H_{2}SO_{4} is a strong acid whereas F_{2} is one of the strongest oxidizing agents. Let us explore the feasibility of the reaction between H_{2}SO_{4} and F_{2}.

**Sulphuric acid (H _{2}SO_{4}) is highly corrosive . On the other hand, fluorine (F_{2}) is one of the most reactive elements which mostly exists in nature in the form of its compounds. **

In this article, we will learn details about the possibility of reaction between H_{2}SO_{4} and F_{2}. There will be no reaction between sulphuric acid and fluorine.

**What is the product of H**_{2}SO_{4} and F_{2}?

_{2}SO

_{4}and F

_{2}?

**The H _{2}SO_{4} + F_{2} reaction fails to take place so there is no product formation. Fluorine is the most powerful oxidizing agent, with oxidation potential of +2.65 V. **

**Conversely, oxidation state of S in H**

_{2}SO_{4}is + 6 which is the highest oxidation state of S. Therefore, H_{2}SO_{4}cannot be further oxidized.**How to balance H**_{2}SO_{4} + F_{2}?

_{2}SO

_{4}+ F

_{2}

**H _{2}SO_{4} + F_{2} reaction cannot be balanced due to no formation of products.**

**H**_{2}SO_{4} + F_{2} titration

**H**titration

_{2}SO_{4}+ F_{2}**A ****titration**** cannot be performed between F_{2} and H_{2}SO_{4} as their chemical reaction is unknown.**

**H**_{2}SO_{4} + F_{2} net ionic equation

**H**net ionic equation

_{2}SO_{4}+ F_{2}**H _{2}SO_{4} + F_{2} reaction has no net ionic equation.**

**H**_{2}SO_{4} + F_{2} conjugate pairs

**H**conjugate pairs

_{2}SO_{4}+ F_{2}**H _{2}SO_{4} + F_{2} reaction has the following conjugate pairs,**

**The****conjugate pair****of H**._{2}SO_{4}is HSO_{4}^{–}**No conjugate pair exists for F**_{2}.

**H**_{2}SO_{4} + F_{2} intermolecular forces

**H**intermolecular forces

_{2}SO_{4}+ F_{2}**H _{2}SO_{4} + F_{2} reaction has the following intermolecular forces,**

**F**_{2}is a non-polar molecule, so a weak Vander wall force of interaction exists between fluorine molecules.**H**_{2}SO_{4}, a polar covalent molecule (structure shown below), exhibits van der Waals dispersion forces, dipole-dipole interactions and hydrogen bonding.

**H**_{2}SO_{4} + F_{2} reaction enthalpy

**H**reaction enthalpy

_{2}SO_{4}+ F_{2}**Since no reaction happens between F_{2} and H_{2}SO_{4}, one cannot calculate **

**enthalpy**

**changes for their chemical reaction.**

**Is ****H**_{2}SO_{4} + F_{2} a buffer solution?

**H**a buffer solution

_{2}SO_{4}+ F_{2}** H_{2}SO_{4} + F_{2} reaction** is not a

**buffer solution**

**because it is neither a combination of a weak acid and its conjugate base nor a weak base and its conjugate acid.**

**Is ****H**_{2}SO_{4} + F_{2} a complete reaction

**H**a complete reaction

_{2}SO_{4}+ F_{2}** H_{2}SO_{4} + F_{2} reaction** cannot be a complete reaction because both of them do not react with each other.

**Is ****H**_{2}SO_{4} + F_{2} an exothermic or endothermic reaction?

**H**an exothermic or endothermic reaction

_{2}SO_{4}+ F_{2}** H_{2}SO_{4} + F_{2} reaction** cannot be an exothermic or endothermic reaction because no reaction occurs between them.

**Is ****H**_{2}SO_{4} + F_{2} a redox reaction

**H**a redox reaction

_{2}SO_{4}+ F_{2}**H _{2}SO_{4} + F_{2} reaction cannot be a redox reaction because both of them do not react with each other.**

**Is ****H**_{2}SO_{4} + F_{2} a precipitation reaction ?

**H**a precipitation reaction

_{2}SO_{4}+ F_{2}**H _{2}SO_{4} + F_{2} reaction** cannot be a

**precipitation**

**reaction because no reaction happens between them.**

**Is ****H**_{2}SO_{4} + F_{2} reversible or irreversible reaction ?

**H**reversible or irreversible reaction

_{2}SO_{4}+ F_{2}**H _{2}SO_{4} + F_{2}** is neither a reversible nor

**irreversible**

**reaction because they do not react with each other.**

**Is ****H**_{2}SO_{4} + F_{2} displacement reaction?

**H**displacement reaction

_{2}SO_{4}+ F_{2}** H_{2}SO_{4} + F_{2} reaction** is not a displacement reaction because there is no reaction between F

_{2}and H

_{2}SO

_{4}.

**Conclusion**

Sulphuric acid and fluorine do not react with each other even though both are highly reactive. This happens because fluorine, in spite of being a strong oxidizing agent, cannot oxidize sulphuric acid to the next higher level. In fact, sulphur atom in H_{2}SO_{4} is already present in its highest oxidation state.

Hello…

I am Debarati Chakraborty. I have completed my Ph.D. in Chemistry from Jadavpur University. I have worked in several chemical industries (R & D) and I am currently attached to a degree college in Kolkata as a State Aided College Teacher. I have been working in the field of organic chemistry for the last 22 years. I am working as a Subject Matter Expert in LambdaGeeks.