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The enolic form of acetone contains:
a) 9-sigma, 1-pi-bond and 2 lone pairs
b) 8-sigma, 2-pi-bond and 2 lone pairs
c) 10-sigma, 1-pi-bond and 1 lone pair
d) 9-sigma, 2 pi-bond and 1 lone pair

Answer
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Hint: Start the question by drawing acetone first and then its enolic form. Then count the number of sigma-bond, pi-bond and lone pairs respectively, taking into account the number of valence shell electrons of each element.

Complete step by step answer:
The chemical formula for acetone is given as -\[{{C}_{3}}{{H}_{6}}O\].
Due to keto-enol tautomerism, acetone exists in two states as shown below –

The first and second structures represent the keto and enol forms respectively. The keto and enol forms are called tautomers of each other.
As we can see from second the structure, there are –
9 sigma bonds – 5 (C-H) + 2 (C-C) + 1 (C-OH) + 1 (O-H)
1 pi bond – 1 (C-C)
2 lone pairs – Oxygen (Alcohol/enol)
Therefore, the answer is option (a) - the enolic form of acetone contains 9-sigma, 1-pi-bond and 2 lone pairs.

Additional information:
Acetone is also known as propanone. It is the smallest and simplest ketone.

Note: Acetone is a colourless, volatile, flammable liquid and has a characteristic odour. It is also miscible with water and therefore, serves as an important solvent in industries, homes, and laboratories. You might be familiar with acetone as it is an active ingredient in nail polish remover and in paint thinner. Not only this, acetone also occurs naturally in the human body as a by-product of metabolism.