In the previous section 3.4, we completed a discussion on classification of elements in the periodic table. In this section, we will see periodic trends
1. Consider any period in the periodic table
• Imagine that, we are moving from left to right along that period
2. We will notice the following points:
• The element at the extreme left end has a very high reactivity
• The 2nd element from the left also has a high reactivity. But not as high as the 1st element
• The 3rd element from the left also has a high reactivity. But not as high as the 2nd element
■ That means, as we move towards the right, the reactivity decreases
3. But this ‘decreasing trend’ will not continue for long
• This ‘decreasing trend’ will continue only up to the middle of the period
• Once we pass the middle point, the reactivity starts to increase
■ Each element will be having a higher reactivity than it’s predecessor
4. This ‘increasing trend’ continues until we reach the ‘last but one’ element
• ‘Last but one’ element in a period is obviously, the element in the 17th group
■ In fact, the element in the 17th group will be having the highest reactivity in a period
5. After the 17th group, we have the 18th group
• Elements in the 18th group are noble gases (inert gases)
• They do not show any reactivity
6. Thus we now know the trend along a period
• A graphical representation of this trend is shown in fig.3.3 below
• Note that, it is only a schematic graph. The actual graph will have a similar shape but will not be so smooth. It will be having many undulations. We will see the actual graph in later sections
• Now let us move from top to bottom along a group
1. Consider group 1
• When we move down along this group, the reactivity increases
2. Consider group 17
• When we move down along this group, the reactivity decreases
• These are shown in figs.3.4 (a) and (b) respectively
• Note that, they are only a schematic graphs. The actual graphs will have a similar shapes but will not be so smooth. They will be having many undulations. We will see the actual graphs in later sections
• Before moving onto further discussions, we need to split the phrase: ‘periodic trend in reactivity‘
• We will split it as: periodic trend + in + reactivity
2. Why do we need to split it like that?
Answer can be written in 4 steps:
(i) By splitting it, we become aware that ‘many periodic trends’ are available for us to learn
• Let us see some examples:
♦ periodic trend + in + atomic radius
♦ periodic trend + in + ionic radius
♦ periodic trend + in + ionization enthalpy
♦ periodic trend + in + electronegativity
(ii) The first two terms remain the same
• Only the last term is changing
(iii) That means, we are considering the phenomenon of ‘periodic trends’
(iv) We are applying this same phenomenon, in relation to various properties like atomic radius, ionic radius etc.,
1. Some of the physical properties of elements are:
(i) Melting point
(ii) Boiling point
(iii) Heat of fusion (The amount of heat energy required by 1 kg of the element to change from solid to liquid)
(iv) Heat of vaporization (The amount of heat energy required by 1 kg of the element to change from liquid to gas)
(v) Atomic radius (Radius of one atom of the element)
(vi) Ionic radius (Radius of one ion of the element)
(vii) Ionization enthalpy (The amount of energy required to remove the outer most electron from the atom)
(viii) Electron gain enthalpy (The amount of energy released by the atom when an electron is gained)
(ix) Electronegativity (Ability of the atom to attract shared pairs of electrons)
• In our present discussion, with regard to physical properties, we will be considering the periodic trends in the last 5 items given above
2. Some of the chemical properties of elements are:
(i) Oxidation states
(ii) Anomalous Properties of Second Period Elements
