Question

Which element has the Lowest Second Ionization Energy?

Answer 1

The ionization energy is the energy required to remove an electron from the outermost shell of an isolated gaseous atom. Less ionization energy is required to remove an electron from a half filled or completely filled orbital.

Complete answer:
Ionization energy depends upon the size of the atom, that is ionization energy is inversely proportional to the size. Ionization energy increases across a period and decreases down the group.
When the first electron or the most loosely bound electron is removed, the amount of energy required is less than the energy required to remove the electron in the next successive shell. Thus, this ionization energy goes on increasing with the number of electrons removed.
The second ionization energy is the energy required to remove the outermost electron from a $1 +$ion of the element. As the positive charge binds electrons more strongly, the second ionization energy of an element is always higher than the first.
Electronic configuration of Beryllium: $1{s^2}2{s^2}$
First ionization energy: $1{s^2}2{s^1}$
After first ionization energy, beryllium attains stable electronic configuration as it has half - filled
$s -$orbital. Therefore, more energy is required to remove an electron from $s -$orbital.
Second Ionization energy: $1{s^2}$
During second ionization energy, beryllium has to lose one more electron in order to attain a noble gas configuration of helium. Hence, it can lose its second electron easily.
Hence, we can conclude that beryllium has Lowest Second Ionization Energy because of its ability to attain noble gas configuration.

Note:
Group $1$ elements have the lowest ionization energy due to its effective nuclear charge than the rest of the elements of the period. It further decreases as we move down the group. Hence, we can say that alkali metals have lowest ionization energy.