Draw the lewis structure of ClO-. Add formal charges, if any, to the atoms.
The Correct Answer and Explanation is :
Lewis Structure of ClO⁻:
To draw the Lewis structure of ( \text{ClO}^- ):
- Count the valence electrons:
- Chlorine (Cl) has 7 valence electrons.
- Oxygen (O) has 6 valence electrons.
- The negative charge adds 1 extra electron.
- Total valence electrons = ( 7 + 6 + 1 = 14 ).
- Create the skeleton structure:
- Place the chlorine (Cl) and oxygen (O) atoms adjacent to each other with a single bond between them.
- Start distributing the remaining electrons.
- Distribute the electrons:
- After the single bond, 12 electrons remain (14 total – 2 used in the bond).
- Assign 6 electrons (3 lone pairs) to oxygen, fulfilling its octet.
- Assign the remaining 6 electrons (3 lone pairs) to chlorine.
- Check octets and formal charges:
- Oxygen has 8 electrons (6 lone pair + 2 bonding), so its octet is satisfied.
- Chlorine has 8 electrons (6 lone pair + 2 bonding), so its octet is satisfied.
- Assign formal charges:
- Formal charge = ( \text{Valence electrons} – (\text{Lone electrons} + \frac{\text{Bonding electrons}}{2}) ).
- For oxygen: ( 6 – (6 + \frac{2}{2}) = -1 ).
- For chlorine: ( 7 – (6 + \frac{2}{2}) = 0 ).
Thus, oxygen has a formal charge of -1, while chlorine has no formal charge.
Explanation:
The Lewis structure of ( \text{ClO}^- ) consists of a single bond between Cl and O, with three lone pairs on each atom. The negative charge is localized on oxygen because it is more electronegative and can stabilize the charge. Chlorine, being less electronegative, remains neutral. This structure minimizes formal charges, adheres to the octet rule, and accurately represents the distribution of electrons in the hypochlorite ion.