(iii) Metallic and non-metallic characters
• We will see them in later sections
Answer can be written in 3 steps:
(i) Learning such trends put us in an advantageous position
• We will be knowing ‘what to expect’ while moving along the periodic table
(ii) For example, while moving down group 1, we will be expecting more and more reactive elements
• So we will not be suddenly confronting with any 'explosive' elements
(iii) Similarly, while moving down group 17, we will be expecting lesser and lesser reactive elements
• So we will not be suddenly confronting with any ‘nearly inert’ elements
• Obviously, it will be very small. It is approximately equal to (1.2 × 10-10) m
2. So it not possible to measure the radius directly
• Moreover, as we have seen in the previous chapter, the electrons exist as clouds around the nucleus
• So there will not be a ‘well defined boundary’ around the atom
3. However, there are practical methods by which we can measure the radius with a good degree of accuracy
• We will see those methods in later sections
• At present we will concentrate on the periodic trends
4. First we will see the trend along a period
• Consider the period 2
• Imagine that, we are moving from left to right along that period
5. We will notice the following points:
• The element at the extreme left end (Li) has an atomic radius of 152 pico meter (pm)
• The second element (Be) has a radius of 111 pm
♦ This is lesser than the radius of Li
• The third element (B) has a radius of 88 pm
♦ This is lesser than the radius of Be
6. The complete table related to periods 1 and 2 is given below:
■ The trend is obvious:
As we move from left to right along a period, the atomic radius decreases
7. The reason can be written in 12 steps
(i) Fig.3.5 below shows the schematic diagrams showing the electrons around the nucleus for two elements A and B
• Let A and B be adjacent elements in the periodic table
♦ B being just to right side of A in the periodic table
(ii) Then, in the two schematic diagrams, there will be only one difference:
• The diagram for B will have one extra electron than A
(Remember that, all elements in a period will be having the same number of main-shells)
(iii) Now consider the force between two items in A:
♦ The nucleus of A
♦ One outermost electron in A
• Let us denote this force as $\mathbf\small{F_A}$
(iv) $\mathbf\small{F_A}$ will be proportional to the 'product of $\mathbf\small{(Z_A\,e)}$ and $\mathbf\small{(e)}$'
• Where
♦ $\mathbf\small{(Z_A)}$ is the atomic number of A
♦ $\mathbf\small{(e)}$ is the 'charge of one electron' which is same as the 'charge of one proton'
(v) So $\mathbf\small{(Z_A\,e)}$ is the total charge of the nucleus
• $\mathbf\small{(e)}$ is the charge of the one electron under consideration
• $\mathbf\small{F_A}$ is the force between the nucleus and that electron under consideration
(vi) We have: $\mathbf\small{F_A \propto \left[(Z_A\,e)(e)\right]}$
$\mathbf\small{\Rightarrow F_A \propto \left[Z_A\,e^2\right]}$
• Each outermost electron in A will be experiencing this same force $\mathbf\small{F_A}$
(vii) Next consider the force between two items in B:
♦ The nucleus of B
♦ One outermost electron in B
• Let us denote this force as $\mathbf\small{F_B}$
(viii) $\mathbf\small{F_B}$ will be proportional to the 'product of $\mathbf\small{(Z_B\,e)}$ and $\mathbf\small{(e)}$'
• Where
♦ $\mathbf\small{(Z_B)}$ is the atomic number of B
♦ $\mathbf\small{(e)}$ is the 'charge of one electron' which is same as the 'charge of one proton'
(ix) So $\mathbf\small{(Z_B\,e)}$ is the total charge of the nucleus of B
• $\mathbf\small{(e)}$ is the charge of the one electron under consideration
• $\mathbf\small{F_B}$ is the force between the 'nucleus of B' and that electron under consideration
(x) We have: $\mathbf\small{F_B \propto \left[(Z_B\,e)(e)\right]}$
