In the previous section 3.7, we completed a discussion on periodic trends in ionization enthalpy. In this section, we will see the periodic trends in electron gain enthalpy
1. We know that, in every atom, the electrons are held in their orbitals around the nucleus
• We can call it an atom only if all the electrons are present and no more electrons are added
• If any electrons are added, it can no longer be called an atom. It will be called an ion (anion)
2. So when we deal with an atom, we need to know these:
• Does the atom has the tendency to readily accept more electrons?
OR
• Does the atom has the tendency to not accept any electrons at all?
■ To answer those questions, we introduce a new property called Electron gain enthalpy
♦ It's symbol is ΔegH
3. Electron gain enthalpy is an energy
• It is defined as the 'change in energy' when the following reaction takes place:
$\mathbf\small{\rm{X(g)+e^-\longrightarrow X^-(g)}}$
4. Recall that, when we defined ionization enthalpy, we said: 'energy that we give'
• That is., when electron is to be removed, we have to give energy
• So, when we add electron, we would expect the opposite
♦ That is., we would expect to receive energy
♦ (opposite of 'give energy' is 'receive energy')
5. But when we add electron, two possibilities arise. We will call them 'case 1' and 'case 2'
Case 1:
In this case, when an electron is added to the atom we receive energy
Case 2:
In this case, if we want to add an electron to the atom, we have to give energy
6. Let us write some examples:
Examples of case 1:
• We will write the example of group 17 elements (halogens)
• We will write it in steps:
(i) Group 17 elements have the general electronic configuration: ns2np5
♦ So if one more electron is added, they will attain the nearest noble gas configuration
(ii) Noble gas configuration is a very stable configuration
(iii)The group 17 elements readily receive one electron to attain this configuration
♦ We do not have to give any energy for this reaction
♦ In fact, we will 'receive energy'
Examples of case 2:
• We will write the example of group 18 elements (noble gases)
• We will write it in steps:
(i) Group 18 elements have the general electronic configuration: ns2np6
♦ We see that, the orbitals are completely filled
♦ It is a very stable configuration
(ii) The noble gases do not want to take in any more electrons
• If at all any new electron comes in, it will have to go to the next higher main-shell
• We will have to 'give a large quantity of energy' to make the noble gas accept an electron
• We will write it in steps:
1. Consider an electron in an atom
• Let the 'quantity of energy' possessed by the 'orbital in which the that electron resides' be 'E' joules
2. Then, we can write two things:
■ Energy of the 'orbital in which the that electron resides' = -E joules
■ Energy of that electron = -E joules
(We will see the 'reason for providing the -ve sign' in higher classes)
3. In this situation, if we give an energy of 'E' joules, to that electron, it's total energy will become:
(-E + E) = 0 joules
• An electron with an energy of '0 joules' has no energy
♦ Such an electron is assumed to have escaped from the influence of the nucleus
♦ Such an electron is assumed to be 'removed from the atom'
4. So, the energy that we give to remove an electron is given a +ve sign
• Thus, the ionization enthalpy (ΔH) that we saw in the previous section, is +ve
5. Ionization enthalpy (ΔH) is the energy related to 'removal of electron'
• But in this section we are discussing 'addition of electron'
• We saw two cases that can arise:
Case 1:
In this case, when an electron is added to the atom we receive energy
Case 2:
In this case, if we want to add an electron to the atom, we have to give energy.
6. So two questions arise:
• Suppose that we receive energy. Is that energy +ve or -ve?
• Suppose that we have to give energy. Is that energy +ve or -ve?
