Chapter 3.7 - Periodic Trends in Ionization Enthalpy

In the previous section 3.6, we completed a discussion on periodic trends in ionic radius. In this section, we will see periodic trends in ionization 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
• If any of the electrons are lost, it can no longer be called an atom. It will be called an ion (cation)
2. So when we deal with an atom, we need to know these:
• Does the atom has the tendency to readily lose the electrons?
OR
• Does the atom has the tendency to keep the electrons tightly bound to itself?
■ To answer those questions, we introduce a new property called ionization enthalpy
3. Ionization enthalpy is an energy
It is measured using an experiment. That experiment can be explained in 3 steps:
(i) We want to measure the ionization enthalpy of an element
• Take one atom of that element
    ♦ Denote this one atom as X
(ii) Remove the outermost electron from X
• Note down the energy required for the removal
(iii) The energy noted in step 2, is the energy required for one atom X
• Calculate the energy required for one mole (6.023 × 10²³ nos. of X atoms)
(This can be done by multiplying the energy value by 6.023 × 10³) 
■ The energy required (in kilo joules) for one mole is recorded as the ionization enthalpy
    ♦ It’s symbol is ΔH
    ♦ It’s unit is kJ/mole
(We will see the practical method for carrying out this experiment in higher classes)
4. For carrying out the experiment, the one atom X must satisfy 3 conditions:
  (i) X must be in the gaseous state
 (ii) X must be isolated from all other atoms
(iii) X must be in the ground state
5. Let us see the reason for imposing the 3 conditions:
(i) We want to pick just 'one atom of the element' for doing the experiment
• This is possible only when the element is in the gaseous state
    ♦ Recall that, in the gaseous state, the atoms will be at a greater distances apart
• So we impose the condition that, the atom X should be in the gaseous state
(ii) If the X is near other atoms, inter molecular forces will be acting
• If those inter molecular forces are acting, our experiment will not give correct values
• So we impose the condition that, the atom X should be isolated from other atoms
(iii) While performing the experiment, all the electrons of X must be in their original orbitals
• If they are in higher orbitals, it would mean that, the electrons already has some 'extra energy'
• In such a situation, the experiment will not give correct values
• So we impose the condition that, the atom X should be in the ground state
6. Now we can write the definition for ionization enthalpy: 
Ionization enthalpy is the energy required to remove an electron from an isolated gaseous atom (X) in it’s ground state
7. Ionization enthalpy can be classified as:
(i) First ionization enthalpy is the energy required to remove the ‘most loosely bound electron’ from X 
• In other words, it is the energy required to carry out the reaction given below:
$\mathbf\small{\rm{X(g)\longrightarrow X^+(g)\,+e^-}}$
(ii) Second ionization enthalpy is the energy required to remove the ‘second most loosely bound electron’ from X 
• In other words, it is the energy required to carry out the reaction given below:
$\mathbf\small{\rm{X^+(g)\longrightarrow X^{2+}(g)\,+e^-}}$
(iii) Third ionization enthalpy is the energy required to remove the ‘third most loosely bound electron’ from X 
• In other words, it is the energy required to carry out the reaction given below:
$\mathbf\small{\rm{X^{2+}(g)\longrightarrow X^{3+}(g)\,+e^-}}$
8. The second ionization enthalpy will be greater than the first ionization enthalpy
• The reason can be written in 3 steps:
(i) When the first electron is removed, the remaining electrons will be held more tightly by the nucleus
(ii) So it is more difficult to remove a second electron
(iii) That means, more energy is required to remove the second electron
9. In the same way:
• Third ionization enthalpy will be greater than the second ionization enthalpy
• Fourth ionization enthalpy will be greater than the third ionization enthalpy
• so on . . .
10. Sometimes we will be given just ‘ionization enthalpy’
• There will not be any prefix: first, second, third etc.,
■ Then it implies that, it is the first ionization enthalpy 
11. Every element has it’s own unique value of ionization enthalpy
• For example, the ionization enthalpy of Li is 520.2 kJ/mol
• A complete list can be seen here

