Draw a Lewis dot structure for sulfuric acid, H2SO4. What is the formal charge on the sulfur atom? Assume that all of the S-O bonds are single bonds.
The Correct Answer and Explanation is :
To draw the Lewis dot structure for sulfuric acid (H₂SO₄), we must first know how to correctly arrange the atoms and place electrons. Here’s how to approach it step by step:
Steps to Draw the Lewis Structure for H₂SO₄:
- Count the total number of valence electrons:
- Hydrogen (H) has 1 valence electron, and there are 2 H atoms: (2 \times 1 = 2).
- Sulfur (S) is in Group 16, so it has 6 valence electrons.
- Oxygen (O) is also in Group 16, so each O atom has 6 valence electrons. Since there are 4 O atoms, (4 \times 6 = 24). Total electrons = 2 (from H) + 6 (from S) + 24 (from O) = 32 electrons.
- Arrange the atoms:
Sulfur (S) is the central atom because it is the least electronegative (except for hydrogen, which is never in the center). Surround the sulfur atom with the oxygen atoms. - Distribute the electrons:
Start by forming single bonds between sulfur and each oxygen. Each single bond will use 2 electrons, so 4 bonds use up (4 \times 2 = 8) electrons. - Complete the octets for oxygen atoms:
Each oxygen atom needs 8 electrons around it (except for hydrogen, which only needs 2). After forming single bonds with sulfur, place lone pairs of electrons on each oxygen atom to complete their octet. - Assign remaining electrons:
The remaining electrons will be placed as lone pairs on the sulfur atom. After completing the octets for oxygen atoms, sulfur will have more than 8 electrons around it, but this is acceptable because sulfur is in the third period and can accommodate more than 8 electrons (expanded octet).
Formal Charge Calculation for Sulfur:
The formal charge formula is:
[
\text{Formal charge} = \text{Valence electrons} – \left( \text{Lone pair electrons} + \frac{1}{2} \times \text{Bonding electrons} \right)
]
For sulfur in H₂SO₄:
- Sulfur has 6 valence electrons.
- It is bonded to 4 oxygen atoms with single bonds, and each bond has 2 electrons. This gives 8 bonding electrons.
- Sulfur has 0 lone pair electrons.
So, the formal charge on sulfur is:
[
\text{Formal charge on S} = 6 – (0 + \frac{1}{2} \times 8) = 6 – 4 = +2
]
Thus, the formal charge on sulfur in H₂SO₄ is +2.
Lewis Structure:
In the Lewis structure, sulfur will be in the center with four oxygen atoms attached via single bonds, and each oxygen will have lone pairs to complete their octet.
Now, let’s generate the image of the Lewis structure for H₂SO₄.
Here is the Lewis dot structure for sulfuric acid (H₂SO₄). The sulfur atom is in the center with four oxygen atoms attached by single bonds. Each oxygen atom has lone pairs to complete its octet. The sulfur atom carries a formal charge of +2, as calculated earlier.
Let me know if you need further clarification!
