Chapter 8.4 - Types Of Redox Reactions

In the previous section we saw how oxidation number can be used to determine whether a reaction is a redox reaction or not. In this section, we will see the different types of redox reactions.

There are four types of redox reactions:
A. Combination reactions
B. Decomposition reactions
C. Displacement reactions
D. Disproportionation reactions

A. Combination reactions
This can be explained in 5 steps:
1. The combination reaction is the reaction in which two or more substances react to form a single new substance.
    ♦ They are also known as synthesis reactions.
2. The general form of a combination reaction is:
A + B → C
3. In order that, a combination reaction is a redox reaction, at least one of the reactants must be in the elemental form. That means, both A and B cannot be compounds.
4. All combustion reactions
    ♦ are combination reactions
    ♦ and at the same time
    ♦ redox reactions also.
• Recall that, in combustion, a substance reacts with elemental oxygen.
    ♦ An example is: $\mathbf\small{\rm{\overset{0}{C}\,(s)+\overset{0}{O}_2\,(g)\rightarrow \overset{+4}{C}\,\overset{-2}{O}_2\,(g)}}$
    ♦ Another example is: $\mathbf\small{\rm{2\,\overset{+2}{C}\,\overset{-2}{O}\,(s)+\overset{0}{O}_2\,(g)\rightarrow 2\,\overset{+4}{C}\,\overset{-2}{O}_2\,(g)}}$
• Here, O gets reduced. It is the oxidant.
5. Combination reactions can be redox reactions, even if elemental oxygen is not one of the reactants.
    ♦ An example is: $\mathbf\small{\rm{3\,\overset{0}{Mg}\,(s)+\overset{0}{N}_2\,(g)\rightarrow \overset{+2}{Mg}_3\,\overset{-3}{N}_2\,(g)}}$ 
• Here, N gets reduced. It is the oxidant.

B. Decomposition reactions
This can be explained in 5 steps:
1. A decomposition reaction is a reaction in which a compound breaks down into two or more simpler substances.
2. The general form of a combination reaction is:
C → A + B
• For this decomposition to take place, energy input (in the form of heat, light or electricity) is required.     
3. In order that, a decomposition reaction is a redox reaction, at least one of the products must be in the elemental form. That means, both A and B cannot be compounds.
4. Let us see some examples:
    ♦ $\mathbf\small{\rm{2\,\overset{+1}{H}_2\,\overset{-2}{O}\,(l)\rightarrow 2\,\overset{0}{H}_2\,(g)+\overset{0}{O}_2\,(g)}}$
    ♦ $\mathbf\small{\rm{2\,\overset{+1}{Na}\,\overset{-1}{H}\,(l)\rightarrow 2\,\overset{0}{Na}\,(g)+\overset{0}{H}_2\,(g)}}$
    ♦ $\mathbf\small{\rm{2\,\overset{+1}{K}\,\overset{+5}{Cl}\,\overset{-2}{O}_3\,(s)\rightarrow 2\,\overset{+1}{K}\,\overset{-1}{Cl}\,(s)+3\,\overset{0}{O}_2\,(g)}}$
5. Let us see a decomposition reaction in which both the products are compounds:
    ♦ $\mathbf\small{\rm{\overset{+2}{Ca}\,\overset{+4}{C}\,\overset{-2}{O}_3\,(s)\rightarrow \overset{+2}{Ca}\,\overset{-2}{O}\,(s)+\overset{+4}{C}\,\overset{-2}{O}_2\,(g)}}$
• This is not a redox reaction because, oxidation numbers remain the same.

C. Displacement reactions
This can be explained in 5 steps:
• A displacement reaction is a reaction in which one element replaces a similar element in a compound.
• The general form of a displacement reaction is:
X + YZ → XZ + Y
   ♦ Here YZ is a compound. The element Y in that compound is replaced by X.
• Note that, only similar elements can do the replacement. That is.,
   ♦ If Y is a metal, X must also be a metal.
   ♦ If Y is a non metal, X must also be a non metal.
• There are three types of displacement reactions:
I. Metal displacement reaction
II. Hydrogen displacement reaction
III. Non metal displacement reaction 

