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Hint: The elements given above belong to the same group. The electronegativity decreases down the group. As electronegativity decreases, the element starts to behave like metal due to increased electropositivity. We know that metals are basic in nature. So, the element with the least electronegativity value is the most basic element from the given options.
Complete step by step solution:
Fluorine, Chlorine, Bromine and Iodine belong to the 17th group of the modern periodic table. The modern periodic table was given by Henry Moseley in the year 1932.
The 17th group is commonly referred to as the halogen group. All the elements of the 17th group act as good oxidising agents. They have a general configuration of $n{{s}^{2}}n{{p}^{5}}$.
Fluorine is the most electronegative element in the periodic table with a value of 4 on the Pauling scale. As we move down the group, the electronegativity value decreases due to a decrease in the charge density and large size of the atom.
As the electronegativity decreases, the lower elements of the group show metal-like properties like positive ion formation. From the above explanation, we can conclude that as the electronegativity of an element decreases, the basic character of the element increases.
Hence the most basic element in the 17th group is Iodine. The correct answer is an option (B).
Note: Although the elements of group 17 are considered to be good oxidising agents, their reactivity rates are not the same. Fluorine's oxidising abilities are so strong that the reaction can easily catch fire and end up not giving favourable products. Another feature of the halogen group is that the element can reduce all the elements below its position in the group. For e.g. Chlorine can oxidise iodine and bromine ions.
Complete step by step solution:
Fluorine, Chlorine, Bromine and Iodine belong to the 17th group of the modern periodic table. The modern periodic table was given by Henry Moseley in the year 1932.
The 17th group is commonly referred to as the halogen group. All the elements of the 17th group act as good oxidising agents. They have a general configuration of $n{{s}^{2}}n{{p}^{5}}$.
Fluorine is the most electronegative element in the periodic table with a value of 4 on the Pauling scale. As we move down the group, the electronegativity value decreases due to a decrease in the charge density and large size of the atom.
As the electronegativity decreases, the lower elements of the group show metal-like properties like positive ion formation. From the above explanation, we can conclude that as the electronegativity of an element decreases, the basic character of the element increases.
Hence the most basic element in the 17th group is Iodine. The correct answer is an option (B).
Note: Although the elements of group 17 are considered to be good oxidising agents, their reactivity rates are not the same. Fluorine's oxidising abilities are so strong that the reaction can easily catch fire and end up not giving favourable products. Another feature of the halogen group is that the element can reduce all the elements below its position in the group. For e.g. Chlorine can oxidise iodine and bromine ions.
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