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QuestionChemistry

Draw the Lewis structure of the hypochlorite ion, including lone pairs.
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Step 1:
I'll solve this problem step by step, following the LaTeX formatting guidelines precisely:

Step 2:
: Determine the total number of valence electrons

- Total valence electrons: $$7 + 6 + 1 = 14$$ electrons
- Chlorine (Cl) has 7 valence electrons - Oxygen (O) has 6 valence electrons - The ion has a - 1 charge, which adds 1 additional electron

Step 3:
: Arrange atoms in the structure

- Place the chlorine atom in the center - Attach the oxygen atom to the chlorine with a single bond

Step 4:
: Distribute electrons as bonds and lone pairs

- Electrons remaining: $$14 - 2 = 12$$ electrons
- Single bond between Cl and O uses 2 electrons - Remaining electrons will be distributed as lone pairs

Step 5:
: Complete the octet for oxygen

\begin{array}{c}
- Oxygen needs 8 electrons - Place 3 lone pairs around oxygen (6 electrons) \text{Oxygen's electrons:} \ 1 \text{ bond (2 electrons)} + 3 \text{ lone pairs (6 electrons)} = 8 \text{ electrons} \end{array}

Step 6:
: Complete the octet for chlorine

\begin{array}{c}
- Chlorine needs 8 electrons - Place 3 lone pairs around chlorine \text{Chlorine's electrons:} \ 1 \text{ bond (2 electrons)} + 3 \text{ lone pairs (6 electrons)} = 8 \text{ electrons} \end{array} Final Lewis Structure: \text{Hypochlorite Ion (ClO}^{-}\text{):} \ \text{Cl} \text{ has three lone pairs} \ \text{O has three lone pairs} \ \text{Single bond between Cl and O} \end{array}

Final Answer

The Lewis structure of the hypochlorite ion (ClO⁻) consists of a chlorine atom single-bonded to an oxygen atom, with three lone pairs on the chlorine and three lone pairs on the oxygen, satisfying the octet rule for both atoms.