The hypochlorite ion, ClO–, consists of a chlorine (Cl) atom bonded to an oxygen (O) atom. To draw the Lewis structure, we start by calculating the total number of valence electrons available for bonding.
- Chlorine has 7 valence electrons.
- Oxygen has 6 valence electrons.
- Since the ion carries a negative charge, we add 1 additional electron.
This gives us a total of 7 + 6 + 1 = 14 valence electrons to work with.
Next, we can determine how to connect these atoms. The more electronegative atom, oxygen, will typically be placed closer to the terminal atom:
- We place Cl and O next to each other, with a single bond between them, initially using 2 electrons.
Now we have 14 – 2 = 12 electrons left. We can place lone pairs of electrons around the atoms to satisfy their octets:
- Place 6 electrons (3 lone pairs) around the oxygen atom. Now oxygen has a total of 8 electrons (2 in the bond with Cl and 6 as lone pairs).
- Place 3 lone pairs of electrons around the chlorine atom. Chlorine now has 5 electrons (2 in the bond with O and 6 as lone pairs).
At this stage, we have used up all 14 electrons. The Lewis structure looks like this:
Next, we need to check the formal charges to ensure the structure is stable:
- For Cl: Number of valence electrons – (non-bonding electrons + 1/2 bonding electrons) = 7 – (6 + 1) = 0.
- For O: Number of valence electrons – (non-bonding electrons + 1/2 bonding electrons) = 6 – (6 + 1) = -1.
The formal charge on Cl is 0 and on O is -1, which sums up correctly to the overall charge of the ion (–1).
This Lewis structure indicates that the hypochlorite ion is relatively stable, with chlorine holding a formal charge of 0 and oxygen bearing the negative charge, consistent with the electronegativity of the atoms involved.