Question

Element having highest I.P. value is:
A) Ne
B) He
C) Be
D) N

Answer 1

Hint:The term used above as I.P. is called ionization energy. The ionization energy or potential is an essential amount of energy that is required to remove an electron from a neutral atom or molecule. The stable that is half-filled and fully filled orbitals of electrons have very high ionization energy.

Complete step by step answer:
1) First of all we will discuss the ionization energy of an element which depends on the electronic configuration, size of the atom, and electric charge of the nucleus. Now we will write the electronic configuration of each given elements in the options as below,
2) Electronic configurations of elements,
\[Ne{\text{ }}:\;\left[ {He} \right]2{s^2}2{p^6}\]
\[He{\text{ }}:\;1{s^2}\]\[1s\]
\[Be:\;1{s^2}2{s^2}\]
\[N{\text{ }}:\;\left[ {He} \right]2{s^2}2{p^3}\]
3) As it can be seen from electronic configuration Ne, He, and N all have fully filled stable configurations. \[1s\] shell being closest to the nucleus makes it quite tough to remove an electron from it, as in the case of He. Thus, the element having the highest I.P value is Helium.
4) As helium is a noble gas, we want to stick to their electrons and retain their stability. Due to less attractive force, it would be difficult for the nucleus to retain an electron that is far away from it (increase distance from the nucleus resulting in weaker binding of the valence electron.
Hence, the element having the highest I.P. value is He which shows option B as the correct choice.

Note:
Ionization potential is an inverse of atomic number and atomic radius; the larger the atomic no. and radius, the smaller the amount of energy required to remove the electron from the outermost orbital. In the periodic table on moving right to left the atomic radius increases, and the ionization energy increases from left to right in the periods and up the groups.