Q. \( \mathrm{CO_2} \) lewis dot diagram.
Answer
\( \mathrm{CO_2} \) lewis dot diagram:
Carbon is in the center. Each oxygen has 2 lone pairs. Carbon has no lone pairs.
Structure with lone pairs:
\[
\begin{aligned}
&:\!\mathrm{O}\;=\;\mathrm{C}\;=\;\mathrm{O}\!:\quad \text{(each \(\mathrm{O}\) has two lone pairs)}\\
&\text{Total valence electrons: }4 + 2(6)=16
\end{aligned}
\]
Lone pairs explicitly:
\[
\begin{aligned}
&\mathrm{..O=C=O..}\quad \text{(“..” on each oxygen = 2 lone pairs)}
\end{aligned}
\]
Detailed Explanation
Goal: Draw the Lewis dot diagram for carbon dioxide, \( \mathrm{CO_2} \).
Step 1: Count valence electrons.
\( \mathrm{CO_2} \) consists of:
- 1 carbon atom, which has \(4\) valence electrons.
- 2 oxygen atoms, each with \(6\) valence electrons, for a total of \(2 \times 6 = 12\).
Total valence electrons:
\[
4 + 12 = 16
\]
Step 2: Place the atoms in the skeleton structure.
In \( \mathrm{CO_2} \), carbon is in the center and oxygen atoms are on both sides:
\( \mathrm{O – C – O} \)
Step 3: Connect carbon and oxygen with bonds.
Carbon needs to satisfy an octet (8 electrons around it) and each oxygen also needs an octet.
If we use double bonds between carbon and each oxygen, we get:
- Each O–C double bond provides \(4\) electrons (two shared pairs).
- Carbon will then have \(2 \times 4 = 8\) bonding electrons around it (an octet).
- Each oxygen will have \(4\) bonding electrons around it from the double bond, and then we add lone pairs to complete the octet.
So the Lewis structure has double bonds on both sides:
\(\mathrm{:O=} \mathrm{C} \mathrm{=O:}\)
Step 4: Add lone pairs to oxygen atoms.
Each oxygen in \( \mathrm{CO_2} \) must have \(8\) electrons around it.
In a double bond, each oxygen already has \(4\) electrons (two shared pairs). That means we must add \(4\) more electrons as lone pairs.
\(4\) electrons corresponds to \(2\) lone pairs per oxygen.
Step 5: Verify electron count.
For the whole molecule:
- Two double bonds total shared electrons: each double bond is \(4\) electrons, so \(2 \times 4 = 8\) shared electrons.
- Each oxygen has \(2\) lone pairs, meaning \(4\) nonbonding electrons per oxygen. With two oxygens: \(2 \times 4 = 8\) nonbonding electrons.
Total electrons:
\[
8 + 8 = 16
\]
This matches the \(16\) valence electrons we counted.
Final Lewis dot diagram (text form).
Carbon dioxide has the structure:
\(\mathrm{O}\) has two lone pairs, double-bonded to carbon, and \(\mathrm{O}\) has two lone pairs on the other side:
\[
\mathrm{\ \ \ :O\ =\ C\ =\ O:\ \ }
\]
And explicitly with lone pairs:
\[
\mathrm{\ \ \ :O:\ } \ \mathrm{=}\ \mathrm{C}\ \mathrm{=}\ \mathrm{\ :O:\ \ }
\]
Each oxygen shows two lone pairs (four dots) around it.
General Chemistry FAQs
How many valence electrons does \( \mathrm{CO_2} \) have?
What is the correct Lewis structure for \( \mathrm{CO_2} \)?
Why does \( \mathrm{CO_2} \) have double bonds?
How many lone pairs are on each oxygen in \( \mathrm{CO_2} \)?
Does carbon have any lone pairs in the Lewis diagram of \( \mathrm{CO_2} \)?
What are the formal charges on \( \mathrm{CO_2} \) in the correct Lewis structure?
What is the molecular geometry and bond angle suggested by the Lewis structure?
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