$\mathbf\small{\Rightarrow F_B \propto \left[Z_B\,e^2\right]}$
• Each outermost electron in B will be experiencing this same force $\mathbf\small{F_B}$
(xi) But $\mathbf\small{Z_B=(Z_A+1)}$
(∵ B is just to the right of A in the periodic table)
• So we get: $\mathbf\small{F_B \propto \left[(Z_A+1)\,e^2\right]}$
(xii) Let us compare the two items:
• $\mathbf\small{F_A \propto \left[Z_A\,e^2\right]}$
• $\mathbf\small{F_B \propto \left[(Z_A+1)\,e^2\right]}$
■ Obviously, $\mathbf\small{F_B}$ is greater
• So each 'outermost electron in B' will be experiencing a greater force than the force in A
• So the 'outermost electrons in B' will be closer to the nucleus of B
• As a result, the 'radius of B' will be lesser than that of A
■ Based on the above discussion, we can write:
When we move from left to right along a period, each element will be having a radius, smaller than the preceding element
8. Now we will see the trend along a group
• Consider the group 1
• Imagine that, we are moving from top to bottom along that group
9. We will notice the following points:
• The top most element (Li) has an atomic radius of 152 pico meter (pm)
• The second element (Na) has a radius of 186 pm
♦ This is greater than the radius of Li
• The third element (K) has a radius of 231 pm
♦ This is greater than the radius of Na
The complete table related to groups is given below:
■ The trend is obvious:
As we move from top to bottom along a group, the atomic radius increases
10. Let us see the reason
(i) Fig.3.6 below shows the schematic diagrams showing the electrons around the nucleus for two elements A and B
• Let A and B be adjacent elements in the periodic table
♦ B being just below A in the periodic table
(ii) Then, in the two schematic diagrams, B will have one more main-shell than A
(Remember that, with each succeeding period, one main-shell is added)
(iii) Now compare the following two items:
• Distance between the nucleus and outermost electron in A
• Distance between the nucleus and outermost electron in B
■ Obviously, the second item will be greater
(iv) It is true that $\mathbf\small{(Z_B\,e)}$ is larger than $\mathbf\small{(Z_A\,e)}$
• We are inclined to think that the greater force from $\mathbf\small{(Z_B\,e)}$ will pull the outer electrons closer
• But the inner main-shell will shield the outer electrons from the nucleus
• So the nucleus will not be able to exert much force on the outer electrons
• The net effect is that, the nucleus in B has a lesser control on the outer electrons
■ Based on the above discussion, we can write:
When we move from top to bottom along a group, each element will be having a radius, larger than the preceding element
1. Consider any period in the periodic table
• Imagine that, we are moving from left to right along that period
2. We will notice the following points:
• The element at the extreme left end has a very high reactivity
• The 2nd element from the left also has a high reactivity. But not as high as the 1st element
• The 3rd element from the left also has a high reactivity. But not as high as the 2nd element
■ That means, as we move towards the right, the reactivity decreases
3. But this ‘decreasing trend’ will not continue for long
• This ‘decreasing trend’ will continue only up to the middle of the period
• Once we pass the middle point, the reactivity starts to increase
■ Each element will be having a higher reactivity than it’s predecessor
4. This ‘increasing trend’ continues until we reach the ‘last but one’ element
• ‘Last but one’ element in a period is obviously, the element in the 17th group
■ In fact, the element in the 17th group will be having the highest reactivity in a period
5. After the 17th group, we have the 18th group
• Elements in the 18th group are noble gases (inert gases)
