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Q9.

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Organic Chemistry
Found in: Page 19

Short Answer

Question: Draw Lewis structures for each molecular formula.

a. C2H4Cl2 ( two isomers)

b. C3H8O (three isomers)

c. C3H6 (two isomers)

Answer

a.

b.

c.

See the step by step solution

Step by Step Solution

Step-by-Step Solution Step 1: Atomic arrangements in Lewis structure

All the carbon atoms are arranged, and any other electropositive elements are at the center.

All the hydrogen and halogen atoms are placed at the periphery because both of these atoms form only one single bond.

The hydrogen atoms are placed around the carbon atoms.

Step 2: Total valence electrons

The total valence electrons is calculated for the whole compound. Then count the total number of electron pairs by dividing the total number of valence electrons by two.

One electron pair as a bond is added between two consecutive atoms. Then, the remaining electron pairs (if any) are added as lone pairs on the more electronegative atoms.

Step 3: Checking the octet

The octet of each atom except hydrogen is checked. If the octet of any of the atoms is not complete, then the lone pair is converted into a bond pair, and a double bond is shown between the two atoms lacking an octet.

Step 4: Forming the Lewis structures of the given compounds

a. All carbon atoms are placed at the center. Then all the halogens and hydrogen atoms are placed at the periphery.

  • The number of valence electrons in the hydrogen atom is one.
  • The number of valence electrons in a carbon atom is four.
  • The number of valence electrons in a chlorine atom is seven.

Therefore,

The total number of valence electron pairs = 13 electron pairs.

These electron pairs are arranged as bonds between the atoms, and the remaining electron pairs are assigned as lone pairs on chlorine atoms.

The octet of each atom is then checked. If the octet of any of the atoms is not complete, then the lone pair is converted into a bond pair of electrons, and a double bond is shown there.

Since the above structure is sterically hindered, two large chlorine atoms are placed side-to-side. So, the sterically less hindered structure is given as:

Lewis structure of a.

The two isomers of this compound are as follows:

Isomers of a.

b. All the carbon atoms are placed side-to-side in the center and then placed one oxygen atom next to the carbon atom such that three isomers are obtained for the given molecular formula.

The hydrogen atoms are placed around all the carbon atoms and on the side of oxygen placed next to the carbon atom.

  • The number of valence electrons in the hydrogen atom is one.
  • The number of valence electrons in a carbon atom is four.
  • The number of valence electrons in the oxygen atom is six

=12e-+8e-+ 6e-

= 26e-

These electron pairs are arranged as bonds between the atoms, and the remaining two electron pairs are assigned as lone pairs on oxygen-atom. The

octet of each atom is now complete.

Lewis structures of the three isomers of b.

Given below are three resonance forms of the given compound.

Isomers of b.

  1. Place all carbon-atoms side-to-side in the center. Then place all the hydrogen atoms around each carbon atom.

  • The number of valence electrons in the hydrogen atom is one.
  • The number of valence electrons in a carbon atom is four.

These electron pairs are arranged as bonds in between the atoms, and the remaining one electron pair is assigned as lone pairs on carbon-atom.

The octet of the middle carbon atom is not complete. So, the lone pair is moved in between the two carbon atoms lacking an octet.

Lewis structures of c

The two isomers of this compound are as follows:

Isomers of b.

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