Arrange the following free radicals in order | vedclass

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(C) The stability of free radicals depends on the degree of substitution and resonance stabilization.$(A)$ is a primary $(1^{\circ})$ alkyl radical: $

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$C > B > A$

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(C) The stability of free radicals depends on the degree of substitution and resonance stabilization.$(A)$ is a primary $(1^{\circ})$ alkyl radical: $R-CH_2•$.$(B)$ is a secondary $(2^{\circ})$ alkyl radical: $R-CH(CH_3)•$. Secondary radicals are more stable than primary radicals due to increased hyperconjugation and inductive effects.$(C)$ is an allylic radical: $Cyclohexylidene-CH-CH_2•$. This radical is resonance-stabilized by the adjacent double bond,making it the most stable among the three.Therefore,the order of decreasing stability is $C > B > A$.

Column-$I$	Column-$II$
$A$. $CH_3-O^{+}=CH-CH_3$	$1$. Stable due to resonance
$B$. $F_3C^{+}$	$2$. Destabilised due to inductive effect
$C$. $(CH_3)_3C^{-}$	$3$. Stabilised by hyperconjugation
$D$. $CH_3-CH^{+}-CH_3$	$4$. $A$ secondary carbocation

Arrange the following free radicals in order of decreasing stability:
$(A)$ Cyclohexyl-CH2-CH2•
$(B)$ Cyclohexyl-$CH$•-CH3
$(C)$ Cyclohexyl=$CH$-CH2•
(Note: The dot represents the radical center.)

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Arrange the following free radicals in order of decreasing stability:$(A)$ Cyclohexyl-CH2-CH2•$(B)$ Cyclohexyl-$CH$•-CH3$(C)$ Cyclohexyl=$CH$-CH2•(Note: The dot represents the radical center.)