Question

Find the chemically most active element
A.Fluorine
B.Chlorine
C.Bromine
D.Iodine
E.Astatine

Answer 1

Hint: We know that Electronegativity is the tendency of an element to attract a shared pair of electrons towards itself and as we move down the group electronegativity decreases.

Complete step by step answer:
We know that the elements Fluorine, chlorine, bromine, iodine, and Astatine are belonging to the same group called the halogen family.
As we move down the group the electronegativity value of the elements decreases. The electronegativity values of the halogen family are,
$
  F - {\text{ }}4.0 \\
  Cl - {\text{ }}3.16 \\
  Br - {\text{ }}2.96 \\
  I - {\text{ }}2.66 \\
  At - {\text{ }}2.2 \\
 $
We are well known for the fact that for nonmetals the chemical activity increases as the electronegativity value increases.
Now, coming to the options (B), (C), (D), and (E) they have lower electronegativity values than fluorine which tells that they are less reactive than fluorine. Hence options (B), (C), (D), and (E) are incorrect.
Now we see fluorine, the electronegativity of fluorine is four which is greatest of all. This means that fluorine is the most chemically active element.
Hence option (A) is correct.

Additional Note:
The atomic number of fluorine is ${\text{9}}{\text{.}}$ the electronic configuration of fluorine is ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{5}}}{\text{.}}$ Fluorine is the most electronegative because it has five electrons in the valence cell. To get stable electronic configuration fluorine needs only one electron which makes it highly attracted to the electron.

Note:
We must know that the most reactive element in the periodic table is fluorine while the most reactive metal in the periodic table is francium which is a radioactive alkali metal.