To draw the Lewis structure of selenium dioxide (SeO2), we first determine the total number of valence electrons available. Selenium (Se) is in group 16 and has 6 valence electrons, while each oxygen (O) also has 6 valence electrons. Therefore, for SeO2, the total number of valence electrons is:
Valence electrons = 6 (from Se) + 2 × 6 (from O) = 18 electrons
Next, we choose selenium as the central atom. It will be surrounded by the two oxygen atoms. We can start by placing a single bond between selenium and each oxygen atom, using 4 of the total valence electrons:
4 electrons used (2 bonds)
This leaves us with:
18 – 4 = 14 electrons remaining
Then, we place 6 electrons (or 3 lone pairs) around each oxygen atom to satisfy the octet rule, using 12 electrons:
12 electrons used (6 for each O)
This leaves us with:
14 – 12 = 2 electrons remaining
Now, we can use the remaining 2 electrons to form a double bond between selenium and one of the oxygen atoms. This results in one oxygen atom being bonded with a double bond, while the other has a single bond and retains 6 electrons (or 3 lone pairs).
The final Lewis structure will look like this:
O = Se – O
where the double bond represents the 2 shared electrons between selenium and one of the oxygen atoms, while the single bond represents the other oxygen atom. The oxygen with the double bond has 4 electrons (2 lone pairs) remaining, while the other oxygen has 6 lone electrons (3 lone pairs).
This structure fulfills the octet rule for all atoms involved and provides an accurate representation of the distribution of electrons in selenium dioxide.