Question

What is the order of the second ionization energies for elements in the second row of the periodic table?

Answer 1

Hint: The ionization energy is the energy required to remove an electron from its orbital around an atom. The ionization energy of an element increases as one moves across a period in the periodic table.

Complete Step By Step Answer:
Ionization energy of an element as the amount of energy needed to remove an electron from the gaseous atom $ E $ In its ground state. The first ionization energy is therefore the energy required for the reaction $ E \to {E^ + } + {e^ - } $ .
The ionization energy is always positive. Larger values of ionization energy mean that the electron is more tightly bound to the atom and is harder to remove. If an atom possesses more than one electron, the amount of energy needed to remove successive electrons increases steadily. We can define second ionization energy to be represented as $ {I_2} $ .
 $ {E^ + } \to {E^{2 + }} + {e^ - } $ (Known as Ionization energy)
The value of ionization energy of lithium and beryllium have the successive ionization energy for an element increase as they go; that is, it takes more energy to remove the second electron from an atom than the first, and so forth.
If the second electron Is being removed from a positively charged species, more energy is required. Second, removing the first electron reduces the repulsive forces among the remaining electrons, so the attraction of the remaining electrons to the nucleus is stronger. The energy required to remove the second electron from lithium is more than ten times greater than the energy needed to remove the first electron.

Note:
ionization energies of the element in the third row of the periodic table exhibit the same pattern as those of the second row. Ionization energy increases steadily. The ionization energy of lithium is the highest. Boron has greater second ionization energy than carbon, thus the order of second ionization energy is: $ Be < C < B < N < F < O < Ne < Li $