Arrange the following carbanions in the decre | vedclass

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(C) The stability of carbanions depends on the $s$-character of the carbon atom bearing the negative charge.As the $s$-character increases,the electro

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$3 > 2 > 1$

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(C) The stability of carbanions depends on the $s$-character of the carbon atom bearing the negative charge.As the $s$-character increases,the electronegativity of the carbon atom increases,making it better at stabilizing the negative charge.$(1)$ $CH_3-CH_2^-$: $sp^3$ hybridized ($25\% \ s$-character).$(2)$ $CH_2=CH^-$: $sp^2$ hybridized ($33.3\% \ s$-character).$(3)$ $HC \equiv C^-$: $sp$ hybridized ($50\% \ s$-character).The order of stability is $sp > sp^2 > sp^3$,which corresponds to $(3) > (2) > (1)$.

Arrange the following carbanions in the decreasing order of their stability:
$(1) CH_3-CH_2^-, (2) CH_2=CH^-, (3) HC \equiv C^-$

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Arrange the following carbanions in the decreasing order of their stability:$(1) CH_3-CH_2^-, (2) CH_2=CH^-, (3) HC \equiv C^-$