Question

Write the full electron configuration of the Period 2 element with the following successive IEs (in \(\mathrm{kJ} / \mathrm{mol}\) ): \(\mathrm{IE}_{1}=801\) \(\mathrm{IE}_{2}=2427\) $$ \mathrm{IE}_{3}=3659 $$ \(\mathrm{IE}_{4}=25,022 \quad \mathrm{IE}_{5}=32,822\)

Short answer

1s^2 2s^2 2p^1

Step-by-step solution

Identify the trend in ionization energies

Examine the given ionization energies (IEs). Notice a significant jump from \( \mathrm{IE}_{3} = 3659 \ \mathrm{kJ} / \mathrm{mol} \ \) to \ \mathrm{IE}_{4} = 25022 \ \mathrm{kJ} / \mathrm{mol}. This large increase suggests the removal of an electron from a stable, filled shell after the third ionization. Thus, the element has three valence electrons.

Identify the element with three valence electrons in Period 2

Identify the Period 2 element with three valence electrons. The elements in Period 2 are: Li, Be, B, C, N, O, F, Ne. The element with three valence electrons (group 13) in Period 2 is boron, B.

Write the full electron configuration for boron

Determine the electron configuration for boron. Boron has 5 electrons. Following the order of filling orbitals \((1s, 2s, 2p)\), the configuration is: \ 1s^2 2s^2 2p^1.

Key concepts

Ionization Energy

Ionization energy (IE) is the energy required to remove an electron from a gaseous atom or ion. It is an important property that reflects how strongly an electron is held within an atom. In the given exercise, the successive ionization energies for a Period 2 element are:

• IE1: 801 kJ/mol
• IE2: 2427 kJ/mol
• IE3: 3659 kJ/mol
• IE4: 25,022 kJ/mol
• IE5: 32,822 kJ/mol

This data shows a significantly high jump from IE3 to IE4. This indicates that the removal of the fourth electron requires much more energy, suggesting it is being removed from a stable, filled electronic shell. This property helps us identify the valence electrons and the possible element.

Valence Electrons

Valence electrons are the outermost electrons of an atom and are involved in forming chemical bonds. Through the ionization energy data, we can infer the number of valence electrons. In the provided exercise, after the removal of three electrons, the enormous jump in ionization energy for the fourth electron implies that the element's stable, filled shell configuration is disrupted. Thus, the element must have three valence electrons, as suggested by the significant IE4 value. In Period 2, elements with this characteristic are part of Group 13, which includes boron (B). Boron, having three valence electrons, aligns with the data provided.

Orbital Filling Order

Orbital filling order refers to the sequence in which electrons occupy the orbitals of an atom. The order follows a specific pattern known as the Aufbau principle, which dictates that electrons fill orbitals starting from the lowest energy level to the highest. The general sequence can be summarized as: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, and so on. Therefore, for boron (B), with its 5 electrons, the electron configuration is determined by filling up the orbitals in this specific order. First, two electrons fill the 1s orbital:

• 1s^2

Then, two electrons fill the 2s orbital:

• 2s^2

Lastly, one electron fills the 2p orbital:

• 2p^1

So, the full electron configuration for boron is 1s^2 2s^2 2p^1. Understanding the orbital filling order is crucial for determining the electron configurations correctly.