I. Metal displacement reaction
• This can be explained in 4 steps:
1. Consider the general form: X + YZ → XZ + Y
   ♦ In a metal displacement reaction, both X and Y will be metals.
2. We have seen some examples in section 8.1
Example 1:
(i) Consider fig.8.1 in section 8.1
• When metallic zinc is placed in the CuSO₄ solution, Zn replaces Cu.
• We can write the reaction as:
Zn + CuSO₄  → ZnSO₄ + Cu
(ii) Comparing with the general form in (1), we get:
   ♦ X is Zn
   ♦ YZ is CuSO₄
   ♦ XZ is ZnSO₄
   ♦ Y is Cu
• So this reaction is a displacement reaction.
(iii) It is also a redox reaction. This will be clear when we write the oxidation numbers in the balanced equation:
$\mathbf\small{\rm{\overset{0}{Zn}\,(s)+\overset{+2}{Cu}\,\overset{6}{S}\,\overset{-2}{O}_4\,(aq)\rightarrow\overset{+2}{Zn}\,\overset{6}{S}\,\overset{-2}{O}_4\,(aq)+\overset{0}{Cu}\,(s)}}$
• We see that:
   ♦ Cu is reduced.
   ♦ Zn is oxidized.

Example 2:
(i) Consider fig.8.2 in section 8.1
• When metallic copper is placed in the AgNO₃ solution, Cu replaces Ag.
• We can write the reaction as:
Cu + 2AgNO₃  → Cu(NO₃)₂ + 2Ag
(ii) Comparing with the general form in (1), we get:
   ♦ X is Cu
   ♦ YZ is AgNO₃
   ♦ XZ is Cu(NO₃)₂
   ♦ Y is Ag
• So this reaction is a displacement reaction.
(iii) It is also a redox reaction. This will be clear when we write the oxidation numbers in the balanced equation:
$\mathbf\small{\rm{\overset{0}{Cu}\,(s)+2\,\overset{+1}{Ag}\,\overset{5}{N}\,\overset{-2}{O}_3\,(aq)\rightarrow \overset{+2}{Cu}\,(\overset{5}{N}\,\overset{-2}{O}_3)_2\,(aq)+2\,\overset{0}{Ag}\,(s)}}$
• We see that:
   ♦ Ag is reduced.
   ♦ Cu is oxidized.

3. A few more examples are given below:
Example 3:
(i) 5Ca (s) + V₂O₅ (s)  → 5CaO (s) + 2V (s)
• Comparing with the general form in (1), we get:
   ♦ X is Ca
   ♦ YZ is V₂O₅
   ♦ XZ is CaO
   ♦ Y is V
• So this reaction is a displacement reaction.
(ii) It is also a redox reaction. This will be clear when we write the oxidation numbers in the balanced equation:
$\mathbf\small{\rm{5\,\overset{0}{Ca}\,(s)+\overset{+5}{V}_2\,\overset{-2}{O}_5\,(s)\rightarrow 5\,\overset{+2}{Ca}\,\overset{-2}{O}\,(s)+2\,\overset{0}{V}\,(s)}}$
• We see that:
   ♦ V is reduced.
   ♦ Ca is oxidized.

Example 4:
(i) 2Mg (s) + TiCl₄ (l)  → 2MgCl₂ (s) + Ti (s)
• Comparing with the general form in (1), we get:
   ♦ X is Mg
   ♦ YZ is TiCl₄
   ♦ XZ is MgCl₂
   ♦ Y is Ti
• So this reaction is a displacement reaction.
(ii) It is also a redox reaction. This will be clear when we write the oxidation numbers in the balanced equation:
$\mathbf\small{\rm{2\,\overset{0}{Mg}\,(s)+\overset{+4}{Ti}\,\overset{-1}{Cl}_4\,(l)\rightarrow 2\,\overset{+2}{Mg}\,\overset{-1}{Cl}_2\,(s)+\overset{0}{Ti}\,(s)}}$
• We see that:
   ♦ Ti is reduced.
   ♦ Mg is oxidized.

