Ionisation energy

The sketch graph shows the 13 successive ionisation energies of aluminium.

a. Define the term 1st ionisation energy.

b. How does the graph provide evidence for the existence of three electron shells in an aluminium atom?

c. Write an equation, including state symbols, to represent the 2nd ionisation energy of aluminium.

d. Write the electronic configuration of an aluminium ion, Al³⁺, using 1s² notation.

a. The 1st ionisation energy of an element is the energy needed to remove one electron from each atom in one mole of atoms of the element in the gaseous state to form one mole of gaseous 1+ ions.

b. From the graph, it is noticed that it is easily ejected out for the first three electrons. We may consider that these electrons belong to the outer shell. 

The fourth electron in the next shell toward the nucleus was eliminated due to a significant rise in ionization energy between the third and fourth electrons.

Just a slow rise in ionization energy for the fourth through eleventh electrons in the same shell.

The removal of the eleventh and twelfth electrons caused a significant rise in ionization energy, leaving the twelveth and thirteenth electrons in the shell near the nucleus.

c. `Al^+(g) → Al^(2+)(g) + e^−`

d. The electronic configuration of Al is `[Ne] 3s² 3p¹`. Thus the the electronic configuration of an aluminium ion, Al³⁺ is `1s^2 2s^2 2p^6`

The table below shows the 1st ionisation energies, ΔH_i1, in `kJ.mol^(–1`, of the elements in Period 3 of the Periodic Table.

a. Explain why there is a general increase in the value of ΔH_i1 across the period.

b. Explain why aluminium has a lower value of ΔH_i1 than magnesium.

c. Write the electronic configuration for argon (Z = 18) using 1s² notation.

a. The increase in nuclear charge from left to right throughout Period 3 is the cause of the overall rise in the value of ΔH_i1. Because of the stronger affinity between the positively charged nucleus and the outermost electron caused by this increase in nuclear charge, it is more difficult to remove the electron, requiring more energy (higher ionization energy) and they are added into the same outer shell, leading to the same amount of shielding.

b. This is due to the presence of an extra electron in the 3p subshell of aluminium, which shields the outermost electron from the nucleus to some extent, reducing the effective nuclear charge experienced by the outermost electron and making it easier to remove, resulting in a lower ionisation energy. Whereas, outermost electron of magnesium in lower energy sublevel which is in s subshell, leading to greater force of attraction between nucleus and s electron than p electron 

c. the electronic configuration for argon is `1s^2 2s^2 2p^6 3s^2 3p^6`

Copy and complete the diagram below for the 15 electrons in phosphorus by 

i. adding labels for the other subshells

ii. showing how the electrons are arranged.

The 1st ionisation energies of several elements with consecutive atomic numbers are shown in the graph below. The letters are not the symbols of the elements.

a. Which of the elements a to i belong to Group 1 in the Periodic Table? Explain your answer.

b. Which of the elements a to i could have the electronic configuration `1s^2 2s^2 2p^6 3s^2`?

c. Explain the rise in 1st ionisation energy between element e and element G.

d. Estimate the 1st ionisation energy of element J.

a. There is big decrease in ionisation energy between B and C. Thus, the outer electron in C is in the next quantum shell.

b. Based on the first ionisation energy, the electronic configuration `1s^2 2s^2 2p^6 3s^2` is D.

c. It is noticed that there is an increased nuclear charge leading to electrons to be added go into the same outer shell. We easily conclude that they are in same amount of shielding. This is based on greater force of attraction across period between (positive) nucleus and (negative) electrons.

d. It will be between 1250 and 2050

The successive ionisation energies of element a are shown in the sketch graph

What information does this graph give about how the electrons are arranged in shells for element A?

There is a gradually increased ionisation energy in the first seven electrons. From this, we can explain this is due to these electrons being removed in the outermost energy level.

However, there is a significant change in energy required when the eighth electron is removed. Therefore, the eighth and ninth electrons are in the next energy level which is closer to the nucleus.

a. Give the equations representing:

    i. the 1st ionisation energy of magnesium

    ii. the 1st ionisation energy of magnesium

b. Which ionisation energies are represented by the equations below?

    i. `Mg^(3+)(g) -> Mg^(4+)(g) + e^–`

    ii. `Al^(5+)(g) -> Al^(6+)(g) + e^–`

a. i. `Mg(g) -> Mg^(+)(g) + e^–`

    ii. `Mg^(2+)(g) -> Mg^(3+)(g) + e^–`

b. i. That is 4th ionisation energy of magnesium

    ii. That is 6th ionisation energy of aluminium

Which property of an atom does not affect its first ionisation energy? 

A. the atomic radius 

B. the number of electron shells 

C. the number of neutrons 

D. the number of protons

The answer is C.

The first ionization energy is the energy required to remove the outermost electron from an atom in the gas phase. The number of neutrons is the characteristic of an atom that has no bearing on its first ionization energy. The attraction between the outermost electron and the nucleus is not directly affected by neutrons.

Which of these elements has the highest fifth ionisation energy?

A. C              

B. N              

C. P              

D. Si

The answer is A.

The electronic configuration of Carbon (C) is `1 s^2 2s^2 2p^2`. Fifth electron is removed from 1s orbital. 

The electronic configuration of Nitrogen (N) is `1s^2 2s^2 2p^3` . Fifth electron is removed from 2s orbital. 

The electronic configuration of Phosphorus (P) is `1s^2 2s^2 2p^6 3s^2 3p^3` . Fifth electron is removed from 3s orbital. 

The electronic configuration of Silicon (Si) is `1s^2 2s^2 2p^6 3s^2 3p^2` .Fifth electron is removed from 2p orbital. 

1s is nearest to nucleus so highest fifth ionization energy will be of carbon.

Why is the first ionisation energy of oxygen less than that of nitrogen?

A. The nitrogen atom has its outer electron in a different subshell. 

B. The nuclear charge on the oxygen atom is greater than that on the nitrogen atom. 

C. The oxygen atom has a pair of electrons in one p orbital that repel one another. 

D. There is more shielding in an oxygen atom.

The answer is C.

The electronic configuration of Oxygen is `1s^2 2s^2 2p^4`.

The electronic configuration of Nitrogen is `1s^2 2s^2 2p^3`.

A pair of electrons in one p orbital of the oxygen atom repel each other. Therefore, oxygen has a lower initial ionization energy than nitrogen.

For the element sulfur, which pair of ionisation energies has the largest difference between them?

A. third and fourth ionisation energies 

B. fourth and fifth ionisation energies 

C. fifth and sixth ionisation energies 

D. sixth and seventh ionisation energies

The answer is D.

The energy required for removing an electron from a gaseous atom is known as ionization energy. The electron configuration for sulfur is `1s² 2s² 2p⁶ 3s² 3p⁴`. One of the 3p electrons is eliminated by the first ionization energy, another by the second, and so forth. When an electron is taken away from a stable, filled shell or subshell, the biggest difference in ionization energy usually happens. The next electron to be removed from sulfur is from the 2p subshell, which is far more stable, following the removal of the sixth electron. Consequently, the sixth and seventh ionization energies for sulfur exhibit the most disparity in ionization energies.