Q. Lewis structure of H\(_2\)SO\(_4\).

We want the Lewis structure of sulfuric acid, \( \mathrm{H_2SO_4} \). A Lewis structure shows which atoms are connected, how many valence electrons are present, and where lone pairs go.

1) Count valence electrons

First find the total number of valence electrons for all atoms.

\( \mathrm{S} \) is in Group 16, so it has \(6\) valence electrons.

\( \mathrm{O} \) is in Group 16, so each \( \mathrm{O} \) has \(6\) valence electrons. There are \(4\) oxygens, so \(4 \times 6 = 24\).

\( \mathrm{H} \) is in Group 1, so each \( \mathrm{H} \) has \(1\) valence electron. There are \(2\) hydrogens, so \(2 \times 1 = 2\).

\[
\text{Total valence electrons} = 6 + 24 + 2 = 32
\]

2) Determine the connectivity (what atoms connect to what)

In \( \mathrm{H_2SO_4} \), sulfur is the central atom. It is bonded to all four oxygens.

There are two \( \mathrm{OH} \)-type oxygens (each bonded to one \( \mathrm{H} \)) and two other oxygens (double-bonded to sulfur in the common Lewis structure).

So the typical connectivity is:

\( \mathrm{S} \) connected to four \( \mathrm{O} \) atoms, and two of those oxygens each connected to one \( \mathrm{H} \).

3) Place electrons to satisfy octets (common correct Lewis structure)

A standard Lewis structure for \( \mathrm{H_2SO_4} \) uses:

• Two double bonds from sulfur to two oxygens
• Two single bonds from sulfur to two oxygens
• Each single-bonded oxygen also bonded to one hydrogen (forming \( \mathrm{-OH} \) groups)
• Lone pairs on all oxygens

4) Assign bonds and lone pairs explicitly

Let’s label the four oxygens as follows:

• Two oxygens are double-bonded to sulfur: \( \mathrm{S{=}O} \)
• Two oxygens are single-bonded to sulfur and also bonded to hydrogen: \( \mathrm{S{-}O{-}H} \)

Now apply octet rules for each oxygen type:

For a double-bonded oxygen \( \mathrm{S{=}O} \):

• The double bond provides 2 electron pairs for the oxygen
• An oxygen in Lewis structure typically has 2 lone pairs to complete the octet

For a single-bonded oxygen in \( \mathrm{S{-}O{-}H} \):

• The single bond to sulfur and the single bond to hydrogen each provide one shared pair
• That gives 2 bonding pairs total around oxygen (2 lines worth of bonding pairs arrive at oxygen)
• Oxygen then has 2 lone pairs to complete the octet

For sulfur, in this structure sulfur has four bonds total (two double bonds count as four “bond skeleton connections” around sulfur in terms of bonding electron pairs). Sulfur in oxoacids like this can expand its octet; the common Lewis structure places sulfur with an expanded valence shell.

5) Verify the total electron count

We start with \(32\) valence electrons and ensure the structure uses them.

Count electrons by counting lone pairs and bonding pairs:

• Two \( \mathrm{S{=}O} \) double bonds: each double bond is 2 bonding pairs, so total bonding pairs from those is \(2 \times 2 = 4\) bonding pairs
• Two \( \mathrm{S{-}O} \) single bonds: total bonding pairs from those is \(2 \times 1 = 2\) bonding pairs
• Two \( \mathrm{O{-}H} \) single bonds: total bonding pairs from those is \(2 \times 1 = 2\) bonding pairs

Total bonding pairs \(= 4 + 2 + 2 = 8\) bonding pairs.

Each bonding pair is 2 electrons, so bonding electrons \(= 8 \times 2 = 16\).

Lone pairs:

• Each oxygen (all 4 oxygens) has 2 lone pairs
• Total lone pairs \(= 4 \times 2 = 8\)

Lone-pair electrons \(= 8 \times 2 = 16\).

Total used \(= 16 + 16 = 32\), which matches the required valence electrons.

6) Final Lewis structure

Here is the standard Lewis structure of \( \mathrm{H_2SO_4} \) with lone pairs:

\[
\mathrm{
\begin{matrix}
\quad & \; \; & \mathrm{O} & \\
& \mathrm{H{-}O} & \;& \mathrm{O} \\
& & \mathrm{S} & \\
& \mathrm{O} & \;& \mathrm{H{-}O}
\end{matrix}
}
\]

To show the bonding clearly in a conventional line format, write it like this (double bonds to two oxygens, single bonds to the two \( \mathrm{-OH} \) oxygens):

\[
\mathrm{
\left(\mathrm{HO{-}S(=O)_2{-}OH}\right)
}
\]

In words:

• The sulfur is bonded to two oxygens by double bonds: \( \mathrm{S{=}O} \)
• The sulfur is bonded to two oxygens by single bonds, and each of those oxygens is bonded to a hydrogen: \( \mathrm{S{-}O{-}H} \)
• Each oxygen has \(2\) lone pairs

If you want the diagram with lone pairs indicated, ensure each oxygen shows two dots (four dots total around each oxygen).