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So now we know the basics of ionization enthalpy (ΔH). We can start the discussion on it’s periodic trends

1. Consider the graph shown in fig.3.7 below
    ♦ Atomic number Z is plotted along the x-axis
    ♦ ΔH is plotted along the y-axis
• The ΔH values of the elements are plotted in serial order of atomic number Z

Fig.3.7

2. We see that, the ΔH of ₂He (2372.3 kJ/mol) is higher than the ΔH of ₁H (1312 kJ/mol)
• This is expected. We can write the explanation in 3 steps:
(i) In ₁H, the outermost electron is pulled by one proton
• In ₂He, each of the outermost electrons are pulled by two protons
(ii) Mathematically, we can write:
• $\mathbf\small{F_H\propto \left[1e^2\right]}$
    ♦ Where $\mathbf\small{F_H}$ is the force with which the electron in ₁H is pulled by it's nucleus
    ♦ '$\mathbf\small{\propto}$' stands for 'is proportional to'
• $\mathbf\small{F_{He}\propto \left[2e^2\right]}$
    ♦ Where $\mathbf\small{F_{He}}$ is the force with which the electron in ₂He is pulled by it's nucleus
(iii) We see that, $\mathbf\small{F_{He}}$ is greater than $\mathbf\small{F_H}$
• That means, the electrons of ₂He are pulled more tightly by the nucleus
• So, when compared to hydrogen, it is more difficult to remove the outermost electron of helium
• In other words ₂He has a greater ΔH value than ₁H
3. After ₂He, we proceed to the next element which is: ₃Li
• We see that, there is a sudden drop in the ΔH value. Li is far below He
• This is not expected because, Li has one proton more than He
4. So what is the reason for the drop?
• We can write the explanation in 3 steps:
(i) The fig.3.8(b) below shows the orbital diagram of Li
• The last electron is present in the 2s orbital

Fig.3.8

(ii) We see that, the 1s orbital is completely filled
• This completely filled 1s orbital (the helium core) is present in between the last electron and the nucleus
(iii) So the 1s electrons form a shield (screen) between the last electron and the nucleus
• As a result, the outermost electron will not experience the full force from the nucleus
• So it is easier to remove the last electron
• In other words, Li will have a lower ΔH value
5. A similar condition occurs in the case of Na
• It’s orbital diagram is shown in fig.3.8(c)
• The neon core acts as a shield between the last electron and the nucleus
• As a result, the outermost electron of Na will not experience the full force from the nucleus
• So it is easier to remove that outermost electron
• Thus, the ΔH of Na will also be very low
6. Similar situation occurs in the case of K, Rb, Cs and Fr  also
    ♦ K has the 4s¹ electron outside the [Ar] core
    ♦ Rb has the 5s¹ electron outside the [Kr] core
    ♦ Cs has the 6s¹ electron outside the [Xe] core
    ♦ Fr has the 7s¹ electron outside the [Rn] core
■ So we can write:
The elements Li, Na, K, Rb, Cs and Fr have very low ΔH values
• Note that, these elements form the group 1 of the periodic table
■ So we can write:
The group 1 elements form the minima of the graph in fig.3.7
7. So we got an advantage:
• While analyzing the reason for the ‘low ΔH value of Li’ in step (3), we came to know why ‘the group 1 elements form the minima’ of the graph
8. The low ΔH values of group 1 elements can be related to their high reactivities
• These elements readily take part in chemical reactions
• This is because, they lose their outermost electron easily
■ Which one among them has the highest reactivity?
• The answer can be written in steps:
(i) The minima in fig.3.7 can be taken out separately
(ii) A graph can be plotted using those 'taken out' values:

Fig.3.9

• We see that there is a decreasing trend:
    ♦ ₁₁Na has a lower ΔH than ₃Li
    ♦ ₁₉K has a lower ΔH than ₁₁Na
    ♦ ₃₇Rb has a lower ΔH than ₁₉K
    ♦ ₅₅Cs has a lower ΔH than ₃₇Rb
• All these elements belong to the same group, which is: group 1 (alkali metals)
(iii) Following this trend, ₈₇Fr will have the lowest ΔH, and hence the highest reactivity
9. After Li, we proceed to the next element which is: Be
• In fig.3.7, we see that, Be has a higher ΔH than Li
• This will become more clear if we enlarge the left portion of the graph in fig.3.7
• Enlargement can be achieved if we decide to discard the atomic numbers beyond 10
• Then the details from atomic numbers 3 to 10 can be drawn in a larger space, and we can add more details
• Such a graph is shown in fig.3.10 below:

Fig.3.10

    ♦ ₃Li has 520.2 kJ/mol
    ♦ ₄Be has 899.5 kJ/mol
10. It is clear that, ΔH of Be is greater
• This is easy to explain. We can do it in 5 steps:
(i) The ΔH depends on two factors:
    ♦ The pull from the nucleus
    ♦ The shielding effect by the core
(ii) We can put this in a mathematical form for both Li and Be: 
    ♦ The ΔH of Li  $\mathbf\small{\propto}$ [Pull from Li nucleus – shielding by 'core in Li atom']
    ♦ The ΔH of Be $\mathbf\small{\propto}$ [Pull from Be nucleus – shielding by 'core in Be atom']
(iii) The ‘shielding by core in Li atom’ is same as ‘shielding by core in Be atom’ 
• Because, both Li and Be have the same [He] core
• So the 'quantity subtracted' is same for both the equations in (ii)
(iv) ‘Pull from Be nucleus’ will be greater than the ‘pull from Li nucleus’
• This is because, Be has one proton more than Li
• So the 'quantity from which subtraction is done' is greater in the second equation in (ii) 
(v) So we can see that 'ΔH of Be' will be greater than the 'ΔH of Li'
(vi) This will become even more clear if we make a bar chart as shown in fig.3.11 below:

Fig.3.11

• We calculate two items:
    ♦ For ΔH of Li: Subtract cyan from yellow
    ♦ For ΔH of Be: Subtract cyan from magenta
• Obviously, the second result will be greater
10. Now we proceed to the next element after Be, which is: B 
• We see that, B has a lower ΔH than Be
    ♦ ₄Be has 899.5 kJ/mol
    ♦ ₅B has 800.6 kJ/mol
11. This is not expected
• B has one proton more than Be. So the electrons in B must be held more tightly than in Be
• Consequently, the ΔH of B must be greater than Be. But this is not what we see from the graph
• So it is an anomaly
('Anomaly' means: Something that deviates from what is standard, normal, or expected. See dictionary meaning here)
12. The explanation can be written in steps:
(i) The orbital diagram of B is shown in fig.3.8(d)
• Comparing it with fig.3.8(c), we can see that:
    ♦ For Be, the electron which is to be removed, is in the 2p orbital
    ♦ For B, the electron which is to be removed, is in the 2s orbital
• Both are in the second main-shell
(ii) Here we make a comparison between two items:
    ♦ The 2s orbital
    ♦ The 2p orbital
• Scientists have proved that, the 2s orbital has 'greater penetration' when compared to the 2p orbital
    ♦ That is, the 2s orbital cloud is ‘closer to the nucleus’ when compared to the 2p orbital cloud
(iii) Now, both the '2s in Be' and '2p in B' will be experiencing the same shielding
    ♦ Because, in both cases, the shielding is caused by the [He] core
(iv) But we see that, the 2s has greater penetration than 2p
• ‘Greater penetration’ results in, ‘lesser shielding’
• That means, the 2s experience a lesser shielding than 2p
• This is same as:
    ♦ 2p experience a greater shielding than 2s
    ♦ So it is easier to ‘remove the 2p electron’ than to ‘remove the 2s electron’
(v) Mathematically we can write:
• The ΔH of Be $\mathbf\small{\propto}$ [Pull from Be nucleus – shielding by 'core of Be atom']
• The ΔH of B $\mathbf\small{\propto}$ [Pull from B nucleus – shielding by 'core of B atom']
(vi) In (v), 'Pull from B nucleus' is higher than 'Pull from Be nucleus'
• So we would expect 'ΔH of B' to be higher than 'ΔH of Be'
(vii) In (v) 'core of Be atom' is same as 'core of B atom', which is the [He] core
(viii) 'shielding by core of Be atom' is very less than the 'shielding by core of B atom' because, Be has greater penetration
(ix) So the net result is that, 'ΔH of B' is a little lower than 'ΔH of Be' 
(x) This will become even more clear if we make a bar chart as shown in fig.3.11 below:

Fig.3.12

• We calculate two items:
    ♦ For ΔH of Be: Subtract cyan from magenta
    ♦ For ΔH of B: Subtract blue from green
• Green is larger than magenta
    ♦ Even then, (green - blue) will be smaller than (magenta - cyan)
    ♦ This is because, cyan is much smaller than blue. The reason is written in (viii) above
• So ΔH of B is smaller than ΔH of Be
13. Now we proceed to the next element after B, which is: C
• We see that, C has a higher ΔH than Be and B
    ♦ ₄Be has 899.5 kJ/mol
    ♦ ₅B has 800.6 kJ/mol
    ♦ ₆C has 1086.5 kJ/mol
14. So the trend is back
■ That is., with increasing atomic number, the ΔH is increasing
• This is because, with the addition of one more proton, the pull towards the nucleus increases
• In the case of ₆C, the pull increases to such a high value that, the shielding effect is out weighted
15. This trend continues upto ₁₀Ne
• However, there is one more anomaly at ₈O
    ♦ ₇N has 1403.3 kJ/mol
    ♦ ₈O has 1313.9 kJ/mol
• Even after the addition of one more proton, the ΔH of O has decreased a bit
16. The explanation can be written in 3 steps:
(i) Fig.3.7(e) shows the orbital diagram of N
• We see that, all the 2p orbitals are half-filled
• This arises due to Hund's rule. We have already seen it in the previous chapter
(ii) Due to the half-filled orbitals, the N has some extra stability
• It is very difficult to remove an electron from the N atom
(iii) But in O, one of the 2p orbitals have two electrons
• This results in an increased repulsion between electrons
• So it is easier (when compared to N) to remove the electron
• Thus the 'ΔH value of O' is slightly lower than the 'ΔH value of N'
17. So we have seen the periodic trend in ionization enthalpy (ΔH) from ₃Li to ₁₀Ne
• This completes the 2nd period
18. The 3rd period starts from ₁₁Na and ends at ₁₈Ar
• In the fig.3.7, compare the following two items:
    ♦ The graph from  ₃Li to ₁₀Ne
    ♦ The graph from ₁₁Na to ₁₈Ar
• We see that, both the graphs have similar shapes
• That means, the same trends can be seen from ₁₁Na to ₁₈Ar also

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We will now see some solved examples

Solved example 3.11
The first ionization enthalpy (∆H) values of the third period elements, Na, Mg and Si are respectively 496, 737 and 786 kJ/mol. Predict whether the first ∆H value for Al will be more close to 575 or 760 kJ/mol ? Justify your answer
Solution:
1. The order of elements is: Na, Mg, Al, Si
    ♦ The electronic configuration of Mg is: 1s²2s²2p⁶3s²
    ♦ The electronic configuration of Al is: 1s²2s²2p⁶3s²3p¹
2. The core (in both the cases) is the first 3 items: [1s²2s²2p⁶]
    ♦ In Mg, this core shields the 3s² electrons from the nucleus
    ♦ In Al, this core shields the (3s² and 3p¹) electrons from the nucleus
3. But the 3s has greater penetration than 3p
• So the 3s in Mg has lesser shielding effect than 3p
• So it is easier to remove the 3p electrons than 3s electrons
• Thus the ΔH of Al will be less than the ΔH of Mg
4. Given that, ΔH of Mg is 737 kJ/mol
• The possible values given for Al are 575 and 760
• Out of these, only 575 is less than 737
■ So we can write:
∆H value for Al will be more close to 575 kJ/mol

