In the previous section 3.12, we completed a discussion on the periodic trends (along periods) in reactivity’. In this section, we will see some examples to demonstrate those trends. Later, we will also see ‘periodic trends (along groups) in reactivity’
■ The chemical reactivity of an element can be demonstrated by two methods:
(i) Analyzing the reaction of that element with oxygen
♦ If the element reacts readily with oxygen, reactivity of that element is high
♦ If the element does not react readily with oxygen, reactivity of that element is low
(ii) Analyzing the reaction of that element with halogens (group 17 elements)
♦ If the element reacts readily with halogen, reactivity of that element is high
♦ If the element does not react readily with halogen, reactivity of that element is low
• In our present discussion, we will consider the reactions with oxygen only. We will write it in steps:
1. Reaction with oxygen:
• When metals react with oxygen, we get metallic oxides
• When non-metals react with oxygen, we get non-metallic oxides
2. These two types of oxides have a fundamental difference:
• Metallic oxides dissolve in water to give alkalies
• Non-metallic oxides dissolve in water to give acids
We have seen those details in our earlier classes (Details here)
3. Let us see some examples:
Example 1:
• Sodium oxide (Na2O) is a metallic oxide
• It reacts with water
• The balanced equation is:
$\mathbf\small{\rm{Na_2O+H_2O\longrightarrow 2NaOH}}$
• The product sodium hydroxide (NaOH) is an alkali
Example 2:
• Dichlorine pentoxide (Cl2O7) is a non-metallic oxide
• It reacts with water
The balanced equation is:
$\mathbf\small{\rm{Cl_2O_7+H_2O\longrightarrow 2HClO_4}}$
• The product perchloric acid (HClO4) is an acid
4. We have seen that, in the periodic table:
♦ Metals reside towards the left
♦ Non-metals reside towards the right
■ So we can write:
• When we move along a period from left to right:
♦ Towards the left, we will get basic oxides
♦ Towards the right, we will get acidic oxides
5. But that is not all. There is a ‘gradation’ also
• That is., we get: stronger acid, weaker acids, etc.,
• Let us elaborate:
♦ ‘Oxide of the element’ on the extreme left will give the strongest base
♦ ‘Oxide of the element’ on the second extreme left will give a less stronger base
Similarly:
♦ ‘Oxide of the element’ on the extreme right will give the strongest acid
♦ ‘Oxide of the element’ on the second extreme right will give a less stronger acid
6. This ‘gradation’ can be schematically shown as in fig.3.21 below:
• Amphoteric oxides have a peculiar property:
♦ They show acidic properties while reacting with bases
♦ They show basic properties while reacting with acids
♦ Some examples are: Al2O3, As2O3
• Neutral oxides do not show any acidic or basic properties at all
♦ Some examples are: CO, NO, N2O
7. In the above discussion, we have considered mainly the 2nd and 3rd periods
• For example:
♦ Na is the left end of the 3rd period
♦ Cl is at the right end of the 3rd period (excluding noble gas)
♦ Al is near the middle of the 3rd period
♦ C and N are near the middle of the 2nd period
• In the 2nd and 3rd periods, we do not have any d-block elements (transition metals)
• So next we will see a separate discussion on 'periodic trends of reactivity along periods' in transition metals
8. Let us travel from left to right along the various periods in the d-block. We will write the analysis in steps:
(i) As far as the d-block is concerned, the ‘first horizontal row’ comes in the 4th period
• In this 4th period,
♦ the first transition element is Sc
♦ the last transition element is Zn
(ii) From Sc to Zn, the inner 3d orbitals are being filled up
• So we will experience only a ‘small reduction in atomic radius’ as we move across the period
(iii) The experience will be still lesser in the case of the 5th period (from Y to Cd)
(iv) The experience will be even lesser in the case of the 6th period because, here, the inner 4f orbitals are being filled up
(v) So in general, for transition elements, we do not find a regular gradation as seen in fig.3.21
• Next we will discuss ‘periodic trends (along groups) of chemical reactivity’
1. Recall the 4 vertical arrows that we saw in fig.3.15 in an earlier section 3.9
• For convenience, that fig.3.15 is shown again below:
2. Out of the four arrows, two are important for our present discussion:
(i) The arrow for ionization enthalpy
(ii) The arrow for electron gain enthalpy
3. Consider the arrow for ionization enthalpy:
■ It indicates that:
In a group, the ionization enthalpy is least for the element at the bottom end
• That means:
In a group, the element at the bottom end can lose it’s electron in the most easiest way
• It follows that:
In a group, the element at the bottom end can become a cation in the most easiest way
• It follows that:
In a group, the element at the bottom left end will have the highest chemical reactivity
4. Consider the arrow for electron gain enthalpy
■ It indicates that:
In a group, the electron gain enthalpy is 'highest negative' for the element at the top end
• That means:
In a group, the element at the top end can gain an electron in the most easiest way
• It follows that:
In a group, the element at the top end can become an anion in the most easiest way
• It follows that:
In a group, the element at the top end will have the highest chemical reactivity
5. So we have two extremes:
• One at the bottom end and the other at the top end
♦ Both have the highest reactivities
6. The 'variation of reactivity' can be written in 3 steps:
(i) The reactivity decreases (initially) as we go from bottom to top along a group
(ii) It reaches the 'lowest reactivity' at the center of the group
(iii) After passing the center, it again increases and reaches the 'highest reactivity' at the top end
7. We can write a note on oxidation and reduction also:
(i) Oxidation:
• 'Losing electrons' is oxidation
♦ That is., 'elements which lose electrons' are said to be oxidised
• The elements at the bottom easily loses electrons
♦ So we can say: Elements at the bottom are easily oxidised
(ii) Reduction:
• 'Gaining electrons' is reduction
♦ That is., 'elements which gain electrons' are said to be reduced
• The elements at the top easily gains electrons
♦ So we can say: Elements at the top are easily reduced
8. We can write a note on metallic and non-metallic character also:
(i) Metallic character:
• Metals are electron donors
♦ That is., metals lose electrons easily
• Elements at the bottom lose electrons easily
♦ So we can say: Elements at the bottom are more metallic
(ii) Non-metallic character:
• Non-Metals are electron acceptors
♦ That is., non-metals accept electrons easily
• Elements at the top accepts electrons easily
♦ So we can say: Elements at the top are more non-metallic
9. A special note must be made about step (4) above:
• It is related to electron gain enthalpy
• Though we indicate it's 'periodic trend' with an upward arrow, it is not a systematic trend. We must always remember the anomalies that we discussed earlier [step (4) below table 3.5] in section 3.9
■ The chemical reactivity of an element can be demonstrated by two methods:
(i) Analyzing the reaction of that element with oxygen
♦ If the element reacts readily with oxygen, reactivity of that element is high
♦ If the element does not react readily with oxygen, reactivity of that element is low
(ii) Analyzing the reaction of that element with halogens (group 17 elements)
♦ If the element reacts readily with halogen, reactivity of that element is high
♦ If the element does not react readily with halogen, reactivity of that element is low
• In our present discussion, we will consider the reactions with oxygen only. We will write it in steps:
1. Reaction with oxygen:
• When metals react with oxygen, we get metallic oxides
• When non-metals react with oxygen, we get non-metallic oxides
2. These two types of oxides have a fundamental difference:
• Metallic oxides dissolve in water to give alkalies
• Non-metallic oxides dissolve in water to give acids
We have seen those details in our earlier classes (Details here)
3. Let us see some examples:
Example 1:
• Sodium oxide (Na2O) is a metallic oxide
• It reacts with water
• The balanced equation is:
$\mathbf\small{\rm{Na_2O+H_2O\longrightarrow 2NaOH}}$
• The product sodium hydroxide (NaOH) is an alkali
Example 2:
• Dichlorine pentoxide (Cl2O7) is a non-metallic oxide
• It reacts with water
The balanced equation is:
$\mathbf\small{\rm{Cl_2O_7+H_2O\longrightarrow 2HClO_4}}$
• The product perchloric acid (HClO4) is an acid
4. We have seen that, in the periodic table:
♦ Metals reside towards the left
♦ Non-metals reside towards the right
■ So we can write:
• When we move along a period from left to right:
♦ Towards the left, we will get basic oxides
♦ Towards the right, we will get acidic oxides
5. But that is not all. There is a ‘gradation’ also
• That is., we get: stronger acid, weaker acids, etc.,
• Let us elaborate:
♦ ‘Oxide of the element’ on the extreme left will give the strongest base
♦ ‘Oxide of the element’ on the second extreme left will give a less stronger base
Similarly:
♦ ‘Oxide of the element’ on the extreme right will give the strongest acid
♦ ‘Oxide of the element’ on the second extreme right will give a less stronger acid
6. This ‘gradation’ can be schematically shown as in fig.3.21 below:
 |
| Fig.3.21 |
♦ They show acidic properties while reacting with bases
♦ They show basic properties while reacting with acids
♦ Some examples are: Al2O3, As2O3
• Neutral oxides do not show any acidic or basic properties at all
♦ Some examples are: CO, NO, N2O
7. In the above discussion, we have considered mainly the 2nd and 3rd periods
• For example:
♦ Na is the left end of the 3rd period
♦ Cl is at the right end of the 3rd period (excluding noble gas)
♦ Al is near the middle of the 3rd period
♦ C and N are near the middle of the 2nd period
• In the 2nd and 3rd periods, we do not have any d-block elements (transition metals)
• So next we will see a separate discussion on 'periodic trends of reactivity along periods' in transition metals
8. Let us travel from left to right along the various periods in the d-block. We will write the analysis in steps:
(i) As far as the d-block is concerned, the ‘first horizontal row’ comes in the 4th period
• In this 4th period,
♦ the first transition element is Sc
♦ the last transition element is Zn
(ii) From Sc to Zn, the inner 3d orbitals are being filled up
• So we will experience only a ‘small reduction in atomic radius’ as we move across the period
(iii) The experience will be still lesser in the case of the 5th period (from Y to Cd)
(iv) The experience will be even lesser in the case of the 6th period because, here, the inner 4f orbitals are being filled up
(v) So in general, for transition elements, we do not find a regular gradation as seen in fig.3.21
