QQuestionChemistry
QuestionChemistry
"There are many Lewis structures you could draw for sulfuric acid, H^2SO^4 (each H is bonded to an O).
Part A: What Lewis structure(2$) would you draw to satisfy the octet rule?
Draw the molecule(2$) by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons."
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Answer
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Step 1:: Identify the number of valence electrons for each atom in H^2SO^4.
- Hydrogen (H): 1 valence electron - Sulfur (S): 6 valence electrons - Oxygen (O): 6 valence electrons
Step 2:: Calculate the total number of valence electrons in the molecule.
- Total valence electrons = (2 * number of Hydrogen atoms) + 6 (Sulfur atom) + (4 * Oxygen atoms) - Total valence electrons = (2 * 2) + 6 + (4 * 4) = 4 + 6 + 16 = 26 valence electrons
Step 3:: Draw a trial structure by placing the atoms on the grid and connecting them with bonds.
- Place S in the center, and then add the four O atoms around it. - Add two H atoms bonded to two of the O atoms. - Distribute the remaining electrons as lone pairs on the O atoms to satisfy the octet rule for all atoms.
Step 4:: Check if the trial structure satisfies the octet rule for all atoms.
- Sulfur atom has 6 valence electrons + 2 electrons from the S-O bonds = 8 electrons (octet rule satisfied) - Each Oxygen atom has 6 valence electrons + 2 electrons from the O-H bonds + 4 electrons from the S-O bonds = 8 electrons (octet rule satisfied) - Each Hydrogen atom has 2 valence electrons + 2 electrons from the H-O bonds = 4 electrons (octet rule satisfied)
Step 5:: Draw the final Lewis structure(2$) for H^2SO^4.
\begin{array}{ccccc}
- There is only one valid Lewis structure for H^2SO^4, as shown below: & & \text{O} & & \ & \text{H} & | & \text{H} & \ \text{H}-\text{S}-\text{O} & & \text{O} & & \text{O} \ & & | & & \ & & \text{O} & & \end{array}
Final Answer
The Lewis structure for H^2SO^4 is shown above, with all atoms satisfying the octet rule and all lone pairs of electrons included.
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