Chapter 4.3 - Modern Periodic table - Solved examples

In the previous section, we saw some solved examples related to the position of elements in the periodic table. In this section we will see a few more solved examples.

Solved example 4.3

Electron configuration of elements P, Q, R and S are given below:

P – 2,2    Q – 2,8,2     R – 2,8,5     S – 2,8

(a) Which among these elements are included in the same period?

(b) Which among these elements are included in the same group?

(c) Which among these elements is a noble gas?

(d) To which group and period does the element R belong?

Solution:

(a) Elements in a same period will have the same number of shells

• In the given problem, P and S have both 2 digits in their electron configuration. So they have both 2 shells. Thus, they both belong to the Period 2

• Q and R have both 3 digits in their electron configuration. So they have both 3 shells. Thus, they both belong to the Period 3

(b) Elements in the same group will have the same number of electrons in their outermost shells

In the given problem, P and Q have both 2 electrons in their outermost shells. So they both belong to Group II

(c) S has an octet configuration in the outermost shell. So it is a noble gas

(d) • R has 5 electrons in it's outermost shell. So it belongs to Group V

• It has 3 digits in it's electron configuration. So it has 3 shells. Thus it belong to the Period 3

Solved example 4.4

An incomplete form of the periodic table is given below.

Write answers to the questions related to the position of elements in it.

(a) Which is the element with the biggest atom in group I?

(b) Which is the element with the lowest ionisation energy in group I?

(c) Which is the element with the smallest atom in period II?

(d) Which are the transition elements?

(e) Which among the two elements L and M, has the smallest electronegativity?

(f) Which among the two elements B and I, has higher metallic nature?

(g) Which are the elements that belong to the halogen family?
(h) Which type of ion will be formed from an atom of 'M'. Cation or anion?

(i) Which is the element that resembles E the most in it's chemical properties?

Solution:

(a) We have discussed the periodic trend related to the 'size of atoms', above in this section. We can write this:

• In a group, from top to bottom, size of atom increases

• In a period, from left to right, size of atom decreases

• In this problem, we have to consider the variation within a group only. The bottom most element will be the biggest. So, in group I, D is the biggest.

(b) We have discussed the periodic trend related to the 'ionisation energy', above in this section. We can write this:

• In a group, from top to bottom, the ionisation energy decreases

• In a period, from left to right, the ionisation energy increases

• In this problem, we have to consider the variation within a group only. The bottom most element will have the lowest ionisation energy. So, in group I, D will be having the lowest ionisation energy.

(c) The variation of size was written in problem (a) above. In this problem, we have to consider the variation within a period only. The element at the extreme right will be the smallest. So, in period 2, M is the smallest.

(d) The transition elements fall between groups II and III. So G and H are transition elements.

(e) We have discussed the periodic trend related to the 'electronegativity', above in this section. We can write this:

• In a group, from top to bottom, the electronegativity decreases

• In a period, from left to right, the electronegativity increases

• In this problem, we have to consider the variation within a period only. This is because, L and M fall within a period. The left most element will have the lower electronegativity. So, among L and M, the element L will be having the lower electronegativity.

(f) We have discussed the periodic trend related to the 'metallic nature'. We can write this:

• In a group, from top to bottom, the metallic nature increases

• In a period, from left to right, the metallic nature decreases

• In this problem, we have to consider the variation within a period only. This is because, B and I fall within a period. The left most element will have the higher metallic nature. So, among B and I, the element B will be having the higher metallic nature.

(g) We have learned that, halogen family is the group VII. So, among all the given elements, M and N are the halogens.
(h) • We have lerned about cations and anions here. Positive ions are Cations and negative ions are anions.
• As we move from left to right in a period, the metallic character decreases.
• That means, the tendency to lose electrons decreases. That means, tendency to gain electron increases
• As M is further to the right in it's period, it will have a tendency to gain electrons. So it will become a negative ion. That is., it will become an anion.

(i) Both B and F are close to E. But F will resemble E more because, F, like E has two electrons in the outermost shell
Solved example 4.5

Give examples for the following:

(a) Two elements that have 1 electron in their outermost shells

(b) Three elements that have 2 electrons in their outermost shells

(c) Three elements that have filled outermost shells

Solution:

(a) All elements in the group I have 1 electron in their outermost shells. We can pick any two from this group. Say lithium (Li) and sodium (Na)

(b) All elements in the group II have 2 electrons in their outermost shells. We can pick any three from this group. Say beryllium (Be), magnesium (Mg) and calcium (Ca)

(c) All elements in the group VIII have filled electrons in their outermost shells. We can pick any three from this group. Say neon (Ne), argon (Ar) and krypton (Kr)

Solved example 4.6

Lithium, sodium and potassium are all metals which react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements?

