Question

Write the full electron configuration of the Period 2 element with the following successive IEs (in kJ/mol):

Answer 1

Answer:

Boron (B) is the element whose IE matches with our data.

Electronic Configuration of boron: $1s^22s^22p^1$

Explanation:

Ionization Energy (IE):

It is the minimum amount of energy which is required to remove the lose electron. If the electron is closer to the nucleus then greater amount of energy is required to remove the electron.

If we look from left to right in a period, ionization energy increases due stability of valance shell.

From the data given to us:

IE₁ = 801

IE₂ = 2427

IE₃ = 3659

IE₄ = 25,022

IE₅ = 32,822

Boron (B) is the element whose IE matches with our data.

Electronic Configuration of boron: $1s^22s^22p^1$

Boron has 5 electrons (3 in valance shell) that's why it has 5 Ionization Energies.