7. To find the answer, we will again consider the examples of halogens and noble gases
A. We will analyze the example of a halogen in steps:
(i) Let the original atom be a halogen atom
• Let the initial energy of that atom be -2 J
(ii) After gaining the electron, it's energy will be less than -2 J (because, the halogen atom releases energy)
• So let us assume a value lesser than -2 J, say -7 J
(iii) Then (final energy - initial energy) = (-7 - -2) = (-7 + 2) = -5 J
(iv) Based on this, we can write:
• When an atom gains an electron, if we receive energy, that energy is -ve
• In other words:
When an atom gains an electron, if we receive energy, then ΔegH is -ve
B. We will analyze the example of a noble gas in steps:
(i) Let original atom be a noble gas atom
• Let the initial energy of that atom be -9 J
(ii) After gaining the electron, it's energy will be greater than -9 J (because, we give energy to the atom)
• So let us assume a value greater than -9 J, say -6 J
(iii) Then (final energy - initial energy) = (-6 - -9) = (-6 + 9) = 3 J
(iv) Based on this, we can write:
• When an atom gains an electron, if we have to give energy, that energy is +ve
• In other words:
When an atom gains an electron, if we have to give energy, then ΔegH is +ve
Note: In the above steps, values like -2 J, -7 J, -9 J etc., are used only for demonstrating the calculations. They have no relation to actual energy values
8. Sometimes we will come across statements like:
■ Electron gain enthalpy of the element X is a large negative value
■ Electron gain enthalpy of the element X is a large positive value
What do those statements mean?
• The answer can be written in steps:
(i) The -ve sign indicates that we receive energy
• So 'a large negative value' indicates that, when an electron is added, 'we receive a large energy'
(ii) The +ve sign indicates that we have to give energy
• So 'a large positive value' indicates that, to add an electron, 'we have to give a large energy'
4. However, there are two anomalies:
(i) Consider the group 16
• The ΔegH of O is less than that of the succeeding S
(ii) Consider the group 17
• The ΔegH of F is less than that of the succeeding Cl
5. The explanation can be written in 8 steps:
(i) The electronic configuration of O is 1s22s22p4
• The electronic configuration of S is 1s22s22p63s23p4
(ii) When a new electron is added to the O atom, that new electron goes to the 2nd main-shell
♦ There, it experiences the repulsion from the 6 electrons already present in the 2nd main-shell
• When a new electron is added to the S atom, that new electron goes to the 3rd main-shell
♦ There, it experiences the repulsion from the 6 electrons already present in the 3rd main-shell
(iii) We see that, in both O and S, the new electron experiences repulsion from 6 electrons which are already present
• Then what is the difference?
The steps below will give the answer:
(iv) In the case of O, the total 7 electrons are present in the 2nd main-shell
• In the case of S, the total 7 electrons are present in the 3rd main-shell
(v) Compare the two items:
♦ Space occupied by the 2nd main-shell
♦ Space occupied by the 3rd main-shell
• Obviously, the latter will be larger
(vi) Compare the two items:
♦ Distances between the 7 electrons in the 2nd main-shell
♦ Distances between the 7 electrons in the 3rd main-shell
• Obviously the latter will be greater. Because, the 7 electrons has greater space to occupy
(vii) Compare the two items:
♦ Repulsion between the 7 electrons in the 2nd main-shell
♦ Repulsion between the 7 electrons in the 3rd main-shell
• Obviously, the latter will be smaller because, when distance increases, both attractive and repulsive forces decreases
(viii) So it is easier to put a new electron in S than in O
• In other words, S is happier than O to accept a new electron
• So we will receive more energy from S when an electron is added
■ We see that, ΔegH of O is -141 kJ/mol and that of S is -200 kJ/mol
6. So we wrote an explanation for the anomaly mentioned in 4(i)
• We wrote the explanation using 8 steps
• In the same way we can write the explanation for the anomaly mentioned in 4(ii) also
♦ The electronic configuration of F is 1s22s22p5
♦ The electronic configuration of Cl is 1s22s22p63s23p5
• The reader may write the steps in his/her own notebook
Solved example 3.16
Which of the following pairs of elements would have a more negative electron gain enthalpy?
(i) O or F (ii) F or Cl
Solution:
Part (i)
1. Both O and F belong to the 2nd period
2. In a period, as we move from left towards the right, the ΔegH becomes more negative
We have seen the reason earlier in this section
3. F is on the right side of O
So, when compared to O, F will have a more negative ΔegH
Part (ii)
1. Both F and Cl belong to the 17th group
2. In a group, as we move from top towards the bottom, the ΔegH becomes less negative
• We have seen the reason earlier in this section
3. Cl is below F
• So, when compared to Cl, we would expect F to have a more negative ΔegH
4. But in reality, when compared to Cl, F has a less negative ΔegH
• This is an anomaly. We have written the explanation above
• So the answer for part (ii) is: When compared to F, Cl will have a more negative ΔegH
Solved example 3.17
Which of the following will have the most negative electron gain enthalpy and which the least negative? P, S, Cl, F.