Solution: All the three elements have one electron in the outermost shell. During chemical reactions, this single electron is readily donated by these metals

Solved example 4.7

In the modern periodic table, which are the metals among the first 10 elements?

Solution:

Lithium (Li) and Beryllium (Be).

Solved example 4.8

By considering their position in the periodic table, which one of the following elements would you expect to have the maximum metallic character?

Ga, Ge, As, Se, Br

Solution:

All the given elements belong to a same period. Among them, Ga is the one which is at the left most position. So It is the one with the maximum metallic character

Solved example 4.9

Which of the following statements is not a correct statement about the trends when going from left to right in any period

(i) The elements become less metallic in nature

(ii) The number of valence electrons increase

(iii) The atoms lose their electrons more easily

Solution:

Statement (iii) is false. As we move from left to right in any period, the metallic character decreases. That is., tendency to lose electrons decreases.

Solved example 4.10

Which element has

(i) Two shells, both of which are completely filled with electrons

(ii) Electron configuration 2,8,2

(iii) A total of 3 shells, with 4 electrons in the valence shell

(iv) A total of 2 shells, with 3 electrons in the valence shell

(v) twice as many electrons in it's second shell as in it's first shell

Solution:

(i) • Given that the element has two shells. So they are K and L

• Given that they are completely filled up. K-shell can hold a maximum of 2×1² = 2×1 = 2 electrons

• L-shell can hold a maximum of 2×2² = 2×4 = 8 electrons

• So the total number of electrons = total number of protons = 2 + 8 = 10

• Thus the atomic number Z = 10, and the element with this atomic number is Neon (Ne)

(ii) • Given that electron configuration is 2,8,2

• So number of electrons = number of protons = 2 + 8 + 2 = 12

• Thus the atomic number Z = 12, and the element with this atomic number is magnesium (Mg)

(iii) • Given that there are a total of 3 shells. So the shells are K,L and M

• Also given that there are 4 electrons in the valence shell. That means K and L are completely filled up. So number of electrons = 2 + 8 + 4 = 14

• Thus the atomic number Z = 14, and the element with this atomic number is silicon (Si)

(iv) • Given that there are a total of 2 shells. So the shells are K and L

• Also given that there are 3 electrons in the valence shell. That means K is completely filled up. So number of electrons = 2 + 3 = 5

• Thus the atomic number Z = 5, and the element with this atomic number is boron (B)

(v) • Here the total number of shells is not given. But that is not necessary to solve the problem

• Given that number of electrons in the second shell is twice that in the first. It is clear that there are some electrons in the second shell

• The second shell will begin to fill only if the first shell is completely filled up. The maximum capacity of the first shell is 2.

• So the number of electrons in the second shell = 2×2 = 4. The second shell is capable to carry more than this 4. So it follows that there are no more shells than K and L.

• So there are no more electrons than the 2 in first shell, and the 4 in the second shell.

• Thus the atomic number Z = 2 + 4 = 6, and the element with this atomic number is carbon (C)

Solved example 4.11

Nitrogen (Z=7) and Phosphorus (Z=15) belong to group V of the periodic table. Write the electron configuration of these elements. Which will be more electronegative? Why?

Solution:

• Electron configuration of nitrogen is 2,5

• Electron configuration of phosphorus is 2,8,5

• Both of them belong to the same group, with nitrogen above phosphorus. We have seen that while moving from top to bottom in a group, electropositive character increases. That is., electronegative character decreases. So, nitrogen, which is above, is more electronegative.

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We have completed the basic discussion on the Modern periodic table. We will learn more details in higher classes. In the next section, we will discuss about Non-metals. 

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Chapter 4.2 - More periodic trends and Solved examples

In the previous section, we saw the periodic trend in the size of atoms. In this section we will see more properties.

Ionisation energy

We have seen reactions in which ions are formed. Consider for example, the formation of NaCl. We have seen the details here. The sodium (Na) atom loses one electron, and becomes Na+ ion. The chlorine (Cl) atom gains one electron, and become Cl- ion.

• Consider the formation of Na⁺ ion. One electron has to be removed from the Na atom, for the formation of Na⁺. Is such a removal easy?

• The negatively charged electrons are attracted towards the central nucleus. This is because of the positively charged protons present in the nucleus.

• So the electron must over come this attractive force, in order to move away from the atom.

• To over come this attractive force, some energy is required.

