Question

The more activeness of fluorine is because
A. $F - F$ bond has less energy
B. ${F_2}$ is a gas at normal temperature
C. Its electronic bond is maximum
D. $F - F$ has more energy

Answer 1

Hint: Fluorine is a p-block element lying in the second period of the periodic table. Due to this, the fluorine is anomalous in many properties. For example, ionisation enthalpy, electronegativity, enthalpy of bond dissociation, and electrode potentials are all higher for the fluorine than the expected trends set up by the halogens.

Complete Step by step answer: In the case of fluorine, the various properties like ionisation enthalpy, bond dissociation enthalpy, etc. are quite higher with respect to the trend set by other halogens. Also, there is an unexpected decrease in melting and boiling point, electron gain enthalpy, and radii with respect to the trend set up by the halogens.
The various factors responsible for the anomalous behavior of fluorine are its small size, highest electronegativity, non-availability of d orbitals in the valence shell, and low F-F bond dissociation enthalpy. Fluorine contains only three subshells i.e. its atomic size is very very small. And it has a total of nine electrons. Due to the small size and a high number of electrons, there will be an inter-electronic repulsion. Due to which fluorine-fluorine bond is very unstable. And it becomes more reactive.

Hence, the option (A) is correct.

Note: Most of the reactions of fluorine are exothermic (due to the small and strong bond formed by it with other elements). It forms only one oxoacid while other halogens form a number of oxoacids. Hydrogen fluoride is a liquid (b.p.\[293\,K\]) due to strong hydrogen bonding. Fluorine is the most electronegative element and cannot exhibit any positive oxidation state.