What is the increasing order of stability for | vedclass

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(B) $1$. The stability of carbanions is primarily governed by resonance and the inductive effect. $2$. Resonance: $Q$ $(C_6H_5\bar{C}H_2)$ is stabiliz

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$R < P < S < Q$

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(B) $1$. The stability of carbanions is primarily governed by resonance and the inductive effect. $2$. Resonance: $Q$ $(C_6H_5\bar{C}H_2)$ is stabilized by resonance with the benzene ring,and $S$ $(H_2\bar{C}^{-}CH=CH_2)$ is stabilized by resonance with the double bond. $Q$ is more stable than $S$ due to the greater delocalization of the negative charge over the aromatic ring. $3$. Inductive Effect: Alkyl groups exhibit a $+I$ effect,which destabilizes carbanions. $P$ $(\bar{C}H_3)$ has no alkyl groups,while $R$ $((CH_3)_2\bar{C}H)$ has two electron-donating methyl groups. Thus,$P$ is more stable than $R$. $4$. Combining these factors,the stability order is $R < P < S < Q$.

List-$I$ (Carbocation)	List-$II$ (Type)
$A$. $CH_3-C^{+}(CH_3)-CH_3$	$I$. Secondary carbocation
$B$. $CH_3-C^{+}H-CH_3$	$II$. Methyl carbocation
$C$. $CH_3-CH_2^+$	$III$. Primary carbocation
$D$. $CH_3^+$	$IV$. Tertiary carbocation

What is the increasing order of stability for the carbanions $P: \bar{C}H_3$,$Q: C_6H_5\bar{C}H_2$,$R: (CH_3)_2\bar{C}H$,and $S: H_2\bar{C}^{-}CH=CH_2$?

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Match the following columns and select the correct option.

List-$I$ (Carbocation) List-$II$ (Type)

$A$. $CH_3-C^{+}(CH_3)-CH_3$ $I$. Secondary carbocation

$B$. $CH_3-C^{+}H-CH_3$ $II$. Methyl carbocation

$C$. $CH_3-CH_2^+$ $III$. Primary carbocation

$D$. $CH_3^+$ $IV$. Tertiary carbocation

Arrange the $(C-H)$ bonds $x, y$ and $z$ in decreasing order of their bond dissociation energies in homolysis.

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