• This energy is called ionisation energy

• But, for the same element, this energy may be different under different conditions. For example, the presence of some other element can cause an increase or decrease in the required energy. Also, more energy is required for an electron in an inner shell, than that in an outer shell. So the conditions are to be specified. Thus the definition of ionisation energy is as follows:

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The amount of energy required to liberate the most loosely bound electron from the outer most shell of an isolated gaseous atom of an element is called it's ionisation energy.

[Note that several conditions are specified:

• The electron must be the one which is most loosely bound

• It must be situated in the outer most shell

• The atom must be isolated. That is., there must not be any presence of any other atoms

• The atom must be in the gaseous state]

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So let us analyse the variation of ionisation energy at different parts of the periodic table:

I Moving from top to bottom in a group:

1. When we move from top to bottom in a group, the size of the atom increases. We have already seen the reason here.

2. When the size increases, the distance from the nucleus to the outer most electron also increases

3. When the distance increases, the force exerted by the nucleus (on the outer most electron) decreases

4. When this force is low, low energy is sufficient to liberate the electron.

5. So we can write: The ionisation energy decreases as we move from top to bottom in a group

II Moving from left to right in a period

1. When we move from left to right in a period, the size of the atom decreases. We have already seen the reason.

2. We have seen that the decrease in size is due to the increase in attractive force from the central nucleus

3. From (2) we can readily reach a conclusion: As the attractive force increases, more energy is required to liberate the electron.

4. So we can write: The ionisation energy increases as we move from left to right in a period

Electronegativity

We have already learned about electronegativity. It is the ability of an atom to attract the shared pairs of electrons in covalent bonds. We have seen the details here. Now we want to understand it's periodic trend.

• Electronegativity is closely related to the size of atom. Let us write an analysis:

I Moving from top to bottom in a group:

1. When we move from top to bottom in a group, the size of the atom increases. We have already seen the reason here.

2. When the size increases, the distance from the nucleus to the outer most electron also increases

3. When the distance increases, the force exerted by the nucleus (on the outer most electron) decreases

4. From (3), we see that, when size increases, distance also increases, and the atom becomes lesser and lesser capable to attract 'it's own outer most electrons'.

5. Then there is nothing more to tell about the 'external shared electrons'. We can straight away say: The atom is even less capable to attract the shared pairs of electrons.

6. So we can write: The electronegativity decreases as we move from top to bottom in a group

II Moving from left to right in a period

1. When we move from left to right in a period, the size of the atom decreases. We have already seen the reason.

2. We have seen that the decrease in size is due to the increase in attractive force from the central nucleus

3. So there is an increase in attraction on the 'outer most electrons'. This increase will be felt by the 'shared pairs of electrons' also. We can straight away say: The atom is capable to exert significant force of attraction on the shared pairs of electrons

4. So we can write: The electronegativity increases as we move from left to right in a period

Metallic and Non-metallic Nature

• Elements with a metallic nature, generally loses electrons in chemical reactions.

• Metals are called electropositive because, in chemical reactions, they lose electrons to become positive ions. (Metals are good conductors of electric current because, they readily lose electrons, and hence have lots of free electrons)

• Non-metals are called electronegative because, in chemical reactions, they gain electrons to become negative ions.

• The metallic or non-metallic nature of an element is closely related to it's ionisation energy. Why?

The reason is simple:

    ♦ If the ionisation energy is low, electrons will be easily liberated. If the electrons are easily liberated, it shows a metallic nature.

   ♦ If the ionisation energy is high, electrons will not be easily liberated. If the electrons are not easily liberated, it shows a non-metallic nature.

• So, from the periodic trend of ionisation energy, we will be able to write the periodic trend for metallic and non-metallic nature also. Thus we get:

I Moving from top to bottom in a group:

1. When we move from top to bottom in a group, ionisation energy decreases

2. Decrease in ionisation energy indicates increase in metallic nature. So we can write:

3. The metallic nature increases (consequently, non-metallic nature decreases) as we move from top to bottom in a group

II Moving from left to right in a period

1. When we move from left to right in a period, ionisation energy increases

2. Increasein ionisation energy indicates increase in non-metallic nature. So we can write:

3. The non-metallic nature increases (consequently, metallic nature decreases) as we move from left to right in a period

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Based on the results in I and II above, we can answer an interesting question:

■ Which is the element with the most metallic character in the periodic table?