Example 5:
(i) 2Al (s) + Cr₂O₃ (s)  → Al₂O₃ (s) + 2Cr (s)
• Comparing with the general form in (1), we get:
   ♦ X is Al
   ♦ YZ is Cr₂O₃
   ♦ XZ is Al₂O₃
   ♦ Y is Cr
• So this reaction is a displacement reaction.
(ii) It is also a redox reaction. This will be clear when we write the oxidation numbers in the balanced equation:
$\mathbf\small{\rm{2\,\overset{0}{Al}\,(s)+\overset{+3}{Cr}_2\,\overset{-2}{O}_3\,(s)\rightarrow \overset{+3}{Al}_2\,\overset{-2}{O}_3\,(s)+2\,\overset{0}{Cr}\,(s)}}$
• We see that:
   ♦ Cr is reduced.
   ♦ Al is oxidized.
4. We see that, metals replace other metals in the compounds. This property is very useful in metallurgy. It can be explained in 3 steps:
(i) Consider the general form that we wrote in (1): X + YZ → XZ + Y
   ♦ YZ is an ore of the metal Y.
         ✰ Y is the useful metal.
         ✰ Z is the impurity which is chemically combined with Y.
• If we can make YZ to react with X, we will get Y in free form
(ii) Recall the series that we wrote in section 8.1: Zn > Cu > Ag
• We saw that, any metal in the series can displace any other metal on it’s right side in the series.
(iii) Scientists have added many more metals to the above series.
• So in metallurgy, to get the metal Y in free form, we select X in such a way that, 'X lies to the left of Y' in the series.

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In the next section, we will see hydrogen displacement reactions.


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Chapter 3.13 - Comparison of Reactivities using Oxides

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:

Fig.3.21

• 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

---

• So we have completed a discussion on ‘periodic trends (along periods) of chemical reactivity’

• 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

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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

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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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Chapter 3.12 - Periodic Trends in Chemical Reactivity

In the previous section 3.11, we completed a discussion on the periodic trends in valence. In this section, we will see Anomalous properties of 2nd period elements. We will also see the periodic trends in chemical reactivity

1. Consider the table 3.8 below:

Table 3.8

• It shows the elements of two periods: Period 2 and Period 3
• We see only the ‘first two elements’ of groups 1, 2, 13, 14, 15, 16, 17 and 18
• Note that, these groups constitute the s and p blocks of the periodic table
2. Take out the first element from each of those groups (except group 18):
    ♦ Li from group 1
    ♦ Be from group 2
    ♦ B from group 13
    ♦ C from group 14
    ♦ N from group 15
    ♦ O from group 16
    ♦ F from group 17
3. Let us analyze each element from the above set:
• Consider the first entry of the above set: Li of group 1
    ♦ We have learnt that all elements of a group will have similar properties
    ♦ But Li differs in many respects from the other members of the group 1    
• Consider the next entry of the above set: Be of group 2
    ♦ We have learnt that all elements of a group will have similar properties
    ♦ But Be differs in many respects from the other members of the group 2
• Consider the next entry of the above set: B of group 13
    ♦ We have learnt that all elements of a group will have similar properties
    ♦ But B differs in many respects from the other members of the group 13
- - - -
- - - -
• Consider the last entry of the above set: F of group 17
    ♦ We have learnt that all elements of a group will have similar properties
    ♦ But B differs in many respects from the other members of the group 13
4. It is clear that, all first elements differ from their group members
■ Let us see some examples:
• Lithium mostly form covalent compounds
    ♦ But all other members of the group 1 forms ionic compounds
• Beryllium also mostly form covalent compounds
    ♦ But all other members of the group 2 forms ionic compounds
5. Here scientist noticed 3 interesting points:
(i) Behavior of Li is more similar to the second element of the following group
• Let us elaborate on this:
• Li belongs to group 1
    ♦ So the ‘following group’ is group 2 
    ♦ The ‘second element’ of the group 2 is Mg
• So Li is more similar to Mg
• This is indicated by the yellow arrow in the table below:

Table 3.9

(ii) Behavior of Be is more similar to the second element of the following group
• Let us elaborate on this:
• Be belongs to group 2
    ♦ So the ‘following group’ is group 3 
    ♦ The ‘second element’ of the group 3 is Al
• So Be is more similar to Al
• This is indicated by the pink arrow in the table above
(iii) In the same way, B is more similar to Si
• Let us elaborate on this:
• B belongs to group 13
    ♦ So the ‘following group’ is group 14 
    ♦ The ‘second element’ of the group 14 is Si
• So B is more similar to Si
• This is indicated by the blue arrow in the table above
6. The diagonal relationship:
• We see that, Li and Mg are diagonal elements
• Also, Be and Al are diagonal elements
• Again, B and Si are diagonal elements
■ So this type of similarity is commonly referred to as diagonal relationship in the periodic properties

