Question

How are electronegativity, electro affinity, ionization energy, atomic radius, and reactivity all interrelated?

Answer 1

Hint: To give the relation between the above given quantities, first of all you have to know about the definition of each and every quantity which are given above. And all of these quantities may be directly or indirectly proportional to each other.

Complete answer:
Definitions of given quantities are as follow:
Electronegativity (E.N.):
Electronegativity of an atom is that parameter which defines the property of that atom to bind an electron strongly or pull an electron from the outside towards itself. Small sized atoms have high electronegativity i.e. they bind an electron more strongly.
Electron affinity (E.A.):
Electron affinity is the just opposite property of an atom from electronegativity, it means electron affinity shows the property of removing an electron easily from the outermost shell. Big sized atoms have high electron affinity i.e. they remove an electron very easily because they have very less nuclear attraction power over valence electrons.
Ionization energy (I.E.):
Generally Ionization energy is also known by the name of first ionization energy and it is that minimum amount of energy which is required to remove an electron from the neutral atom. And ionization energy of small sized atoms is high as they have great nuclear attraction over outermost electrons.
Atomic radius (r):
Atomic radius is the length between the nucleus or center of an atom and outermost shell. And when the size of an atom increases, the atomic radius of that atom will also increase.
Reactivity:
Reactivity of an atom means how easily any species will react with other species. So an atom with very high and very low electronegative or electron affinity will show great reactivity. Electronegativity and electron affinity are vice – versa of each other.
Relation between the above mention quantities are shown in the following manner:
$\text{Atomic size} \propto \text{E.A.} \propto \dfrac{1}{\text{E.N.}} \propto \dfrac{1}{\text{I.E.}} \propto r$

Note:
Here some of you may think that after first ionization energy, second and third ionization energy is also high, but that assumption is not correct. Because it is not compulsory that energy required to remove the second & third electron is more than or equal to the first electron.