Solution: 1. From II above we have: In any period, the left most element is the most metallic in nature

2. So the element that we are seeking, is some where in the first group

3. From I above we have: In any group, the bottom most element is the most metallic in nature
4. So the bottom most element in the first group (francium) is the one which is most metallic in character

• However, francium is a man made element. Among the natural elements, caesium is the one which is the most metallic in character. Look at the position of caesium. It is just above francium.

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■ In the above discussion, we derived the periodic trend of metallic and non-metallic character based on the 'trend of ionisation energy'
■ We can derive the periodic trend of metallic and non-metallic character based on the 'trend of electronegativity' also. The reader is advised to write such an analysis by him/herself.

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Now we will see some solved examples

Solved example 4.1

Complete the table given below:

Solution:

1. Consider lithium. It’s electron configuration is given as 2,1. So the atomic number Z = 2 + 1 = 3

2. Consider oxygen. Z is given as 8. So the electron configuration is 2,6

• From the electron configuration, we get: No. of electrons in the outer most shell = 6. So oxygen belongs to Group VI

• There are 2 digits in the electron configuration. So No. of shells = 2. Thus, oxygen belongs to the Period 2

See rules 7 and 8

3. Consider argon. Z is given as 18. So the electron configuration is 2,8,8

• From the electron configuration, we get: No. of electrons in the outer most shell = 8. So argon belongs to Group VIII

• There are 3 digits in the electron configuration. So No. of shells = 3. Thus, argon belongs to the Period 3

4. Consider calcium. It’s electron configuration is given as 2,8,8,2. So Z = 2 + 8 + 8 + 2 = 20

• From the electron configuration, we get: No. of electrons in the outer most shell = 2. So calcium belongs to Group II

• There are 4 digits in the electron configuration. So No. of shells = 4. Thus, oxygen belongs to the Period 4

The completed table is given below:

Solved example 4.2

There are 3 shells in the atom of element 'X'. 6 electrons are present in it's outermost shell.

(a) Write the electron configuration of the element

(b) What is it's atomic number Z?

(c) In which period does this element belong?

(d) In which group does this element belong?

(e) Write the name and symbol of this element

(f) To which family of elements does this element belong?

Solution:

(a) Given that the element has 3 shells. So the element has K, L, and M shells

• Given that the M shell has 6 electrons. So the L and M would be filled up to their full capacities.

• That means there are 2 electrons in K, and 8 electrons in L

• So the total number of electrons = 2 + 8 + 6 = 16

The electron configuration is 2,8,6

(b) Total number of electrons = 16 = Total number of protons

So atomic number Z = 16

(c) There are 3 shells for the atom. So element belongs to the Period 3

(d) There are 6 electrons in the outermost shell. So the element belongs to Group VI

(e) The name of the element with Z = 16 is Sulfur. It's symbol is S

(f) The family is Oxygen family

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In the next section, we will see a few more solved examples. 

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Chapter 3.6 - Oxidation and Reduction

In the previous section, we saw how to determine the chemical formula of any compound from the valency of it's component elements. In this section, we will see Oxidation and Reduction.

Oxidation and Reduction

We have seen the reaction between magnesium and chlorine. (Details here). The electron dot diagram is shown again below:

• After the reaction, Mg has become a positive ion. This is due to the loss of electrons. Two electrons are lost. So the charge is 2+. 
    ♦ The process of losing the two electrons can be written as: Mg → Mg²⁺ + 2e^-
• Each Cl has become a negative ion. This is due to the gain of electron. One electrons is gained by each Cl. So the charge is 1- for each.
    ♦ The process of gaining electrons can be written as: 2Cl + 2e^- → 2Cl^-
• In the reaction, Mg donates electrons, and, Cl accepts electrons.
■ Oxidation is the process of donation of electrons
■ Reduction is the process of accepting electrons 

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Is the word 'oxidation' related to 'oxygen' in any way?
To find the answer, we must look into some history. 
• The term 'oxidation' was first used by Lavoisier to mean 'reaction of a substance with oxygen'. 
• Much later, it was realized that, when any element 'X' reacts with oxygen (that is., when 'X' is oxidised), electrons are donated by 'X'
• So later, the term 'oxidation' began to be applied to all elements which donates electrons in a reaction. Regardless of whether it's reaction is with oxygen, or not.

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An analogy to easily remember the 'difference between oxidation and reduction' is as follows:

When electrons are accepted, negative charge increases. 'Increasing negative' can be considered as 'decreasing (reducing) value'. So the element which accepts electrons can be considered to be 'reduced'.
And the opposite is applicable to oxidation:
The element which donates electrons can be considered to be 'oxidised'

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• So in the above reaction, chlorine is reduced. Reduced by whom?

