Write Lewis formulas for the following ion ClO⁻.

The Lewis structure for the hypochlorite ion (ClO^–) can be drawn by following a few simple steps:

1. Determine the total number of valence electrons: Chlorine has 7 valence electrons and oxygen has 6. Since we have one negative charge, we add one more electron. So, the total is 7 + 6 + 1 = 14 valence electrons.
2. Place the atoms: Place the chlorine atom in the center and the oxygen atom connected to it. The usual bonding between chlorine and oxygen involves a single bond initially.
3. Draw the bond: Connect chlorine and oxygen with a single bond, which uses 2 electrons from the total. Now we have 14 – 2 = 12 electrons remaining.
4. Distribute the remaining electrons: Start by completing the octets for the outer atoms first. Oxygen needs 6 more electrons to complete its octet, so we place 6 electrons (as 3 lone pairs) around the oxygen atom. This consumes 6 electrons, leaving us with 12 – 6 = 6 electrons.
5. Distributing remaining electrons: Now, we have 6 electrons left. Place 2 lone pairs (4 electrons) around the chlorine atom, and that uses 4 of the remaining electrons, leaving us with 2 electrons.
6. Final adjustments: To accommodate the ion’s charge and stabilize the structure, place a single bond between chlorine and oxygen, as Cl already has 4 or 5 electrons, it can share more to stabilize the form, contributing the remaining electrons as lone pairs or bonds as required.

Thus, the final Lewis structure for ClO^– will have chlorine with two lone pairs and a single bond to oxygen, which has three lone pairs. This visually represents the sharing and distribution of electrons that stabilize the ion. You should see a formal charge of -1 distributed across the oxygen atom, maintaining the charge of the hypochlorite ion securely.