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Hi, in this question we are asked to draw the levy structure of ketene that has molecular formula c2h2o.
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It is given that in the structure there is no oh bond.
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We are asked to find out the hybridization of carbon atoms.
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Also, we need to draw a separate diagram to show sigma and pi bonds.
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The first step to draw the levy structure is to count the total number of...
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Valence electrons in carbon hydrogen and oxygen.
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Here we can say that total number of valence electrons are equal to 2 into 4 plus 2 into 1 plus 1 6.
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Here we are having 2 carbon atoms each carbon atom is having 4 valence electron we are having two hydrogen atom.
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Each hydrogen atom is having one valence electron.
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And we are having one oxygen atom and it is having six valence electron.
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2 into 4 for carbon, 2 into 1 for hydrogen and 1 into 6 for oxygen.
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On solving this we are getting 16 valence electrons.
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Now in the next step we will draw the skeletal structure of carbon, hydrogen and oxygen.
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Before drawing this, we need to keep in mind that there is no oxygen hydrogen bond.
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Therefore, we can draw the skeletal structure as carbon atom, which is attached with two hydrogen atom, this carbon atom is further attached with carbon atom, and this carbon is further attached with oxygen atom.
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We know that each bond is made up of two electrons.
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Therefore, we need to count how many electrons from 16 are already used in the skeletal structure.
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Let us start to count from here.
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1 .2.
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3.
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4.
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5.
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6.
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7.
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8.
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We have used 8 electron till now...