How do group \[7\] elements react?

Hint: We have to ensure that the periodic table is the tabular display of chemical elements. These elements are arranged in the increasing order of atomic number, electronic configuration and the chemical properties. And the periodic table contains vertical columns and horizontal rows and it is known as groups and periods respectively. The reaction of elements mainly depends on the number of electrons present in the outermost orbital, which are known as valence electrons.

Complete answer:
The group\[7\] metals are known as halogens. And the halogen elements are fluorine $\left( F \right)$, chlorine $\left( {Cl} \right)$, bromine $\left( {Br} \right)$, iodine $\left( I \right)$ and astatine $\left( {At} \right)$.
The reactivity of group \[7\] elements decreases when moving top to down in a group. Among the halogens, fluorine is the most reactive element due to its small size and high electronegativity. The reactivity decreases when increasing the atomic number. The halogens are non- metal. When it is reacting with metals, the halogens will gain the electron. The outer electronic configuration of halogen is \[n{s^2}n{p^5}\]. Hence, it is easy to gain one electron from the other element to attain octet electronic configuration. The accumulation of electrons is harder for the larger atoms due to the weaker attraction. The group\[7\]elements rapidly react with the elements present in the group one like sodium and potassium and there is a formation of metal halides. Because, it needs only one electron to complete the octet.

Note:
 The elements present in group seven are known as halogens. These can easily react with other elements by sharing the one electron present in its outermost orbital to complete the octet. It vigorously reacts with the metals like sodium and potassium and there is a formation of metal halides like sodium fluoride, potassium chloride, potassium iodide, sodium bromide etc.