The enolic form of acetone contains:

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(A) The keto form of acetone is $CH_3COCH_3$. The enolic form is obtained by the migration of an $\alpha$-hydrogen to the oxygen atom,resulting in $CH

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$9\sigma, 1\pi$ bonds and $2$ lone pairs

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(A) The keto form of acetone is $CH_3COCH_3$. The enolic form is obtained by the migration of an $\alpha$-hydrogen to the oxygen atom,resulting in $CH_3-C(OH)=CH_2$.In $CH_3-C(OH)=CH_2$:$1$. Total $\sigma$ bonds: $3$ (in $CH_3$) + $1$ ($C$-$C$) + $1$ ($C$-$O$) + $1$ ($O$-$H$) + $2$ ($C$-$H$ in $CH_2$) + $1$ ($C$=$C$) = $9\sigma$ bonds.$2$. Total $\pi$ bonds: $1$ (in $C=C$ double bond) = $1\pi$ bond.$3$. Lone pairs: Oxygen has $2$ lone pairs in the keto form,but in the enolic form,the oxygen atom involved in the $C-OH$ group retains $2$ lone pairs. However,looking at the structure $CH_3-C(OH)=CH_2$,the oxygen atom has $2$ lone pairs. Re-evaluating the options,the standard representation for this specific question often considers the lone pairs on the oxygen atom. Given the structure,there are $2$ lone pairs on the oxygen atom.

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