• So we have completed a discussion on ‘periodic trends (along periods) of chemical reactivity’
1. Recall the 4 vertical arrows that we saw in fig.3.15 in an earlier section 3.9
• For convenience, that fig.3.15 is shown again below:
(i) The arrow for ionization enthalpy
(ii) The arrow for electron gain enthalpy
3. Consider the arrow for ionization enthalpy:
■ It indicates that:
In a group, the ionization enthalpy is least for the element at the bottom end
• That means:
In a group, the element at the bottom end can lose it’s electron in the most easiest way
• It follows that:
In a group, the element at the bottom end can become a cation in the most easiest way
• It follows that:
In a group, the element at the bottom left end will have the highest chemical reactivity
4. Consider the arrow for electron gain enthalpy
■ It indicates that:
In a group, the electron gain enthalpy is 'highest negative' for the element at the top end
• That means:
In a group, the element at the top end can gain an electron in the most easiest way
• It follows that:
In a group, the element at the top end can become an anion in the most easiest way
• It follows that:
In a group, the element at the top end will have the highest chemical reactivity
5. So we have two extremes:
• One at the bottom end and the other at the top end
♦ Both have the highest reactivities
6. The 'variation of reactivity' can be written in 3 steps:
(i) The reactivity decreases (initially) as we go from bottom to top along a group
(ii) It reaches the 'lowest reactivity' at the center of the group
(iii) After passing the center, it again increases and reaches the 'highest reactivity' at the top end
7. We can write a note on oxidation and reduction also:
(i) Oxidation:
• 'Losing electrons' is oxidation
♦ That is., 'elements which lose electrons' are said to be oxidised
• The elements at the bottom easily loses electrons
♦ So we can say: Elements at the bottom are easily oxidised
(ii) Reduction:
• 'Gaining electrons' is reduction
♦ That is., 'elements which gain electrons' are said to be reduced
• The elements at the top easily gains electrons
♦ So we can say: Elements at the top are easily reduced
8. We can write a note on metallic and non-metallic character also:
(i) Metallic character:
• Metals are electron donors
♦ That is., metals lose electrons easily
• Elements at the bottom lose electrons easily
♦ So we can say: Elements at the bottom are more metallic
(ii) Non-metallic character:
• Non-Metals are electron acceptors
♦ That is., non-metals accept electrons easily
• Elements at the top accepts electrons easily
♦ So we can say: Elements at the top are more non-metallic
9. A special note must be made about step (4) above:
• It is related to electron gain enthalpy
• Though we indicate it's 'periodic trend' with an upward arrow, it is not a systematic trend. We must always remember the anomalies that we discussed earlier [step (4) below table 3.5] in section 3.9
Now we will see a solved example based on the above discussion
Solved example 3.23
Considering the elements F, Cl, O and N, the correct order of their chemical reactivity
in terms of oxidizing property is :
(a) F > Cl > O > N (b) F > O > Cl > N (c) Cl > F > O > N (d) O > F > N > Cl
Solution:
1. All the given options are in the decreasing order
• So the first element should have the maximum capacity to oxidize other elements
2. 'Oxidize other elements' means: The other element gets oxidized
• That means, the 'other elements' loses electrons
• That means, the elements given in the question gains electrons
3. We know that, in the periodic table, the right most elements readily gains electrons
• Also, we know that F is the most electronegative element
• So F should be written first
4. O and N falls in the same period as F
• So the three elements can be compared easily
• We get: F > O > N
5. But Cl belongs to a different period. So we have to compare the actual values of electronegativity
• From table 3.6 in section 3.9 , we get:
F = 4.0, O = 3.5, N = 3 and Cl = 3.0
• So the correct order is: F > O > Cl > N
• Thus option (b) is correct
Solved example 3.23
Considering the elements F, Cl, O and N, the correct order of their chemical reactivity
in terms of oxidizing property is :
(a) F > Cl > O > N (b) F > O > Cl > N (c) Cl > F > O > N (d) O > F > N > Cl
Solution:
1. All the given options are in the decreasing order
• So the first element should have the maximum capacity to oxidize other elements
2. 'Oxidize other elements' means: The other element gets oxidized
• That means, the 'other elements' loses electrons
• That means, the elements given in the question gains electrons
3. We know that, in the periodic table, the right most elements readily gains electrons
• Also, we know that F is the most electronegative element
• So F should be written first
4. O and N falls in the same period as F
• So the three elements can be compared easily
• We get: F > O > N
5. But Cl belongs to a different period. So we have to compare the actual values of electronegativity
• From table 3.6 in section 3.9 , we get:
F = 4.0, O = 3.5, N = 3 and Cl = 3.0
• So the correct order is: F > O > Cl > N
• Thus option (b) is correct
So we have completed this chapter on Classification of elements and periodic trends. In the next chapter, we will see Chemical bonding and molecular structure
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