To draw the Lewis structure for OCl2, we start by counting the total number of valence electrons. Oxygen has 6 valence electrons, and each chlorine has 7 valence electrons. Therefore, the total is:
6 (from O) + 2 × 7 (from Cl) = 20 valence electrons.
Next, we place oxygen in the center as it is less electronegative than chlorine. We then surround it with the two chlorine atoms. Each Cl atom forms a single bond with the O atom, using 4 electrons (2 for each bond).
This leaves us with:
20 – 4 = 16 valence electrons.
Now we distribute the remaining electrons to complete the octets of the chlorine atoms first. Each Cl atom will take 6 electrons (3 lone pairs), using 12 of the remaining electrons:
16 – 12 = 4 valence electrons left.
We place these 4 remaining electrons as 2 lone pairs on the oxygen, giving us the final Lewis structure:
This structure shows two single bonds between oxygen and each chlorine, and oxygen has two lone pairs.
Now, to determine the molecular shape, we look at the number of bonding pairs and lone pairs around the oxygen atom. Oxygen has 2 bonding pairs (for the O-Cl bonds) and 2 lone pairs. According to the VSEPR theory, the arrangement of these electron pairs corresponds to a tetrahedral electron geometry.
However, since we are interested in the molecular shape, we only consider the bonding pairs. With two bonding pairs and two lone pairs, the shape of the molecule is classified as ‘bent’.
Therefore, the correct option for the molecular shape of OCl2 is:
- f) bent