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• Let us see the reason for this type of a similarity. We will write it in steps:

1. Imagine that, we move along the period 2
• Let us move from Li to Be
2. We know that, the size of the atom decreases
• So Be will be smaller than Li
3. Now imagine that, from Be, we move downwards along the group 2
• Let us move from Be to Mg
4. We know that, the size of the atom increases
• Mg will be larger than Be
5. So we have two effects:
(i) ‘Decreasing size’ when moving horizontally towards the right from Li to Be
(ii) ‘Increasing size’ when moving downwards from Be to Mg
6. These two effects are opposite to each other and so cancel out
• The result of this ‘cancelling effect’ can be written in 3 steps:
(i) Imagine that we move in a ‘L-shaped’ path from Li to Mg (through Be)
(ii) We will find that the size decreases and then ‘increases back’ to nearly the original size
(iii) That means, the atomic size of Li and Mg are nearly the same
• We find that, it is indeed true:
    ♦ Metallic radii of Li, Be and Mg are 152, 111 and 160 pm respectively
    ♦ Note that, 160 is close to 152
7. Consider the next set: Be and Al
• Let us move in a ‘L-shaped’ path
    ♦ We start from Be and reach Al (through B)
    ♦ When B is reached, the size becomes small
    ♦ But when Al is reached, the size ‘increases back’ to nearly the size of Be
8. A similar analysis can be written in the case of the next set 'B and Si' also:
• Let us move in a ‘L-shaped’ path
    ♦ We start from B and reach Si (through C)
    ♦ When C is reached, the size becomes small
    ♦ But when Si is reached, the size ‘increases back’ to nearly the size of B
9. We will see more about 'diagonal relationship' when we discuss in detail about s and p block elements in later chapters

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Chemical reactivity

So far in this chapter, we have seen the periodic trends in the following items:

A. Physical properties:
  (i) atomic radii
 (ii) Ionic radii
(iii) Ionization enthalpy
(iv) Electronegativity
B. Chemical properties:
 (i) Valence
(ii) Oxidation state
■ By now, we know that, the periodic trends is a result of this:
Systematic arrangement of elements on the basis of electronic configuration
• We have one more item to learn:
The chemical reactivity
• The four physical properties and the two chemical properties that we have seen, can be used to explain chemical reactivity
• We will write it in steps:
1. Recall the 4 horizontal arrows that we saw in fig.3.14 in an earlier section 3.9
• For convenience, that fig.3.14 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 period, the ionization enthalpy is least for the element at the extreme left end
• That means:
In a period, the element at the extreme left end can lose it’s electron in the most easiest way
• It follows that:
In a period, the element at the extreme left end can become a cation in the most easiest way
• It follows that:
In a period, the element at the extreme left end will have the highest chemical reactivity
    ♦ Chemical Reactivity is the ‘readiness to  take part in chemical reactions’
4. Consider the arrow for electron gain enthalpy
■ It indicates that:
In a period, the electron gain enthalpy is 'highest negative' for the element at the extreme right end
• That means:
In a period, the element at the extreme right end can gain an electron in the most easiest way
• It follows that:
In a period, the element at the extreme right end can become an anion in the most easiest way
• It follows that:
In a period, the element at the extreme right end will have the highest chemical reactivity
5. So we have two extremes:
• One at the left end and the other at the right 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 left to right across a period
 (ii) It reaches the 'lowest reactivity' at the center of the periodic table
(iii) After passing the center, it again increases and reaches the 'highest reactivity' at the extreme right 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 left easily loses electrons
    ♦ So we can say: Elements at the left are easily oxidised
(ii) Reduction:
• 'Gaining electrons' is reduction
    ♦ That is., 'elements which gain electrons' are said to be reduced
• The elements at the right easily gains electrons
    ♦ So we can say: Elements at the right 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 left lose electrons easily
    ♦ So we can say: Elements at the left are more metallic
(ii) Non-metallic character:
• Non-Metals are electron acceptors
    ♦ That is., non-metals accept electrons easily 
• Elements at the right accepts electrons easily
    ♦ So we can say: Elements at the right are more non-metallic

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So we have completed a discussion on ‘periodic trends (along periods) in reactivity’. In the next section, we will see some examples to demonstrate those trends. We will also see ‘periodic trends (along groups) in reactivity’

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