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 stability of the following carbocations follows the order:

Which is the correct order of stability of the following three carbonium ions?
$(I)$ $CH_2 = CH - CH^+ - CH_3$
$(II)$ $CH_2 = C(CH_3) - CH_2^+$
$(III)$ $CH_3 - CH = CH - CH_2^+$

The correct order of increasing stabilities of the following carbocations is:
$(i)$ Allyl carbocation
$(ii)$ Ethyl carbocation
$(iii)$ Benzyl carbocation
$(iv)$ Isobutyl carbocation

What is the decreasing order of stability (most stable $\to$ least stable) of the following carbocations?

The lower stability of ethyl anion compared to methyl anion and the higher stability of ethyl radical compared to methyl radical,respectively,are due to

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