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+
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
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 O2- ion
• The O2- ion has 10 electrons
• It’s electronic configuration is 1s22s22p6
3. From this configuration, it is clear that, O2- 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 O2- = $\mathbf\small{(Z_O\,e)}$ = 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
(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
1. The forces on outermost electrons
• In O2- 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
♦ O2- 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'
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 :
N3-, O2-, 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- < O2- < N3-
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
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+
Let us see some details (not related to periodic trends) about ions:
• 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
Given below are four cases related to ions:
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 O2- ion
• The O2- ion has 10 electrons
• It’s electronic configuration is 1s22s22p6
3. From this configuration, it is clear that, O2- 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 O2- = $\mathbf\small{(Z_O\,e)}$ = 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
In the above 4 cases, we notice the following 4 points:
(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
Let us write more details about them:
• In O2- 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
♦ O2- 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'
Now we will see some solved examples:
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 :
N3-, O2-, 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- < O2- < N3-
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
In the next section, we will see ionization enthalpy
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