• They do not show any reactivity
6. Thus we now know the trend along a period
• A graphical representation of this trend is shown in fig.3.3 below
 |
| Fig.3.3 |
• In the above discussion, we moved from left to right along a period
1. Consider group 1
• When we move down along this group, the reactivity increases
2. Consider group 17
• When we move down along this group, the reactivity decreases
• These are shown in figs.3.4 (a) and (b) respectively
 |
| Fig.3.4 |
1. Both the figs.3.3 and 3.4 show ‘periodic trend in reactivity‘
• We will split it as: periodic trend + in + reactivity
2. Why do we need to split it like that?
Answer can be written in 4 steps:
(i) By splitting it, we become aware that ‘many periodic trends’ are available for us to learn
• Let us see some examples:
♦ periodic trend + in + atomic radius
♦ periodic trend + in + ionic radius
♦ periodic trend + in + ionization enthalpy
♦ periodic trend + in + electronegativity
(ii) The first two terms remain the same
• Only the last term is changing
(iii) That means, we are considering the phenomenon of ‘periodic trends’
(iv) We are applying this same phenomenon, in relation to various properties like atomic radius, ionic radius etc.,
■ The items can be classified as physical properties and chemical properties
(i) Melting point
(ii) Boiling point
(iii) Heat of fusion (The amount of heat energy required by 1 kg of the element to change from solid to liquid)
(iv) Heat of vaporization (The amount of heat energy required by 1 kg of the element to change from liquid to gas)
(v) Atomic radius (Radius of one atom of the element)
(vi) Ionic radius (Radius of one ion of the element)
(vii) Ionization enthalpy (The amount of energy required to remove the outer most electron from the atom)
(viii) Electron gain enthalpy (The amount of energy released by the atom when an electron is gained)
(ix) Electronegativity (Ability of the atom to attract shared pairs of electrons)
• In our present discussion, with regard to physical properties, we will be considering the periodic trends in the last 5 items given above
2. Some of the chemical properties of elements are:
(i) Oxidation states
(ii) Anomalous Properties of Second Period Elements
(iii) Metallic and non-metallic characters
• We will see them in later sections
■ Why do we want to learn about such periodic trends?
(i) Learning such trends put us in an advantageous position
• We will be knowing ‘what to expect’ while moving along the periodic table
(ii) For example, while moving down group 1, we will be expecting more and more reactive elements
• So we will not be suddenly confronting with any 'explosive' elements
(iii) Similarly, while moving down group 17, we will be expecting lesser and lesser reactive elements
• So we will not be suddenly confronting with any ‘nearly inert’ elements
First we will consider the physical properties
Periodic trends in atomic radius
1. The radius of an atom is the distance from the nucleus to the outer most electron• Obviously, it will be very small. It is approximately equal to (1.2 × 10-10) m
2. So it not possible to measure the radius directly
• Moreover, as we have seen in the previous chapter, the electrons exist as clouds around the nucleus
• So there will not be a ‘well defined boundary’ around the atom
3. However, there are practical methods by which we can measure the radius with a good degree of accuracy
• We will see those methods in later sections
• At present we will concentrate on the periodic trends
4. First we will see the trend along a period
• Consider the period 2
• Imagine that, we are moving from left to right along that period
5. We will notice the following points:
• The element at the extreme left end (Li) has an atomic radius of 152 pico meter (pm)
• The second element (Be) has a radius of 111 pm
♦ This is lesser than the radius of Li
• The third element (B) has a radius of 88 pm
♦ This is lesser than the radius of Be
6. The complete table related to periods 1 and 2 is given below:
 |
| Table 3.3 |
As we move from left to right along a period, the atomic radius decreases
7. The reason can be written in 12 steps
(i) Fig.3.5 below shows the schematic diagrams showing the electrons around the nucleus for two elements A and B
 |
| Fig.3.5 |
♦ B being just to right side of A in the periodic table
(ii) Then, in the two schematic diagrams, there will be only one difference:
• The diagram for B will have one extra electron than A
