Which of the following species (A),(B),(C),an | vedclass

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(D) To determine the hybridization of the carbon atom in each species:$1$. $CH_3^+$ (Carbocation): The carbon atom has $3$ bond pairs and $0$ lone pai

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$A, C$ and $D$

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(D) To determine the hybridization of the carbon atom in each species:$1$. $CH_3^+$ (Carbocation): The carbon atom has $3$ bond pairs and $0$ lone pairs. Steric number $= 3 + 0 = 3$,which corresponds to $sp^2$ hybridization.$2$. $CH_3^-$ (Carbanion): The carbon atom has $3$ bond pairs and $1$ lone pair. Steric number $= 3 + 1 = 4$,which corresponds to $sp^3$ hybridization.$3$. $\dot{C}H_3$ (Free radical): The carbon atom has $3$ bond pairs and $1$ unpaired electron. The geometry is planar,and the carbon is $sp^2$ hybridized.$4$. $:CH_2$ (Carbene): In singlet carbene,the carbon atom has $2$ bond pairs and $1$ lone pair. Steric number $= 2 + 1 = 3$,which corresponds to $sp^2$ hybridization.Therefore,species $(A)$,$(C)$,and $(D)$ contain $sp^2$ hybridized carbon atoms.

Which of the following species $(A)$,$(B)$,$(C)$,and $(D)$ contain an $sp^2$ hybridized carbon atom? $(A) CH_3^+$,$(B) CH_3^-$,$(C) \dot{C}H_3$,$(D) :CH_2$

Similar Questions

The order of decreasing stability of the carbanions is:
$1.$ $(CH_3)_3C^-$
$2.$ $(CH_3)_2CH^-$
$3.$ $CH_3CH_2^-$
$4.$ $C_6H_5CH_2^-$

Arrange the following carbocations in decreasing order of their stability:
$\stackrel{+}{C}H_3, (CH_3)_3\stackrel{+}{C}, CH_3\stackrel{+}{C}H_2, (CH_3)_2\stackrel{+}{C}H$

Among the following carbocations:
$(I)$ $Ph_{2}\stackrel{+}{C} CH_{2} Me$
$(II)$ $Ph CH_{2} CH_{2} \stackrel{+}{C} HPh$
$(III)$ $Ph_{2} CH \stackrel{+}{C} HMe$
$(IV)$ $Ph_{2} C(Me) \stackrel{+}{C} H_{2}$
The order of stability is:

The correct order of stability of the given carbocations is:

Carbocation as an intermediate is likely to be formed in the reaction:

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