Solved example 3.12
Energy of an electron in the ground state of the hydrogen atom is  -2.18 × 10^-18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of J/mol
Hint: Apply the idea of mole concept to derive the answer
Solution:
1. Given that the energy of one electron in the ground state of the hydrogen atom = -2.18 × 10^-18 J
2. So, if we supply this much energy, that electron will get removed from the atom
• That means, to remove one electron in the ground state of the hydrogen atom, we have to supply 2.18 × 10^-18 J of energy
3. This is the energy required to remove the electron from one atom
• So for 6.023 × 10^-18 nos. of atoms, we have to supply (2.18 × 10^-18 × 6.023 × 10²³) = 1.313 × 10⁶ J of energy
4. But the 'energy required to remove the electron from one mole of atoms' is ionization enthalpy
■ So we can write:
The ionization enthalpy of atomic hydrogen in terms of J/mol is 1.313 × 10⁶ J/mol    

Solved example 3.13
Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne.
Explain why
(i) Be has higher ∆H than B
(ii) O has lower ∆H than N and F?
Solution:
Part (i):
See steps (10) to (12) above
Part (ii):
See steps (15) and (16) above  

Solved example 3.14
How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Solution:
1. Let us write the electronic configurations:
• The electronic configuration of Na is [1s²2s²2p⁶]3s¹
• The electronic configuration of Mg is [1s²2s²2p⁶]3s²
2. In both cases, the first electron will be removed from the 3s orbital
• The Mg has greater number of protons
• So it's electrons will be held more tightly
• Consequently, it's ΔH value will be higher than that of Na
3. Once the first electrons are removed,
• The electronic configuration of Na+ will be [1s²2s²2p⁶]
• The electronic configuration of Mg+ will be [1s²2s²2p⁶]3s¹
4. We see that,
• In Na⁺, there is no third main-shell
    ♦ The configuration is same as that of the noble gas Ne
    ♦ It is very difficult to remove an electron from a noble gas configuration
• In Mg+, the last electron is present in the third main-shell
    ♦ That electron can be removed easily to attain noble gas configuration
5. That means,
• In Na⁺, the last electron is more tightly held to the nucleus than in Mg⁺
• So the second ionization enthalpy of Na will be higher than the second ionization enthalpy of Mg

Solved example 3.15
What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
Solution:
We can write 2 factors:
1. Increasing number of main-shells. This can be explained in 4 steps:
(i) When we move down the group, the number of main-shells increases
(ii) So the valence electrons are not tightly held to the nucleus
(iii) As a result, the valence electrons can be easily removed from the atom
(iv) In other words, the ΔH values become lower and lower
2. Increasing shielding effect. This can be explained in 6 steps:
(i) When we move down the group, the number of main-shells increases
(ii) So the number of inner shells also increases as we go down the group
(iii) This increases the shielding effect
• That is., when we move down the group, the valence electrons are more effectively shielded from the nucleus 
(iv) So the valence electrons are not tightly held to the nucleus
(v) As a result, the valence electrons can be easily removed from the atom
(vi) In other words, the ΔH values become lower and lower

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In the next section, we will see electron gain enthalpy

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Chapter 3.6 - Periodic Trends in Ionic Radius

In the previous section 3.5, we completed a discussion on periodic trends in atomic radius. In this section, we will see periodic trends in ionic radius

1. When an electron is removed from an atom, we get a +ve ion (cation)
• When an electron is added to an atom, we get a -ve ion (anion)
2. The ionic radii of elements shows the same trend as the atomic radii
• So we can apply the information that we learnt in the case of atomic radii
Some examples:
• The atomic radii increases down the group
    ♦ So Na⁺ will be larger than Li⁺
• Atomic radii decreases towards the right along a period
    ♦ So Na⁺ will be larger than Mg²⁺

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Let us see some details (not related to periodic trends) about ions:

1. In the case of cations, the ‘number of electrons’ will be less than that in the parent atom
• The charge in the nucleus remains the same
• So the nucleus is now able to pull the outermost electrons more tightly
• As a result, the size of the cation will be less than the size of the parent atom
• An example:
    ♦ Atomic radius of Na = 186 pm
    ♦ Ionic radius of Na⁺ = 95 pm
2. In the case of anions, the ‘number of electrons’ will be more than that in the parent atom
    ♦ This will increase the repulsive forces among the electrons
• The charge in the nucleus remains the same
• But this full charge is now not effective because, there is increased repulsive forces between electrons
• So the nucleus is now not able to pull the outermost electrons as tightly as it did before
• As a result, the size of the anions will be greater than the size of the parent atom
• An example:
    ♦ Atomic radius of F = 64 pm
    ♦ Ionic radius of F^- = 136 pm

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Given below are four cases related to ions:

Case 1:
1. Consider the oxygen atom (O)
• It has 8 electrons
• It’s electronic configuration is 1s²2s²2p⁴
2. If the O atom gains 2 electrons, it will become O^2- ion
• The O^2- ion has 10 electrons
• It’s electronic configuration is 1s²2s²2p⁶
3. From this configuration, it is clear that, O^2- ion has 8 electrons in the outermost main-shell
4. There is no change in the number of protons. So the nuclear charge remains the same
• We have:
Nuclear charge of O = Nuclear charge of O^2- = $\mathbf\small{(Z_O\,e)}$ = 8e

Case 2:
1. Consider the fluorine atom (F)
• It has 9 electrons
• It’s electronic configuration is 1s²2s²2p⁵
2. If the F atom gains 1 electron, it will become F^- ion
• The F^- ion has 10 electrons
• It’s electronic configuration is 1s²2s²2p⁶
3. From this configuration, it is clear that, F^- ion has 8 electrons in the outermost main-shell
4. There is no change in the number of protons. So the nuclear charge remains the same
• We have:
Nuclear charge of F = Nuclear charge of F^- = $\mathbf\small{(Z_F\,e)}$ = 9e

Case 3:
1. Consider the sodium atom (Na)
• It has 11 electrons
• It’s electronic configuration is 1s²2s²2p⁶3s¹
2. If the Na atom loses 1 electron, it will become Na⁺ ion
• The Na⁺ ion has 10 electrons
• It’s electronic configuration is 1s²2s²2p⁶
3. From this configuration, it is clear that, Na⁺ ion has 8 electrons in the outermost main-shell
4. There is no change in the number of protons. So the nuclear charge remains the same
• We have:
Nuclear charge of Na = Nuclear charge of Na⁺ = $\mathbf\small{(Z_{Na}\,e)}$ = 11e

Case 4:
1. Consider the sodium atom (Mg)
• It has 12 electrons
• It’s electronic configuration is 1s²2s²2p⁶3s²
2. If the Na atom loses 2 electrons, it will become Mg²⁺ ion
• The Mg²⁺ ion has 10 electrons
• It’s electronic configuration is 1s²2s²2p⁶
3. From this configuration, it is clear that, Mg²⁺ ion has 8 electrons in the outermost main-shell
4. There is no change in the number of protons. So the nuclear charge remains the same
• We have:
Nuclear charge of Mg = Nuclear charge of Mg²⁺ = $\mathbf\small{(Z_{Mg}\,e)}$ = 12e

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In the above 4 cases, we notice the following 4 points:

(i) All the 4 ions have the same number of electrons
(ii) All the 4 ions have the same electronic configuration
(iii) All the 4 ions have the same number of electrons in the outermost main-shell
(iv) But all the 4 ions have different nuclear charges
■ Atoms and ions having the same number of electrons are called isoelectronic species
• So the above 4 ions are isoelectronic species

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Let us write more details about them:

1. The forces on outermost electrons
• In O^2- ion, the force ($\mathbf\small{F_O}$) on each outermost electron will be such that $\mathbf\small{F_O\propto \left[8e^2\right]}$
    ♦ Where '$\mathbf\small{\propto}$' stands for 'is proportional to'
• In F^- ion, the force ($\mathbf\small{F_F}$) on each outermost electron will be such that $\mathbf\small{F_F\propto \left[9e^2\right]}$
• In Na⁺ ion, the force ($\mathbf\small{F_{Na}}$) on each outermost electron will be such that $\mathbf\small{F_{Na}\propto \left[11e^2\right]}$
• In Mg²⁺ ion, the force ($\mathbf\small{F_{Mg}}$) on each outermost electron will be such that $\mathbf\small{F_{Mg}\propto \left[12e^2\right]}$
2. We see that:
• The largest force is $\mathbf\small{F_{Mg}}$
• The smallest force is $\mathbf\small{F_O}$
3. We know that, the 'number of electrons in the outermost main-shell' is the same for all the 4 ions
■ So, the ion in which the force is the largest will be the smallest ion
■ Also, the the ion in which the force is the smallest will be the largest ion
4. Thus we get:
• Among the given 4 isoelectronic species:
    ♦ Mg²⁺ will be having the smallest ionic radius
    ♦ O^2- will be the largest ionic radius
5. So in general, we can write:
■ Among the given isoelectronic species,
• The cation with the largest positive charge will be having the smallest ionic radius
• The anion with the largest negative charge will be having the largest ionic radius
6. We can write it in the form of a pattern:
(i) We have:
• The order from the smallest to the largest ionic radius is:
X^(+max) < X^(-max)
    ♦ Where 'X' is any element
(ii) By 'writing the species in the order from (+max) to (-max)', we get the 'increasing order of ionic radius'

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Now we will see some solved examples:

Solved example 3.8
What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.
(i) F^- (ii) Ar (iii) Mg²⁺ (iv) Rb⁺
Solution:
Part (a):
Atoms and ions having the same number of electrons are called isoelectronic species.
Part (b):
(i) F has 9 electrons
    ♦ So F^- has 10 electrons
• Ne also has 10 electrons
• So we can write: F^- and Ne are isoelectronic species
(ii) Ar has 18 electrons
• Cl has 17 electrons
    ♦ So Cl^- has 18 electrons
• So we can write: Ar and Cl^- are isoelectronic species
(iii) Mg has 12 electrons
    ♦ So Mg²⁺ has 10 electrons
• Na has 11 electrons
    ♦ So Na⁺ also has 10 electrons
• So we can write: Mg²⁺ and Na⁺ are isoelectronic species
(iv) Rb has 37 electrons
    ♦ So Rb⁺ has 36 electrons
• Kr also has 36 electrons
• So we can write: Rb⁺ and Kr are isoelectronic species

Solved example 3.9
Consider the following species :
N^3-, O^2-, F^-, Na⁺, Mg²⁺ and Al³⁺
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii
Solution:
Part (a):
(i) All the 6 ions have the same number of electrons, which is: 10
(ii) All the 6 ions have the same electronic configuration, which is: 1s²2s²2p⁶ 
(iii) All the 6 ions have the same number of electrons in the outermost main-shell, which is: 8
Part (b):
1. The given 6 ions are isoelectronic species
■ We know the rule 'for writing the increasing order':
Write the species in the order from (+max) to (-max)
2. Thus we get:
Al³⁺ < Mg²⁺ < Na⁺ < F^- < O^2- < N^3-

Solved example 3.10
Which of the following species will have the largest and the smallest size?
Mg, Mg²⁺, Al, Al³⁺
Solution:
1. First we compare the atoms: Mg and Al
• Mg and Al are in the same period
• We know how to arrange such atoms in the increasing order of atomic radii
• The 'atom of the element' on the right side in the periodic table will be smaller
• So we get: Al < Mg
2. Next we compare the ions: Mg²⁺ and Al³⁺
They are isoelectronic species
■ We know the rule 'for writing the increasing order' of ionic radius:
Write the species in the order from (+max) to (-max)
• Thus we get:
Al³⁺ < Mg²⁺
3. So we have two sets: [Al < Mg] and [Al³⁺ < Mg²⁺]
4. We will compare the smallest among the two sets in (3)
• The smallest are: Al and Al³⁺
• We know that Al³⁺ < Al
• So Al³⁺ is the smallest among all the given 4 species
5. Next, we will compare the largest among the two sets in (3)
• The largest are: Mg and Mg²⁺
• We know that Mg²⁺ < Mg
• So Mg is the largest among all the given 4 species

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In the next section, we will see ionization enthalpy

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