Q. Lewis structure of \( \mathrm{SO_4^{2-}} \)

We want the Lewis structure of the sulfate ion, \( \text{SO}_4^{2-} \). That means we will: (1) count valence electrons, (2) choose a central atom, (3) place initial bonds, (4) add lone pairs, (5) add the charge by adjusting electrons, and (6) verify the options and resonance.

1) Count total valence electrons

Determine the number of valence electrons from each atom:

• Sulfur (S) is in Group 16, so it has \(6\) valence electrons.
• Oxygen (O) is in Group 16, so each O has \(6\) valence electrons. There are \(4\) oxygens, so \(4 \times 6 = 24\).
• The ion has charge \(2-\), so we add \(2\) extra electrons: \(+2\) electrons.

Total valence electrons:

\[
6 + 24 + 2 = 32
\]

2) Choose the central atom

The least electronegative atom is usually central. Sulfur is less electronegative than oxygen, so sulfur is the central atom.

Thus, sulfur is the center, surrounded by four oxygens.

3) Make initial bonding (skeleton structure)

Start by connecting sulfur to all four oxygens with single bonds. That gives a “cross” structure: one S–O bond to each O.

Each bond is two electrons. With four single bonds:

\[
4 \text{ bonds} \times 2 = 8 \text{ electrons used}
\]

Remaining electrons:

\[
32 – 8 = 24
\]

4) Place lone pairs on the oxygens

Distribute the remaining electrons as lone pairs on the outer atoms (the oxygens first).

Each oxygen typically gets \(3\) lone pairs when it is single-bonded (because oxygen has 6 valence electrons; a single bond uses 2 of them, leaving 4 electrons = two lone pairs? Let’s do it correctly using the electron count approach):

After placing single bonds, the remaining \(24\) electrons correspond to lone pairs on four oxygen atoms:

\[
24 \text{ electrons} \div 2 = 12 \text{ lone pairs}
\]

Dividing equally among four oxygens:

\[
12 \div 4 = 3 \text{ lone pairs per oxygen}
\]

So each O gets \(3\) lone pairs.

5) Check formal charges and adjust (resonance with one double bond location)

Now we check what this structure implies:

• With four single bonds from S to O, sulfur has only 4 bonds (8 bonding electrons around it).
• But in sulfate, sulfur typically forms 5 or 6 bonds in resonance forms (expanded octet), and the ion has a \(2-\) charge.

The electron arrangement that best matches the overall charge and typical sulfate resonance is: two S–O bonds are double bonds and two S–O bonds are single bonds, with the single-bonded oxygens bearing \(3\) lone pairs and the double-bonded oxygens bearing \(2\) lone pairs.

One resonance form can be described like this:

• Choose two oxygens to be double-bonded to sulfur.
• The other two oxygens remain single-bonded to sulfur.
• Single-bonded oxygens each have \(3\) lone pairs.
• Double-bonded oxygens each have \(2\) lone pairs.
• The overall ion charge is \(2-\).

6) Write the Lewis structure with formal charges

In the resonance form:

• Each double-bonded oxygen typically has formal charge \(0\).
• Each single-bonded oxygen typically has formal charge \(-1\).

So the total formal charge is:

\[
(-1) + (-1) + 0 + 0 = -2
\]

which matches \( \text{SO}_4^{2-} \).

7) Final Lewis structure (one resonance form)

Below is one valid Lewis structure resonance form. Any of the four oxygens can be the “single-bonded” ones, so multiple equivalent resonance structures exist.

One resonance form:

Structure description:

• S connected to four O atoms.
• Two S–O are double bonds.
• Two S–O are single bonds.
• Single-bonded O atoms each have 3 lone pairs and formal charge \(-1\).
• Double-bonded O atoms each have 2 lone pairs and formal charge \(0\).

Lewis structure written in text form (choose any two oxygens to be double-bonded):

\[
\text{O}=\text{S}(\text{O}^{-})-\text{O}
\]

This shorthand is meant to indicate the pattern: two \( \text{S=O} \) and two \( \text{S-O}^{-}\) in the same resonance form.

8) Resonance conclusion (important)

Because the sulfate has 4 oxygens and the double bonds are interchangeable in resonance, there are multiple equivalent resonance Lewis structures. The structure actually represents an average where all S–O bonds are equivalent in length (each bond order averages to \(1.5\)).

Final Answer

The Lewis structure of \( \text{SO}_4^{2-} \) is one where sulfur is central and bonded to four oxygens, with two S–O double bonds and two S–O single bonds. Each single-bonded oxygen has 3 lone pairs and formal charge \( -1 \), while each double-bonded oxygen has 2 lone pairs and formal charge \(0\). Multiple resonance forms exist where the positions of the double bonds swap among the oxygens.