Explain your answer
Solution:
A. To find the most negative electron gain enthalpy:
1. P, S and Cl falls in a period
• Within a period, the right most element will have the most negative electron gain enthalpy
• So, among P, S and Cl, the Cl will be having the most negative electron gain enthalpy
2. To find the one from among all the four, we compare Cl and F
• Both F and Cl belong to the 17th group
• In a group, as we move from top towards the bottom, the ΔegH becomes less negative
• We have seen the reason earlier in this section
3. Cl is below F
• So, when compared to Cl, we would expect F to have a more negative ΔegH
4. But in reality, when compared to Cl, F has a less negative ΔegH
• This is an anomaly. We have written the explanation above
• So, when compared to F, Cl will have a more negative ΔegH
5. Thus we can write:
Cl has the most negative ΔegH among the given four elements
B. To find the least negative electron gain enthalpy:
1. P, S and Cl falls in a period
• Within a period, the left most element will have the least negative electron gain enthalpy
• So, among P, S and Cl, the P will be having the least negative ΔegH
2. To find the one from among all the four, we compare P and F
P is below N in the 15th group
• In a group, as we move from top towards the bottom, the ΔegH becomes less negative
• So ΔegH of P is less negative than that of N
3. N and F belong to the 2nd period
• Within a period, the left most element will have the least negative ΔegH
• So ΔegH of N is less negative than that of F
4. But from (2) we have:
ΔegH of P is less negative than that of N
• So obviously
ΔegH of P is less negative than that of F
• This can be shown pictorially as in fig.3.13 below:
5. Thus we can write:
P has the least negative ΔegH among the given four elements
Solved example 3.18
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer
Solution:
1. Consider the first electron gain enthalpy of O
• It is the energy involved in the reaction:
$\mathbf\small{\rm{O(g)+e^-\longrightarrow O^-(g)}}$
• In this reaction, we will receive energy
• That means, in this reaction, ΔegH is negative
2. Consider the second electron gain enthalpy of O
• It is the energy involved in the reaction:
$\mathbf\small{\rm{O^-(g)+e^-\longrightarrow O^{2-}(g)}}$
3. When the second electron approaches the O-, there will be a strong electrostatic repulsion
• This repulsion is between two items: (i) O- (ii) e-
4. That means, the O- is reluctant to receive the second electron
• So we will have to give energy
• The energy that we give is positive
5. Thus we can write:
The second electron gain enthalpy of oxygen is positive
1. We know that, in every atom, the electrons are held in their orbitals around the nucleus
• We can call it an atom only if all the electrons are present and no more electrons are added
• If any electrons are added, it can no longer be called an atom. It will be called an ion (anion)
2. So when we deal with an atom, we need to know these:
• Does the atom has the tendency to readily accept more electrons?
OR
• Does the atom has the tendency to not accept any electrons at all?
