Draw the Lewis structure of OSF4 where the formal charge is zero on each atom.

Draw the Lewis structure of OSF4 where the formal charge is zero on each atom. The sulfur atom is the central atom in the structure, and it is bonded to the oxygen atom and each of the four fluorine atoms. Be as specific as possible, please. Use the bond energies in the table to estimate the enthalpy of reaction for the formation of sulfur tetrafluoride monoxide. Bond Energy (kJ/mol) S-F: 310 S-O: 265 S=O: 523 O-O: 142 O=O: 498 O-F: 180 Î”?rxn = Î”Hrxn = kJ/mol

The Correct Answer and Explanation is:

To draw the Lewis structure of OSF₄ (sulfur tetrafluoride monoxide) where the formal charge on each atom is zero, follow these steps:

1. Count the valence electrons:

Sulfur (S): Sulfur is in group 16, so it has 6 valence electrons.
Oxygen (O): Oxygen is in group 16, so it has 6 valence electrons.
Fluorine (F): Each fluorine atom has 7 valence electrons, and there are 4 fluorine atoms.

Total valence electrons:
6 (S)+6 (O)+4×7 (F)=6+6+28=40 electrons6 \, (\text{S}) + 6 \, (\text{O}) + 4 \times 7 \, (\text{F}) = 6 + 6 + 28 = 40 \, \text{electrons}6(S)+6(O)+4×7(F)=6+6+28=40electrons.

2. Structure construction:

Sulfur (S) is the central atom, bonded to one oxygen (O) atom and four fluorine (F) atoms.
Place a single bond between sulfur and each of the fluorine atoms (S-F) and between sulfur and oxygen (S-O).
Each single bond represents two electrons, so we use 5 bonds (4 S-F and 1 S-O), which accounts for 5×2=105 \times 2 = 105×2=10 electrons.

3. Distribute the remaining electrons:

After placing the 5 bonds, we have 40−10=3040 – 10 = 3040−10=30 electrons left to be distributed. These remaining electrons should be placed around the atoms to fulfill their octet rule, with the exception of sulfur, which can accommodate more than 8 electrons.

The 4 fluorine atoms will each need 6 electrons to complete their octet, so we place 6 electrons around each fluorine, using 4×6=244 \times 6 = 244×6=24 electrons.
The remaining 6 electrons go to the oxygen atom, completing its octet as well.

4. Double-check the structure and formal charges:

Sulfur: It has 5 bonds (one to oxygen and four to fluorine), and each bond involves 2 electrons, totaling 10 electrons. It also has a lone pair of electrons (since sulfur can exceed the octet rule), so it has 12 electrons in its valence shell, which is acceptable for sulfur.
Oxygen: It has 1 bond with sulfur and 2 lone pairs, so it has 8 electrons in its valence shell, fulfilling the octet rule.
Fluorine: Each fluorine has 1 bond with sulfur and 3 lone pairs, satisfying the octet rule.

No atom has a formal charge in this structure, as each atom satisfies the conditions for formal charge calculation:Formal charge=Valence electrons−Non-bonding electrons−12×Bonding electrons\text{Formal charge} = \text{Valence electrons} – \text{Non-bonding electrons} – \frac{1}{2} \times \text{Bonding electrons}Formal charge=Valence electrons−Non-bonding electrons−21×Bonding electrons

5. Estimate the enthalpy of reaction (ΔHₓₙ):

We can now calculate the enthalpy change for the formation of OSF₄ using bond energies. The reaction for the formation of OSF₄ from elemental sulfur, oxygen, and fluorine can be written as:S+O₂+4F₂→OSF₄\text{S} + \text{O₂} + \text{4F₂} \rightarrow \text{OSF₄}S+O₂+4F₂→OSF₄

The enthalpy change is given by:ΔHrxn=∑Bond Energies of Bonds Broken−∑Bond Energies of Bonds Formed\Delta H_{\text{rxn}} = \sum \text{Bond Energies of Bonds Broken} – \sum \text{Bond Energies of Bonds Formed}ΔHrxn=∑Bond Energies of Bonds Broken−∑Bond Energies of Bonds Formed

Bonds broken:
- S-F bonds: 4 bonds × 310 kJ/mol = 1240 kJ/mol
- O=O bonds (in O₂): 1 bond × 498 kJ/mol = 498 kJ/mol
- F-F bonds: 2 bonds × 158 kJ/mol = 316 kJ/mol

Total energy required to break bonds = 1240+498+316=2054 kJ/mol1240 + 498 + 316 = 2054 \, \text{kJ/mol}1240+498+316=2054kJ/mol

Bonds formed:
- S-F bonds: 4 bonds × 310 kJ/mol = 1240 kJ/mol
- S=O bond: 1 bond × 523 kJ/mol = 523 kJ/mol

Total energy released by bonds formed = 1240+523=1763 kJ/mol1240 + 523 = 1763 \, \text{kJ/mol}1240+523=1763kJ/mol

6. Calculate ΔHₓₙ:

ΔHrxn=2054 kJ/mol−1763 kJ/mol=291 kJ/mol\Delta H_{\text{rxn}} = 2054 \, \text{kJ/mol} – 1763 \, \text{kJ/mol} = 291 \, \text{kJ/mol}ΔHrxn=2054kJ/mol−1763kJ/mol=291kJ/mol

So, the enthalpy of reaction for the formation of sulfur tetrafluoride monoxide (OSF₄) is 291 kJ/mol.

Conclusion:

The Lewis structure of OSF₄ satisfies the octet rule for all atoms except sulfur, which can accommodate more than 8 electrons.
The enthalpy change for the formation of OSF₄ from elemental sulfur, oxygen, and fluorine is 291 kJ/mol.