Question

Fluorine is the most reactive among all the halogens, because of its:
(A) sSmall size
(B) Low dissociation energy of $F - F$
(C) Large size
(D) High dissociation energy of $F - F$

Answer 1

Hint:Fluorine is the first member of the Halogen family. Its atomic number is $9$. It has $7$ electrons in its valence shell. All the members of the halogens family are highly reactive in the whole periodic table due to their high electronegativity. They all are non-metals and highly toxic in nature.

Complete step by step answer:
Fluorine lies in the second period and group $17$ of the periodic table. Its electronic configuration is $1{s^2}2{s^2}2{p^5}$. Fluorine exists as a gas at room temperature. The outermost shell is a p-subshell which contains $5$ electrons. To achieve the Noble gas configuration of Neon it needs one more electron. Fluorine has $5$ electrons in its \[2p\] shell. The ideal configuration of the \[p\] orbital has 6 electrons, since Fluorine is so close to attaining the noble gas configuration, the electrons are held very strongly to the nucleus. Thus it makes the fluorine most electronegative because of its small radius as the positive protons hold a very strong attraction to the electrons. Due to the short bond length, the repulsion between non-bonding electrons is very high. Thus it makes the fluorine most electronegative because of its small radius as the positive protons hold a very strong attraction to the electrons. Being an element of the group $17$ fluorine tends to form polar covalent bonds with non-metal. Eg: $H - F$. Having a $ - 1$ oxidation state fluorine can also form an ionic bond.
Hence, Fluorine is the most reactive of all the halogens due to its small size.
Therefore option (A) is correct.

Note:
Electronegativity is defined as the tendency of an atom in a molecule to draw the shared pair of electrons towards itself. The electronegativity of an element increases from moving down the group and along the period. Fluorine has a high electron affinity and high ionization energy.