■ To answer those questions, we introduce a new property called Electron gain enthalpy
♦ It's symbol is ΔegH
3. Electron gain enthalpy is an energy
• It is defined as the 'change in energy' when the following reaction takes place:
$\mathbf\small{\rm{X(g)+e^-\longrightarrow X^-(g)}}$
4. Recall that, when we defined ionization enthalpy, we said: 'energy that we give'
• That is., when electron is to be removed, we have to give energy
• So, when we add electron, we would expect the opposite
♦ That is., we would expect to receive energy
♦ (opposite of 'give energy' is 'receive energy')
5. But when we add electron, two possibilities arise. We will call them 'case 1' and 'case 2'
Case 1:
In this case, when an electron is added to the atom we receive energy
Case 2:
In this case, if we want to add an electron to the atom, we have to give energy
6. Let us write some examples:
Examples of case 1:
• We will write the example of group 17 elements (halogens)
• We will write it in steps:
(i) Group 17 elements have the general electronic configuration: ns2np5
♦ So if one more electron is added, they will attain the nearest noble gas configuration
(ii) Noble gas configuration is a very stable configuration
(iii)The group 17 elements readily receive one electron to attain this configuration
♦ We do not have to give any energy for this reaction
♦ In fact, we will 'receive energy'
Examples of case 2:
• We will write the example of group 18 elements (noble gases)
• We will write it in steps:
(i) Group 18 elements have the general electronic configuration: ns2np6
♦ We see that, the orbitals are completely filled
♦ It is a very stable configuration
(ii) The noble gases do not want to take in any more electrons
• If at all any new electron comes in, it will have to go to the next higher main-shell
• We will have to 'give a large quantity of energy' to make the noble gas accept an electron
• Next we will learn about the sign (+ve or -ve) to be assigned to the energy
1. Consider an electron in an atom
• Let the 'quantity of energy' possessed by the 'orbital in which the that electron resides' be 'E' joules
2. Then, we can write two things:
■ Energy of the 'orbital in which the that electron resides' = -E joules
■ Energy of that electron = -E joules
(We will see the 'reason for providing the -ve sign' in higher classes)
3. In this situation, if we give an energy of 'E' joules, to that electron, it's total energy will become:
(-E + E) = 0 joules
• An electron with an energy of '0 joules' has no energy
♦ Such an electron is assumed to have escaped from the influence of the nucleus
♦ Such an electron is assumed to be 'removed from the atom'
4. So, the energy that we give to remove an electron is given a +ve sign
• Thus, the ionization enthalpy (ΔH) that we saw in the previous section, is +ve
5. Ionization enthalpy (ΔH) is the energy related to 'removal of electron'
• But in this section we are discussing 'addition of electron'
• We saw two cases that can arise:
Case 1:
In this case, when an electron is added to the atom we receive energy
Case 2:
In this case, if we want to add an electron to the atom, we have to give energy.
6. So two questions arise:
• Suppose that we receive energy. Is that energy +ve or -ve?
• Suppose that we have to give energy. Is that energy +ve or -ve?
7. To find the answer, we will again consider the examples of halogens and noble gases
A. We will analyze the example of a halogen in steps:
(i) Let the original atom be a halogen atom
• Let the initial energy of that atom be -2 J
(ii) After gaining the electron, it's energy will be less than -2 J (because, the halogen atom releases energy)
• So let us assume a value lesser than -2 J, say -7 J
(iii) Then (final energy - initial energy) = (-7 - -2) = (-7 + 2) = -5 J
(iv) Based on this, we can write:
• When an atom gains an electron, if we receive energy, that energy is -ve
• In other words:
When an atom gains an electron, if we receive energy, then ΔegH is -ve
B. We will analyze the example of a noble gas in steps:
(i) Let original atom be a noble gas atom
• Let the initial energy of that atom be -9 J
(ii) After gaining the electron, it's energy will be greater than -9 J (because, we give energy to the atom)
• So let us assume a value greater than -9 J, say -6 J
(iii) Then (final energy - initial energy) = (-6 - -9) = (-6 + 9) = 3 J
(iv) Based on this, we can write:
• When an atom gains an electron, if we have to give energy, that energy is +ve
• In other words:
When an atom gains an electron, if we have to give energy, then ΔegH is +ve
Note: In the above steps, values like -2 J, -7 J, -9 J etc., are used only for demonstrating the calculations. They have no relation to actual energy values
8. Sometimes we will come across statements like:
■ Electron gain enthalpy of the element X is a large negative value
■ Electron gain enthalpy of the element X is a large positive value
What do those statements mean?
