Which is formed when fluorine reacts with hot and concentrated sodium hydroxide?
(A) \[{{O}_{2}}\]
(B) \[{{O}_{3}}\]
(C) \[NaO\]
(D) \[HF\]

Hint: Interaction between ionic molecule \[\left( NaOH \right)\] and nonpolar molecules (fluorine) is due to ion-induced dipole which is similar to the dipole-induced dipole. In this type of interaction, the opposite charge of ions (ionic molecules) induces a dipole in non-polar molecules.

Complete Step by Step Answer:
A bond between two fluorine atoms to form fluorine is a covalent bond as both fluoride atoms are non-metal. Now covalent bonds are of two types, one is polar, and the other is non-polar. Fluorine is non-polar as there is no electronegativity difference between two fluorine atoms.

And, sodium hydroxide is most polar or can be called ionic as sodium is metal and the oxygen atom of a hydroxyl group is non-metal. The electronegativity of oxygen is \[3.44\] and electronegativity of sodium is \[0.93\]. The electronegativity difference between sodium and oxygen is greater than \[1.7\] due to which it is an ionic compound.

When \[{{F}_{2}}\] reacts with hot and concentrates \[NaOH\], the positive ion of \[NaOH\]\[\left( N{{a}^{+}} \right)\], induce an opposite temporary charge on fluorine, negative and on the other pole of fluorine cloud rest with temporary positive charge or vice versa. As fluoride is a very electronegative element so it tends to form bonds with less electronegativity element of another compound \[\left( Na\text{ }of\text{ }NaOH \right)\]during reaction and oxygen will liberate such as

Thus, the correct option is A.

Note: As the bond between fluorine atoms is non-polar covalent so it is difficult to break its bond because both are same atoms so, there is no electronegativity difference between them. In ionic compounds oxygen has good electronegativity as compared to sodium. So, fluorine molecules tend to form bonds with sodium whose electronegativity is less but not with oxygen atoms of \[NaOH\].