(Remember that, all elements in a period will be having the same number of main-shells)
(iii) Now consider the force between two items in A:
♦ The nucleus of A
♦ One outermost electron in A
• Let us denote this force as $\mathbf\small{F_A}$
(iv) $\mathbf\small{F_A}$ will be proportional to the 'product of $\mathbf\small{(Z_A\,e)}$ and $\mathbf\small{(e)}$'
• Where
♦ $\mathbf\small{(Z_A)}$ is the atomic number of A
♦ $\mathbf\small{(e)}$ is the 'charge of one electron' which is same as the 'charge of one proton'
(v) So $\mathbf\small{(Z_A\,e)}$ is the total charge of the nucleus
• $\mathbf\small{(e)}$ is the charge of the one electron under consideration
• $\mathbf\small{F_A}$ is the force between the nucleus and that electron under consideration
(vi) We have: $\mathbf\small{F_A \propto \left[(Z_A\,e)(e)\right]}$
$\mathbf\small{\Rightarrow F_A \propto \left[Z_A\,e^2\right]}$
• Each outermost electron in A will be experiencing this same force $\mathbf\small{F_A}$
(vii) Next consider the force between two items in B:
♦ The nucleus of B
♦ One outermost electron in B
• Let us denote this force as $\mathbf\small{F_B}$
(viii) $\mathbf\small{F_B}$ will be proportional to the 'product of $\mathbf\small{(Z_B\,e)}$ and $\mathbf\small{(e)}$'
• Where
♦ $\mathbf\small{(Z_B)}$ is the atomic number of B
♦ $\mathbf\small{(e)}$ is the 'charge of one electron' which is same as the 'charge of one proton'
(ix) So $\mathbf\small{(Z_B\,e)}$ is the total charge of the nucleus of B
• $\mathbf\small{(e)}$ is the charge of the one electron under consideration
• $\mathbf\small{F_B}$ is the force between the 'nucleus of B' and that electron under consideration
(x) We have: $\mathbf\small{F_B \propto \left[(Z_B\,e)(e)\right]}$
$\mathbf\small{\Rightarrow F_B \propto \left[Z_B\,e^2\right]}$
• Each outermost electron in B will be experiencing this same force $\mathbf\small{F_B}$
(xi) But $\mathbf\small{Z_B=(Z_A+1)}$
(∵ B is just to the right of A in the periodic table)
• So we get: $\mathbf\small{F_B \propto \left[(Z_A+1)\,e^2\right]}$
(xii) Let us compare the two items:
• $\mathbf\small{F_A \propto \left[Z_A\,e^2\right]}$
• $\mathbf\small{F_B \propto \left[(Z_A+1)\,e^2\right]}$
■ Obviously, $\mathbf\small{F_B}$ is greater
• So each 'outermost electron in B' will be experiencing a greater force than the force in A
• So the 'outermost electrons in B' will be closer to the nucleus of B
• As a result, the 'radius of B' will be lesser than that of A
■ Based on the above discussion, we can write:
When we move from left to right along a period, each element will be having a radius, smaller than the preceding element
8. Now we will see the trend along a group
• Consider the group 1
• Imagine that, we are moving from top to bottom along that group
9. We will notice the following points:
• The top most element (Li) has an atomic radius of 152 pico meter (pm)
• The second element (Na) has a radius of 186 pm
♦ This is greater than the radius of Li
• The third element (K) has a radius of 231 pm
♦ This is greater than the radius of Na
The complete table related to groups is given below:
 |
| Table 3.4 |
As we move from top to bottom along a group, the atomic radius increases
10. Let us see the reason
(i) Fig.3.6 below shows the schematic diagrams showing the electrons around the nucleus for two elements A and B
 |
| Fig.3.6 |
♦ B being just below A in the periodic table
(ii) Then, in the two schematic diagrams, B will have one more main-shell than A
(Remember that, with each succeeding period, one main-shell is added)
(iii) Now compare the following two items:
• Distance between the nucleus and outermost electron in A
• Distance between the nucleus and outermost electron in B
■ Obviously, the second item will be greater
(iv) It is true that $\mathbf\small{(Z_B\,e)}$ is larger than $\mathbf\small{(Z_A\,e)}$
• We are inclined to think that the greater force from $\mathbf\small{(Z_B\,e)}$ will pull the outer electrons closer
• But the inner main-shell will shield the outer electrons from the nucleus
• So the nucleus will not be able to exert much force on the outer electrons
• The net effect is that, the nucleus in B has a lesser control on the outer electrons
■ Based on the above discussion, we can write:
When we move from top to bottom along a group, each element will be having a radius, larger than the preceding element
In the next section, we will see the periodic trends in ionic radius
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