• The answer can be written in steps:
(i) The -ve sign indicates that we receive energy
• So 'a large negative value' indicates that, when an electron is added, 'we receive a large energy'
(ii) The +ve sign indicates that we have to give energy
• So 'a large positive value' indicates that, to add an electron, 'we have to give a large energy'
• Based on the above discussion this, we can write a note on 'periodic trends in electron gain enthalpy'
• The electron gain enthalpy does not follow a systematic pattern
• However, we can write some general rules
• First we will see the trend along a period. We will write it in steps:
1. Imagine that, we are moving from left to right along a period
• We know that the three items will occur:
(i) Atoms will become smaller and smaller
(ii) As a result, the outermost electrons will be nearer and nearer to the nucleus
(iii) The effective nuclear charge becomes greater and greater
2. So it will become more and more easier to add an electron
• That means the atoms towards the right are happier to accept electrons
• That means, as we move towards the right, we will receive more and more energies while adding electrons
3. The 'energy received' is negative
■ So we can write the final statement:
When we move from left to right along a period, the ΔegH becomes more and more negative
• Note that, this trend will continue only up to the 17th period. The noble gases in the 18th period have positive ΔegH
1. Imagine that, we are moving from top to bottom along a group
• We know that three items will occur:
(i) Atoms will become larger and larger
(ii) As a result, the outermost electrons will be further and further away from the nucleus
(iii) The effective nuclear charge becomes smaller and smaller
2. So it will become more and more difficult to add an electron
• That means the atoms towards the bottom becomes less and less happier to accept electrons
• That means, as we move towards the bottom, we will receive less and less energies while adding electrons
3. The 'energy received' is negative
■ So we can write the final statement:
When we move from top to bottom along a group, the ΔegH becomes less and less negative
• This trend can be seen in the table 3.5 given below:
• The electron gain enthalpy does not follow a systematic pattern
• However, we can write some general rules
• First we will see the trend along a period. We will write it in steps:
1. Imagine that, we are moving from left to right along a period
• We know that the three items will occur:
(i) Atoms will become smaller and smaller
(ii) As a result, the outermost electrons will be nearer and nearer to the nucleus
(iii) The effective nuclear charge becomes greater and greater
2. So it will become more and more easier to add an electron
• That means the atoms towards the right are happier to accept electrons
• That means, as we move towards the right, we will receive more and more energies while adding electrons
3. The 'energy received' is negative
■ So we can write the final statement:
When we move from left to right along a period, the ΔegH becomes more and more negative
• Note that, this trend will continue only up to the 17th period. The noble gases in the 18th period have positive ΔegH
• Next we will see the trend along a group. We will write it in steps:
• We know that three items will occur:
(i) Atoms will become larger and larger
(ii) As a result, the outermost electrons will be further and further away from the nucleus
(iii) The effective nuclear charge becomes smaller and smaller
2. So it will become more and more difficult to add an electron
• That means the atoms towards the bottom becomes less and less happier to accept electrons
• That means, as we move towards the bottom, we will receive less and less energies while adding electrons
3. The 'energy received' is negative
■ So we can write the final statement:
When we move from top to bottom along a group, the ΔegH becomes less and less negative
• This trend can be seen in the table 3.5 given below:
 |
| Table 3.5 |
(i) Consider the group 16
• The ΔegH of O is less than that of the succeeding S
(ii) Consider the group 17
• The ΔegH of F is less than that of the succeeding Cl
5. The explanation can be written in 8 steps:
(i) The electronic configuration of O is 1s22s22p4
• The electronic configuration of S is 1s22s22p63s23p4
(ii) When a new electron is added to the O atom, that new electron goes to the 2nd main-shell
♦ There, it experiences the repulsion from the 6 electrons already present in the 2nd main-shell
• When a new electron is added to the S atom, that new electron goes to the 3rd main-shell
♦ There, it experiences the repulsion from the 6 electrons already present in the 3rd main-shell
(iii) We see that, in both O and S, the new electron experiences repulsion from 6 electrons which are already present
• Then what is the difference?
