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In a double bond, two pairs of electrons are shared between two atoms. This means that a total of four electrons are involved in a double bond.

To write the Lewis structure of a molecule containing a double bond, follow these steps: 1. Determine the total number of valence electrons in the molecule. 2. Choose a central atom, which is usually the element that needs the greatest number of electrons to complete its octet (except Hydrogen). 3. Place the remaining atoms around the central atom. 4. Add single bonds between the central atom and each surrounding atom. 5. Distribute the remaining valence electrons as lone pairs to complete the octet of each atom (except hydrogen, which needs only a pair). 6. If any atoms still lack a complete octet, move the lone pairs to form a double bond (creating a shared pair of electrons). As an example, let's draw the Lewis structure for a molecule of oxygen (O2).

Each oxygen atom has 6 valence electrons. So, in the case of O2, there is a total of 12 valence electrons (6+6).

Since there are only two oxygen atoms in O2, they will both be central atoms.

As there are only two atoms, they will be bonded together.

Draw a single bond between the two oxygen atoms. This will use 2 of the 12 valence electrons.

Each oxygen atom needs 6 more electrons to complete its octet. Distribute the remaining electrons as lone pairs. There are two pairs of electrons for each oxygen atom currently and 10 more valence electrons to be distributed \$ (12 - 2 = 10) \$.

Since the octet for each oxygen atom has not been completed, we move two lone pairs on each oxygen atom to be shared between them, creating a double bond (two shared pairs). Now, we have a double bond between the two oxygen atoms, and each oxygen atom has a completed octet. The final Lewis structure for the oxygen molecule containing a double bond is: \[ \mathrm{O} = \mathrm{O} \] Each oxygen atom has two bond pairs (represented by the double bond) and two lone pairs.