Q. \[ \mathrm{ClO_3^-}\ \text{Lewis structure (summary)} \]

Goal: Draw the Lewis structure for the species \(\text{ClO}_3^-\).

Step 1: Identify the charge and total valence electrons

Chlorine is in group 17, so it has 7 valence electrons.

Oxygen is in group 16, so each oxygen has 6 valence electrons. There are 3 oxygens.

Because the ion has a \(-1\) charge, we add one extra electron.

Total valence electrons:

\[
7 + 3(6) + 1 = 7 + 18 + 1 = 26
\]

So you start with 26 electrons.

Step 2: Choose the central atom

In most oxyanions like this, the central atom is the element less electronegative than oxygen.

Here, chlorine is less electronegative than oxygen, so place Cl in the center and connect three O atoms to it.

Initial skeleton shape (single bonds):

You place one bond from Cl to each O. That uses 1 pair of electrons per bond.

There are 3 Cl–O single bonds, so that uses \(3\) electron pairs \(= 6\) electrons.

Remaining electrons after bonds:

\[
26 – 6 = 20
\]

Step 3: Complete octets on the oxygen atoms

Each oxygen should have an octet (8 electrons around it).

Right now, each oxygen has only 2 electrons from its single bond (so it has a partial octet).

To reach an octet, each oxygen needs 3 more lone pairs.

For each oxygen, assign 3 lone pairs (that is 6 electrons) plus the 2 electrons in the bond, giving 8 total.

This uses:

\[
3 \text{ O atoms} \times 3 \text{ lone pairs} = 9 \text{ lone pairs} = 18 \text{ electrons}
\]

Electrons used: \(18\)

Electrons left:

\[
20 – 18 = 2
\]

So after giving each oxygen 3 lone pairs, there are 2 electrons left.

Step 4: Place remaining electrons on chlorine

The remaining 2 electrons form one lone pair on the central chlorine.

Now chlorine has:

3 bonds (each bond contributes 2 electrons) plus 1 lone pair.

So chlorine has a total of \(6 + 2 = 8\) electrons around it, which is an octet.

Step 5: Confirm formal charges (and final structure)

In \(\text{ClO}_3^-\), the most common Lewis structure is the one with one negative charge distributed by resonance, but the typical drawing shows:

• One Cl–O is a double bond
• The other two Cl–O are single bonds
• The oxygen atoms have appropriate lone pairs
• There is no net mismatch in overall charge

To be explicit, the correct Lewis structure for \(\text{ClO}_3^-\) is represented by resonance among three equivalent forms.

Step 6: Write one resonance form (common Lewis structure)

Draw this form:

• Cl in the center
• One O is double-bonded to Cl
• Two O are single-bonded to Cl
• Single-bonded oxygens have 3 lone pairs each
• Double-bonded oxygen has 2 lone pairs
• Chlorine has no lone pair in the double-bond resonance form

One resonance structure:

Written in words: \(\text{O}=\text{Cl}(\text{O}^-)_2\) with resonance.

In Lewis-notation form, you can draw it as: one \(\text{Cl}=\text{O}\) and two \(\text{Cl}-\text{O}\), with a negative charge on one of the single-bonded oxygens (and the double bond position shifts in resonance).

Step 7: Provide the final answer (set of resonance structures)

Final Lewis structure of \(\text{ClO}_3^-\):

1. \[
   \text{O}=\text{Cl}-\text{O}^- \quad \text{with }\text{O} \text{ also bonded to Cl}
   \]
   where the double bond can be on any of the three oxygens.
2. The double bond position is interchangeable, giving three resonance forms.

Equivalent best depiction: three resonance structures where one \(\text{Cl}=\text{O}\) double bond and two \(\text{Cl}-\text{O}\) single bonds exist, with the \(-1\) formal charge located on one of the single-bonded oxygens in each resonance form.