    ♦ By magnesium. Magnesium reduces chlorine by 'forcing chlorine to accept electrons'. So magnesium is the reducing agent

The reverse can also be written:

• Magnesium is oxidised. Oxidised by whom?

    ♦ By chlorine. Chlorine oxidises magnesium by 'forcing magnesium to donate electrons'. So chlorine is the oxidising agent.

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Now let us see one more example. That of Sodium fluoride (NaF):
• NaF is produced by the reaction between sodium (Na) and fluorine (F)
• Na, with atomic number 11, has the electronic configuration 2,8,1  (See here)
• F, with atomic number 9, has the electronic configuration 2,7
• So during the reaction, Na donates one electron, and, F accepts one electron
• Thus we say: Na is oxidised, and, F is reduced
• Na is oxidised by F. Because, F forces Na, to donate one electron. So F is oxidising agent
• F is reduced by Na. Because, Na forces F, to accept one electron. So Na is the reducing agent

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Oxidation number

Every element in a compound will get a particular number. What is this number? How is it determined? What is it's significance? Does this number help us in any way? Let us see the answers:

• This number is called the oxidation number

• Upon seeing the oxidation number of an element in a compound, we will immediately be able to say how many electrons that element has gained (or lost)

• For example, in NaCl, the Na atom has lost one electron. So it is given an oxidation number: +1

(Why 'positive' one? The answer is simple: losing electrons means gaining positive charge)

• Similarly, the Cl atom has gained one electron. So it is given an oxidation number: -1
• Together, they are represented as: Na⁺¹Cl^-1

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So when we see Na⁺¹Cl^-1 , we can write the following details:
• Na is in an oxidation state of +1. That is., it has lost one electron
• Cl is in an oxidation state of -1. That is., it has gained one electron

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Let us consider some more examples:
Consider MgO. We have seen the details here. The diagram is shown again below:

We can write the oxidation state as: Mg⁺²O^-2
• Mg is in an oxidation state of +2. That is., it has lost two electrons
• O is in an oxidation state of -2. That is., it has gained two electrons

Consider Aluminium chloride (AlCl₃)
• Al, with atomic number 13, has the electronic configuration 2,8,3  (See here)
• Cl, with atomic number 17, has the electronic configuration 2,8,7
• So Al needs to lose 3 electrons and Cl needs to gain one electron
• Thus 3 chlorine atoms combine with one aluminium atom. Each chlorine atom accept one of the three electrons donated by the aluminium atom
• One aluminium atom loses 3 electrons. So it's oxidation state is +3
• Each chlorine atom gains one electron so oxidation state of each is -1
• So we can write: Al⁺³Cl₃^-1 

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When we see Al⁺³Cl₃^-1, we can write the following:
• Aluminium is in an oxidation state of +3. That is., it has lost 3 electrons
• Chlorine is in an oxidation state of -1. That is., each chlorine atom has gained one electron

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For covalent compounds, there is no transfer of electrons. Only sharing of electrons takes place. That is., there is no donating or accepting of electrons. In such a case, how do we write the oxidation number?
Consider HCl. We have seen the details here. The diagram is shown again below:

In such cases, it is assumed that, the shared pairs of electrons are completely displaced towards the more electronegative atom. Let us see the application of this assumption in our present case.
• Electronegativity of H = 2.20 (see the chart here)
• Electronegativity of Cl = 3.16 
• Chlorine is more electronegative. So the shared pair of electrons move towards the chlorine atom. 
• That is., two electrons move towards the chlorine atom
• Out of those two, one was already possessed by Cl. So the number of extra electrons is only 1
• Thus the chlorine is given an oxidation number of -1
• What about hydrogen?
• The hydrogen is now left with zero electrons. Because the pair as a whole, has moved away from the hydrogen atom
• Out of the two electrons that has moved away, one belonged to the hydrogen atom. Now it is gone.
• So in effect, hydrogen has lost one electron, and so it is given an oxidation number of +1
• We can write the equation as:  H₂⁰ + Cl₂⁰ → 2H⁺¹Cl^-1 

Let us see another example. Consider carbon dioxide CO₂
Two oxygen atoms forms a covalent bond with carbon. We saw the details here. It is shown again below:

• Electronegativity of C = 2.55 
• Electronegativity of O = 3.44 
• Oxygen is more electronegative. So the shared pair of electrons move towards the oxygen atom. 
• That is., in each bond, four electrons move towards the oxygen atom
• Out of those four, two were already possessed by O. So the number of extra electrons is only 2
• Thus each oxygen is given an oxidation number of -2
• What about carbon?
• The carbon is now left with zero electrons. Because both the pairs have moved away from the carbon atom
• Out of the eight electrons that has moved away, four belonged to the carbon atom. Now they are gone.
• So in effect, carbon has lost four electrons, and so, it is given an oxidation number of +4
• We can write the equation as:  C⁰ + 2O⁰ → C⁺⁴O₂^-2 

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So we see that for both ionic and covalent compounds, the component elements will have oxidation numbers. But it is very important to remember an exception to this rule:

• Consider the molecules of elements. In such molecules, the component atoms are all of the same element.

