Fluorine reacts with dilute NaOH and forms a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The reaction of fluorine with dilute $NaOH$ is given by: $2F_2 + 2NaOH \rightarrow 2NaF + OF_2 + H_2O$ Thus,the gaseous product $A$ is oxygen difl

Vedclass · Accepted Answer

$103^{\circ}$

Vedclass · Answer

(B) The reaction of fluorine with dilute $NaOH$ is given by: $2F_2 + 2NaOH \rightarrow 2NaF + OF_2 + H_2O$ Thus,the gaseous product $A$ is oxygen difluoride $(OF_2)$. In $OF_2$,the central oxygen atom is $sp^3$ hybridized with two bond pairs and two lone pairs. Due to the high electronegativity of fluorine atoms,the lone pair-lone pair repulsion is significant,which compresses the bond angle. The bond angle in $OF_2$ is $103^{\circ}$.

Fluorine reacts with dilute $NaOH$ and forms a gaseous product $A$. The bond angle in the molecule of $A$ is

Similar Questions

What is the geometry of the $H_2O$ molecule?

The linear structure is assumed by

Which one is correct for bond angle?

Which of the following species is planar?

Which species is planar?

Vedclass Test Series

Exam Paper Generator

Online Exam Module