Question

Draw the Lewis structure of ${\mathbf{\text{O}}}_{\mathbf{\text{2}}}{\mathbf{\text{F}}}_{\mathbf{\text{2}}}$ . Assign oxidation states and formal charges to the atoms in${\mathbf{\text{O}}}_{\mathbf{\text{2}}}{\mathbf{\text{F}}}_{\mathbf{\text{2}}}$ . The compound${\mathbf{\text{O}}}_{\mathbf{\text{2}}}{\mathbf{\text{F}}}_{\mathbf{\text{2}}}$ is a vigorous and potent oxidizing and fluorinating agent. Are oxidation states or formal charges more useful in accounting for these properties of${\mathbf{\text{O}}}_{\mathbf{\text{2}}}{\mathbf{\text{F}}}_{\mathbf{\text{2}}}$ ?

Short answer

The oxidation state is not favourable for such an electro negative element. Then, it is said that ${\text{O}}_{\text{2}}{\text{F}}_{\text{2}}$ is an unstable compound.

Step-by-step solution

Define Electronegative:

Electronegativity is a chemical property that details the tendency of an atom or a functional group to attract electrons toward itself.

Are oxidation states or formal charges more useful in accounting for these properties of O2F2 ?

With the help of given Lewis structure, we can observe that each oxygen atom has a tetrahedral arrangement of electron pairs (two bonds and two lone pairs). Therefore, bond angles are said to be $109.5°$and each oxygen is${\text{sp}}^{\text{3}}$ hybridised. The formal charges are marked in the colour blue, and oxidation states are marked in colour red

In this case the oxidation states are more important than the formal charges.

Oxygen has a$+1$oxidation state which is not very favourable for such an electro negative element . Due to this,${\text{O}}_{\text{2}}{\text{F}}_{\text{2}}$is an unstable compound.

Therefore, oxidation state is not very favourable for such an electro negative element . So, ${\text{O}}_{\text{2}}{\text{F}}_{\text{2}}$is an unstable compound.