• All the component atoms will be having the same power to attract the shared electrons. We have seen examples of such bonding here. Some are shown again below:

• So we see that, when the component atoms are the same, the shared electron pairs will be attracted equally, and so they will remain exactly midway between the atoms

• No atom will get any extra electrons, and so the oxidation numbers of all component atoms will be zero

• We write them as: F₂⁰, Cl₂⁰ etc.,

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Telling 'whether it is an oxidation or a reduction' by using oxidation number

So we have seen the oxidation number of elements in both ionic and covalent compounds. We will now see a practical application of this number:
Consider the formation of HCl that we saw above. If we examine the equation H₂⁰ + Cl₂⁰ → 2H⁺¹Cl^-1 , we will see that:
• The oxidation number of H has increased from 0 to +1
• The oxidation number of Cl has decreased from 0 to -1
■ The element, whose oxidation number increases, is oxidised. In other words, the element, whose oxidation number increases, has undergone oxidation.
    ♦ In our example, hydrogen is oxidised. In other words, hydrogen has undergone oxidation
■ The element, whose oxidation number decreases, is reduced. In other words, the element, whose oxidation number decreases, has undergone reduction.
    ♦ In our example, chlorine is reduced. In other words, chlorine has undergone reduction.

Another example: Consider the formation of carbon dioxide. If we examine the equation C⁰ + 2O⁰ → C⁺⁴O₂^-2 , we will see that:
• The oxidation number of C has increased from 0 to +4. So C has undergone oxidation
• The oxidation number of O has decreased from 0 to -2. So O has undergone reduction

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We see that, in a reaction, one element is oxidised and the other element is reduced. That is., oxidation and reduction takes place simultaneously. So the overall process is known as a redox process.

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We see that, in a reaction, the element which increases in the oxidation number is the one which gets oxidised.
• That element  gets oxidised by whom?
• It is oxidised by the other element. We know that, this 'other element' which causes the oxidation is called the 'oxidising agent'.
• Now, what is the state of the 'oxidation number' of this 'other element', which is the 'oxidising agent'?
• The 'oxidation number' of this 'other element' decreases.
■ So we can write: The element whose oxidation number decreases, is the oxidising agent.
■ Similarly, The element whose oxidation number increases is the reducing agent

This can be explained using a general example. Consider the reaction between two elements 'P' and 'Q' (symbols are not real):
• P^x1 + Q^y1 → P^x2Q^y2   [(x2 > x1) and (y2 <y1)]
• We see that, after the reaction, P has an increased oxidation number.
• So P has undergone oxidation
• Obviuosly, this oxidation was caused by Q, and so, we call Q, the oxidising agent
• Oxidation number of Q has decreased
■ So we can write: The element whose oxidation number decreases, is the oxidising agent.
And the reverse can also be worked out:
• We see that, after the reaction, Q has a decreased oxidation number.
• So Q has undergone reduction
• Obviuosly, this reduction was caused by P, and so, we call P, the reducing agent
• Oxidation number of P has increased
• So we can write: The element whose oxidation number increases, is the reducing agent. 

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We will now see an example:

• The equation for the reaction between hydrochloric acid (HCl) and zinc (Zn), to form zinc chloride and hydrogen is given below:

• Zn + 2HCl → ZnCl₂ + H₂

• Note that, it is a balanced equation. The modified form of this equation, which shows the oxidation number of each element is given below:

• Zn⁰ + 2H⁺¹Cl^-1→ Zn⁺²Cl₂^-1 + H₂⁰

• From this modified equation, we can write the following details:

■ Oxidation number of Zn increases from 0 to +2. So, Zn undergoes oxidation. And also, it is the reducing agent

■ Oxidation number of H decreases from +1 to 0. So H undergoes reduction. And also, it is the oxidising agent

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• So now we are able to find which element gets oxidised, and which element gets reduced