The steps below will give the answer:
(iv) In the case of O, the total 7 electrons are present in the 2nd main-shell
• In the case of S, the total 7 electrons are present in the 3rd main-shell
(v) Compare the two items:
♦ Space occupied by the 2nd main-shell
♦ Space occupied by the 3rd main-shell
• Obviously, the latter will be larger
(vi) Compare the two items:
♦ Distances between the 7 electrons in the 2nd main-shell
♦ Distances between the 7 electrons in the 3rd main-shell
• Obviously the latter will be greater. Because, the 7 electrons has greater space to occupy
(vii) Compare the two items:
♦ Repulsion between the 7 electrons in the 2nd main-shell
♦ Repulsion between the 7 electrons in the 3rd main-shell
• Obviously, the latter will be smaller because, when distance increases, both attractive and repulsive forces decreases
(viii) So it is easier to put a new electron in S than in O
• In other words, S is happier than O to accept a new electron
• So we will receive more energy from S when an electron is added
■ We see that, ΔegH of O is -141 kJ/mol and that of S is -200 kJ/mol
6. So we wrote an explanation for the anomaly mentioned in 4(i)
• We wrote the explanation using 8 steps
• In the same way we can write the explanation for the anomaly mentioned in 4(ii) also
♦ The electronic configuration of F is 1s22s22p5
♦ The electronic configuration of Cl is 1s22s22p63s23p5
• The reader may write the steps in his/her own notebook
Now we will see some solved examples
Which of the following pairs of elements would have a more negative electron gain enthalpy?
(i) O or F (ii) F or Cl
Solution:
Part (i)
1. Both O and F belong to the 2nd period
2. In a period, as we move from left towards the right, the ΔegH becomes more negative
We have seen the reason earlier in this section
3. F is on the right side of O
So, when compared to O, F will have a more negative ΔegH
Part (ii)
1. Both F and Cl belong to the 17th group
2. In a group, as we move from top towards the bottom, the ΔegH becomes less negative
• We have seen the reason earlier in this section
3. Cl is below F
• So, when compared to Cl, we would expect F to have a more negative ΔegH
4. But in reality, when compared to Cl, F has a less negative ΔegH
• This is an anomaly. We have written the explanation above
• So the answer for part (ii) is: When compared to F, Cl will have a more negative ΔegH
Solved example 3.17
Which of the following will have the most negative electron gain enthalpy and which the least negative? P, S, Cl, F.
Explain your answer
Solution:
A. To find the most negative electron gain enthalpy:
1. P, S and Cl falls in a period
• Within a period, the right most element will have the most negative electron gain enthalpy
• So, among P, S and Cl, the Cl will be having the most negative electron gain enthalpy
2. To find the one from among all the four, we compare Cl and F
• Both F and Cl belong to the 17th group
• In a group, as we move from top towards the bottom, the ΔegH becomes less negative
• We have seen the reason earlier in this section
3. Cl is below F
• So, when compared to Cl, we would expect F to have a more negative ΔegH
4. But in reality, when compared to Cl, F has a less negative ΔegH
• This is an anomaly. We have written the explanation above
• So, when compared to F, Cl will have a more negative ΔegH
5. Thus we can write:
Cl has the most negative ΔegH among the given four elements
B. To find the least negative electron gain enthalpy:
1. P, S and Cl falls in a period
• Within a period, the left most element will have the least negative electron gain enthalpy
• So, among P, S and Cl, the P will be having the least negative ΔegH
2. To find the one from among all the four, we compare P and F
P is below N in the 15th group
• In a group, as we move from top towards the bottom, the ΔegH becomes less negative
• So ΔegH of P is less negative than that of N
3. N and F belong to the 2nd period
• Within a period, the left most element will have the least negative ΔegH
• So ΔegH of N is less negative than that of F
4. But from (2) we have:
ΔegH of P is less negative than that of N
• So obviously
ΔegH of P is less negative than that of F
• This can be shown pictorially as in fig.3.13 below:
 |
| Fig.3.13 |
P has the least negative ΔegH among the given four elements
Solved example 3.18
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer
Solution:
1. Consider the first electron gain enthalpy of O
• It is the energy involved in the reaction:
$\mathbf\small{\rm{O(g)+e^-\longrightarrow O^-(g)}}$
• In this reaction, we will receive energy
• That means, in this reaction, ΔegH is negative
2. Consider the second electron gain enthalpy of O
• It is the energy involved in the reaction:
$\mathbf\small{\rm{O^-(g)+e^-\longrightarrow O^{2-}(g)}}$
3. When the second electron approaches the O-, there will be a strong electrostatic repulsion
• This repulsion is between two items: (i) O- (ii) e-
4. That means, the O- is reluctant to receive the second electron
• So we will have to give energy
• The energy that we give is positive
5. Thus we can write:
The second electron gain enthalpy of oxygen is positive
In the next section, we will see electronegativity
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