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  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-39

  • question_answer
    With a change in hybridization of the carbon bearing the charge, the stability of a carb anion decreases in the order:

    A)  \[sp<s{{p}^{2}}<s{{p}^{3}}\]             

    B)  \[sp<s{{p}^{3}}<s{{p}^{2}}\]

    C)  \[s{{p}^{3}}<s{{p}^{2}}<sp\]             

    D)  \[s{{p}^{2}}<sp<s{{p}^{3}}\]

    Correct Answer: C

    Solution :

     More the s-character, more is the stability of the carbanion. hence the correct order is sp > sp2 > sp3.


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