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# Successive Ionisation Enthalpy - First, Second, Third Ionisation Enthalpies, Examples, Successive Ionisation Enthalpies of Second period Elements, Practice Problems and FAQ

Once upon a time, there was a pinchpenny who was extremely hesitant to part with his money. His three grandchildren, Kim, Kom, and Kyle, decided to approach him individually and ask for some treats.

Kim approached him first and requested money. He considered it for a moment before handing him four cents. Kom was up next. When Kom put forth a similar demand, he became enraged and almost denied it. Kom had to work extra hard to persuade him of the same. Finally, when Kyle came in and asked for money to get some ice cream, 'Mr. Miser' almost rebuked him and refused to give him even a penny. Kyle was the youngest and she almost fought with him and snatched some for her share!

I wonder what would Mr Miser have done if he would have had more grandchildren to demand!

Well, this case is quite similar to that of an atom from whom multiple electrons can be tried to take out!

Ionisation Enthalpy is the amount of energy needed to remove an electron from an isolated gaseous atom.

What if we want to remove one more electron from that species? And what if we wish to do it another time, and so on..?

Let's find out !!

• Successive Ionisation Enthalpies - Definition
• First Ionisation Enthalpy
• Second Ionisation Enthalpy
• Third Ionisation Enthalpy
• Example of Successive Ionisation Enthalpy
• Successive Ionisation Enthalpies of Second period Elements
• Determining the Group of an Element using Successive Ionisation Enthalpies
• Practice Problems
• Frequently Asked Questions - FAQ

## Successive Ionisation Enthalpies - Definition

Successive ionisation enthalpy is defined as the energy needed for the removal of second electron or third electron and so on, from the unipositive ion which represents the formation of dispositive or tripositive cations respectively.

• The successive ionisation enthalpies of an atom always increase in the order:
• The removal of electrons from a cation is more difficult than removing electrons from a neutral gaseous atom.
• As the first outermost electron is removed from the isolated gaseous atom, it gains a positive charge and the ratio of p/e (p is the number of protons and e is the number of electrons) increases. As a result, the effective nuclear charge experienced by an electron increases. Therefore, it becomes difficult to remove the next electron. Thus, the successive ionisation energy values increase.

## First Ionisation Enthalpy

It is defined as the energy required to remove the most loosely held electron or outermost electron from an isolated gaseous atom in its ground state to form the isolated gaseous unipositive ion.

;

## Second Ionisation Enthalpy

It is defined as the energy required to remove the most loosely held electron from an isolated unipositive gaseous ion in its ground state to form an isolated dispositive gaseous ion.

;

## Third Ionisation Enthalpy

It is defined as the energy required to remove the most loosely held electron from an isolated dipositive gaseous ion in its ground state to form an isolated tripositive gaseous ion.

;

The third ionisation enthalpy will be higher than the second ionisation enthalpy, which in turn will be higher than the first ionisation enthalpy because it is much more difficult to remove an electron from a positively charged ion than from a gaseous neutral atom.

Therefore, and so on.

## Example of Successive Ionisation Enthalpies

The first three successive ionisation enthalpies of aluminium are as follows:

Exemplary Question: The first ionisation enthalpy values of the 3rd period elements namely Na, Mg and Si are respectively 496, 737 and 786 . Justify whether the first ionisation enthalpy value for Al will be closer to 575 or 760

Answer: It will be closer to 575 . The value for Al should be lower than that of Mg owing to effective shielding of 3p electrons from the nucleus by the 3s-electrons.

## Successive Ionisation Enthalpies of Second Period Elements

Let us first take a look at the values of ionisation enthalpies of the 2nd period elements. The red-coloured values in the table show the removal of the electron from the fully filled subshell, half-filled subshell, or noble gas configuration, thus having exceptionally higher values. Now we shall try to decode the trends that we observe from the given table.

• In , the first electron is removed from the subshell. However, after removing the first electron, attains a noble gas configuration (He). Hence, the second ionisation energy of is exceptionally high (). After this, the third electron is removed from the 1s orbital, and we know that the successive ionisation energy always increases. Hence, it has a higher value of ionisation energy () than the second ionisation energy.
• In , the first electron is removed from 2s subshell that is fully filled and thus, it has a higher first ionisation energy as compared to that of . Now, as the next electron is also removed from the subshell, it attains the noble gas configuration. Therefore, the third ionisation energy of Be is exceptionally high (14850 ), as it is difficult to remove the electron from a stable noble gas configuration.
• In, the first electron is removed from the subshell, and we know that the s-orbital is more penetrating than the p-orbital. Due to this, an electron is removed easily from the p-orbital as compared to that of the s-orbital. Thus, has lower first ionisation energy (801 ) as compared to Be (899). After this, the second electron is removed from the fully filled 2s orbital, so it has a high second ionisation energy (2427 ). The third electron is also removed from the 2s orbital, and we know that the successive ionisation energy always increases. Hence, it has a higher value than that of the second ionisation energy (3660 ).
• In , the first and second electrons are removed from the subshell. The third electron is removed from the fully filled orbital and hence, it requires much higher energy. Thus, it has a high value of the third ionisation energy, 4620 .
• In , the first electron is removed from the half-filled 2p subshell, and we know that the electronic configuration also affects the ionisation energy and so, it has a higher first ionisation energy (1402 ). The second and third electrons are also removed from the 2p subshell, and we know that the successive ionisation energy always increases. Hence, the third ionisation energy (4578 ) is greater than the second ionisation energy (2856 ).
• In O, the first electron is removed from the 2p orbital but it has lower value of the first ionisation energy (1314 ) than N, and this is because N has half-filled 2p subshell and a stable electronic configuration (first I.E., 1402 ). However, on losing one electron, oxygen attains the stable half-filled electronic configuration, thus, the second ionisation energy is higher (3388 ). The third ionisation energy is greater than the second ionisation energy as the successive I.E.'s are higher (5300 ).
• In , the removal of three electrons take place from the subshell, and the successive ionisation energies increases in the order as follows:

I.E.1 (1681 ) < I.E.2 (3374 ) < I.E.3 (6050 )

• For , the first electron is removed from the fully filled 2p subshell, having an ionisation energy of 2080 . Ne has the highest first ionisation energy in the second period. The successive ionisation energies always increase, so the third ionisation energy (6122 ) is greater than the second ionisation energy (3952 ).

## Determining the Group of an Element using Successive Ionisation Enthalpy Values

Whenever there arises a huge difference between the successive ionisation energies of the same element, then the group prediction can be done easily.

For example, if any element has a huge difference in the values of I.E.1 and I.E.2, i.e., I.E.2 has a much higher value than I.E.1, it means that it is very difficult to remove the second electron from a stable noble gas electronic configuration obtained after the removal of the first electron (I.E.1). Therefore, the group number for such an element will be 1 (Group 1: alkali metals) having one valence electron ().

Similarly, if the I.E.3 value for an element is extremely high as compared to I.E.1 and I.E.2, then the group number for an element will be 2 (Group 2: alkaline earth metals), having two valence electrons ).

• Non-metals have higher values of I.E.1, I.E.2, and I.E.3 as compared to metals.
• Alkaline earth metals and pnictogens (elements belonging to the nitrogen family) have higher values of I.E.1 as compared to the succeeding groups (up to the 4th period).
• I.E.3 values of alkaline earth metals show an exceptional jump. Noble gases have the highest I.E.1 values in any period.
• The knowledge of ionisation energy can be used to find the number of valence electrons in an atom.

For example, in the case of Li, the value of I.E.1 and I.E.2 = and respectively. This indicates that the first electron is much more readily removed than the second electron. Thus, there is only one electron in the valence shell of a lithium atom.

## Practice Problems

Q 1. The first four ionisation energy values of an element are 191, 578, 872, 5962 , respectively. Find the number of valence electrons in the element.

a. 1
b. 2
c. 3
d. 4

Answer: From the given data, we can see that there is a large ionisation energy difference between I.E.3

and I.E.4, i.e., the I.E.4 value is extremely high. It is difficult to remove the 4th electron from the stable noble gas electronic configuration of an atom. As a result, it has three valence electrons in the outermost shell. So, option C) is the correct answer.

Q 2. Choose the correct order of I.E. for the following:

a.
b.
c.
d.

Answer: All the given species () have the same number of protons. Now, as the number of electrons increases or the negative charge increases, the p/e ratio (where p is the number of protons and e is the number of electrons) decreases and thus, the effective nuclear charge decreases. So, the decreasing order of effective nuclear charge is . As the effective nuclear charge decreases, it becomes easier to remove the outermost electron and thus, the ionisation energy decreases. So, option D) is the correct answer.

Q 3. Which of the following elements has the highest second ionisation energy?

a.
b.
c.
d.

Answer: To understand the second ionisation energy, we need to write the individual electronic configurations for each atom.

A) =

On the removal of the first electron, we obtain = . Now, for the removal of the second electron, we will have to remove an electron from the fully filled p-orbital (). Hence, this process needs higher energy. Therefore, is very high.

B) =

On the removal of the first electron, we obtain = . Now, removing one electron from the orbital will result in a stable noble gas configuration (). Therefore, it requires less energy to form .

C) = .

On the removal of the first electron, we obtain, = . Now, removing one electron from the 4s orbital will result in a stable noble gas configuration (). Therefore, it requires less energy.

D) =

On the removal of the first electron, we obtain, = . A very high amount of

energy will be required for this, as we know that removal of electrons from stable electronic

configurations (half-filled or fully filled) is very difficult. The removal of the second electron will be from the incompletely filled p-subshell (), which will require less energy as compared to the first ionisation energy of argon.

Hence, Na has the highest second ionisation energy among the given elements. So, option A) is the correct answer.

Q 4. How many electrons can be lost by Aluminium if its fourth ionisation energy is greatest in magnitude than the rest?

a. 1
b. 2
c. 3
d. 4

Answer: Aluminium can lose up to three electrons, as removing three electrons will make it achieve a stable noble gas configuration and hence, the fourth electron removal will have the greatest value. So, option C) is the correct answer.

## Frequently Asked Questions - FAQ

Q 1. Why is the second ionisation potential of lower than that of ?
The electronic configuration of is and that of is . After the removal of 1 electron from each orbital (since it is outermost), the electronic configuration becomes:

= and =
From the electronic configuration, it is clear that the complete removal of 4th shell electron in results in a significant decrease in the size of the atom as compared to . Ionisation energy is inversely proportional to size. Since the size of the atom decreases significantly in , it has higher second ionisation energy.

Q 2. Why do noble gases have the highest ionisation enthalpies among all elements of a particular period?
This is due to their stable octet electronic configuration. This is why it is very difficult to remove electrons from the outermost shell of noble gases and hence, requires a great amount of energy.

Q 3. Why does gallium have higher ionisation energy than aluminium?