Question

What is the second ionization energy of sodium?

Answer 1

Ionization energy is the minimum amount of energy required to remove an electron from the outermost orbital or shell from its gaseous or atomic state can be called as first ionization energy. The amount of energy required to remove an electron from its positive ion is called second ionization energy.

Complete answer:
Periodic table is the representation of chemical elements arranged in the increasing order of atomic numbers arranged in the vertical columns and horizontal rows. These vertical columns are called groups and the horizontal rows are called periods. There are a total of $7$ periods and $18$ groups.
The elements belonging to group $1$ are called alkali metals.
Alkali metals are electropositive, and can easily lose the electrons.
Hydrogen, lithium, sodium, potassium, Rubidium and caesium are the elements belonging to alkali metals.
Sodium is an element with atomic number $11$ and has the valence electrons of $1$ .
The electronic configuration of sodium is $1{s^2}2{s^2}2{p^6}3{s^1}$
Ionization energy is the minimum amount of energy required to remove an electron from the outermost shell. The first ionization energy of sodium is $496kJ{\left( {mole} \right)^{ - 1}}$
When one electron is removed from sodium it attains the noble gas configuration.
Thus, to remove an extra electron from the sodium ion is very difficult and has the value of $4562kJ{\left( {mole} \right)^{ - 1}}$ .
As the nuclear attraction on the outermost electron is greater, second ionization energies of atoms will be higher than the first ionization energy.

Note:
Since the removal of one electron from the atoms results in a greater number of protons than electrons the nuclear attraction will be high on the outermost electrons which leads to the higher values of second ionization energy.