• Also, we can find which is the oxidising agent, and which is the reducing agent
• We are able to find the above because, the 'oxidation number' of one element increases, and that of the other element decreases.
• But there is some thing more than just 'increase and decrease':
■ The sum of oxidation numbers in a compound will be always zero. 
Let us see an example. Consider Zn⁺²Cl₂^-1:
• Total oxidation number of Zn:
    ♦ There is only one Zn atom, and it has an oxidation number of +2. Thus:
    ♦ Total oxidation number of Zn = number of Zn atoms × oxidation number of each = 1 × +2 = +2 
• Total oxidation number of Cl:
    ♦ There are two Cl atoms, each with an oxidation number -1. Thus:
    ♦ Total oxidation number of Cl = number of Cl atoms × oxidation number of each = 2 × -1 = -2
■ Total sum of the oxidation numbers of all the atoms in the molecule = 2 + -2 = 0 

Another example: Consider Al⁺³Cl₃^-1:
• Total oxidation number of Al:
    ♦ There is only one Al atom, and it has an oxidation number of +3. Thus:
    ♦ Total oxidation number of Al = number of Al atoms × oxidation number of each = 1 × +3 = +3 
• Total oxidation number of Cl:
    ♦ There are three Cl atoms, each with an oxidation number -1. Thus:
    ♦ Total oxidation number of Cl = number of Cl atoms × oxidation number of each = 3 × -1 = -3
■ Total sum of the oxidation numbers of all the atoms in the molecule = 3 + -3 = 0

Another example: Consider Mg⁺²O^-2:
• Total oxidation number of Mg:
    ♦ There is only one Mg atom, and it has an oxidation number of +2. Thus:
    ♦ Total oxidation number of Mg = number of Mg atoms × oxidation number of each = 1 × +2 = +2 
• Total oxidation number of O:
    ♦ There is only one O atom, and it has an oxidation number of -2. Thus:
    ♦ Total oxidation number of O = number of O atoms × oxidation number of each = 1 × -2 = -2 
■ Total sum of the oxidation numbers of all the atoms in the molecule = 2 + -2 = 0

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Thus we find that the sum of oxidation numbers in any molecule is equal to zero. In the next section, we will see a practical application of this property 

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Chapter 3.3 - Electronegativity and Polar compounds

In the previous section, we completed the discussion on covalent bond.  In this section, we will learn about Electronegativity.

We have seen that in covalent bonding, pairs of electrons are shared by atoms. The atoms apply an attractive force towards the shared electrons. That is:
■ Each atom in the molecule will try to pull the shared pairs of electrons towards it
• If the molecule has all the atoms same (that is., molecule of an element), the attraction will be equal. 
• Let us see an example: 
    ♦ Consider F₂ molecule. Both the atoms in that molecule are same: that of fluorine
    ♦ So the 'magnitude of the attractive force' on the shared electrons will be equal
    ♦ So the shared electrons will remain exactly midway between the two fluorine atoms
If the molecules has different atoms (that is., molecule of a compound), the attraction will not be equal. 
• Let us consider an example:
    ♦ Consider HCl molecule. It has one atom of hydrogen, and one atom of chlorine. 
    ♦ Out of these two, the chlorine atom has a greater capacity to exert attractive force on the shared electrons
    ♦ So the shared electrons will not remain exactly at midmay between the hydrogen and chlorine atoms. 
    ♦ They will move towards the chlorine atom

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So we want to know the following:

• Which atoms have greater capacity to attract electrons?
• Which atoms have lesser capacity to attract electrons?
■ It is essential to have such a knowledge because, only then will we be able to tell the position of the shared electrons. 
■ The shared electrons will be located near that atom which has the greater attractive capacity, and we want to be able to tell which.

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We can find the attractive capacity of the elements by doing various experiments. Consider the following analogy: We have a large number of magnets in hand. The capacity of each is different. We can find the capacity of each by doing various experiments on them. 

In the same way, the capacity of each of the elements can also be determined. But we do not have to do any experiments. The required experiments have been already done by scientists, and the capacity of each element is now known. The values given by American scientist Linus Pauling is widely accepted. We will also follow it. It is known as Pauling's electronegativity scale. It is in the form of a chart as shown here. A few readings from the chart are given below:

• Electronegativity of fluorine = 3.98
• Electronegativity of manganese = 1.55
• Electronegativity of hydrogen = 2.20  

• Among the above three elements, fluorine has the highest electronegativity. In fact, fluorine has the highest electronegativity among all the elements in the chart
• Manganese has the least electronegativity among the above three
• From the values, we can infer that, in a compound formed between hydrogen and fluorine, the shared electrons will be located nearer to fluorine. This is because, it's electronegativity 3.98 is greater than the electronegativity 2.20 of hydrogen

