation:
1. Neutral structures are more stable than charged ones. Therefore, structure (I) is the most stable as it is neutral.
2. Among the charged structures, a positive charge (+) on a less electronegative atom (like carbon) is more stable than a positive charge on a more electronegative atom (like oxygen).
- Hence, structure (II), where C+ is present, is more stable than (III), where O+ is present.
Order: \( I > II > III.\)
Final Answer: Option (2).
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