The application of the chart will become more clear from the following table: 

We will now see each of the compound in the above table in detail:
■ Consider carbon monoxide (CO):
• It's constituent elements are: 
    ♦ Carbon with electronegativity 2.55
    ♦ Oxygen with electronegativity 3.44
• CO is a covalent compound. The shared pairs of electrons will be located nearer to Oxygen
• Difference in electronegativity values = 3.44 - 2.55 = 0.89
■ Consider sodium chloride (NaCl):
• It's constituent elements are: 
    ♦ Sodium with electronegativity 0.93
    ♦ Chlorine with electronegativity 3.16
• NaCl is an ionic compound
• Difference in electronegativity values = 3.16 - 0.93 = 2.23
■ Consider methane (CH₄):
• It's constituent elements are: 
    ♦ Carbon with electronegativity 2.55
    ♦ Hydrogen with electronegativity 2.2
• CH₄ is a covalent compound. The shared pairs of electrons will be located nearer to carbon
• Difference in electronegativity values = 2.55 - 2.2 = 0.35
■ Consider magnesium chloride (MgCl₂):
• It's constituent elements are: 
    ♦ Magnesium with electronegativity 1.31
    ♦ Chlorine with electronegativity 3.16
• MgCl₂ is an ionic compound
• Difference in electronegativity values = 3.16 - 1.31 = 1.85
■ Consider sodium oxide (Na₂O):
• It's constituent elements are: 
    ♦ Sodium with electronegativity 0.93
    ♦ Oxygen with electronegativity 3.44
• Na₂O is an ionic compound
• Difference in electronegativity values = 3.44 - 0.93 = 2.51 

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In the above table, some compounds are ionic, while others are covalent. 
■ If the difference in elecronegativity values of elements in a compound is 1.7, or greater than 1.7, the compound generally shows ionic character. 
■ If the difference is less than 1.7, the compound generally shows covalent character 

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Polar nature

We have seen that, in covalent compounds, the shared pairs of electrons will be located nearer to the atom which has greater electronegativity. Consider the case of Hydrogen chloride molecule (HCl).

• It's constituent elements are: 
    ♦ Hydrogen with electronegativity 2.20
    ♦ Chlorine with electronegativity 3.16

• HCl is a covalent compound. The shared pairs of electrons will be located nearer to the nucleus of chlorine

■ As a result, a small negative charge will develop in the chlorine atom

• This small negative charge is denoted as δ^- (read as Delta negative)

■ A small positive charge will develop in the hydrogen atom

• This small positive charge is denoted as δ⁺ (read as Delta positive)

■ The whole HCl molecule can be represented as:

Such compounds having partial electrical charge separation within the molecule are called polar compounds. HF, HBr, H₂O are examples of polar compounds.

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We will now see a solved example.
Solved example 3.5
Electronegativity values of some elements are given below:
Ca = 1.0, O = 3.5, C = 2.5, S = 2.58, H = 2.2, F = 3.98
Determine whether the following compounds are ionic or covalent:
(i) Sulphur dioxide (SO₂), (ii) Water (H₂O), (iii) Calcium fluoride (CaF₂), (iv) Carbon dioxide (CO₂)
Solution:
(i) Consider SO₂:
• It's constituent elements are: 
    ♦ Sulphur with electronegativity 2.58
    ♦ Oxygen with electronegativity 3.5
• Difference in electronegativity values = 3.5 - 2.58 = 0.92
• Difference is less than 1.7. So SO₂ is a covalent compound
(ii) Consider H₂O:
• It's constituent elements are: 
    ♦ Hydrogen with electronegativity 2.2
    ♦ Oxygen with electronegativity 3.5
• Difference in electronegativity values = 3.5 - 2.2 = 1.3
• Difference is less than 1.7. So H₂O is a covalent compound
(iii) Consider CaF₂:
• It's constituent elements are: 
    ♦ Calcium with electronegativity 1
    ♦ Fluorine with electronegativity 3.98
• Difference in electronegativity values = 3.98 - 1 = 2.98
• Difference is greater than 1.7. So CaF₂ is an ionic compound
(iv) Consider CO₂:
• It's constituent elements are: 
    ♦ Carbon with electronegativity 2.5
    ♦ Oxygen with electronegativity 3.5
• Difference in electronegativity values = 3.5 - 2.5 = 1.0
• Difference is less than 1.7. So CO₂ is a covalent compound.

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In the next section, we will learn about Valency.

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