Chapter 3.7 - Calculation of Oxidation number

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

Method of determining the oxidation number

Consider the following table:

It gives the oxidation number of some common elements. Look at the values carefully. Do the values look familiar? Indeed they do. The values are same as the valencies of the corresponding elements. But we must put the appropriate sign ('+' or '-') before the valency values. Let us check for a few elements in the table:
We have seen the details about 'valency calculation' in a previous section here.
■ Consider sodium (Na): It has atomic number = 11. It's electronic configuration is 2,8,1
• The outer most shell has 1 electron. So valency = 1
• Now check the oxidation number corresponding to Na in the table
• It is also 1. With a '+' sign in front. Why this '+' sign?
• Because sodium always donates one electron, and so, it's oxidation number is always positive.
■ Consider aluminium (Al): It has atomic number = 13. It's electronic configuration is 2,8,3
• The outer most shell has 3 electron. So valency = 3
• Now check the oxidation number corresponding to Al in the table
• It is also 3. With a '+' sign in front. Why this '+' sign?
• Because aluminium always donates three electrons, and so, it's oxidation number is always positive.
■ Consider oxygen (O): It has atomic number = 8. It's electronic configuration is 2,6
• The outer most shell has 6 electron. So valency = 8-6 =2
• Now check the oxidation number corresponding to O in the table
• It is also 2. With a '-' sign in front. Why this '-' sign?
• Because oxygen always accepts two electrons, and so, it's oxidation number is always negative.

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Once we become familiar with the values in the table, we can think about a more advanced case:
■ What about those elements which are not included in the table? What are their oxidation numbers?
The answer is that, we can find the oxidation number of any component element in a molecule. For that, we have to use two things:
• The oxidation numbers of the common elements given in the table above    
• The property that the sum of oxidation numbers in any molecule is equal to zero

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With the knowledge of the above two, the procedure is simple, involving basic algebra. Let us see an example:
Find the oxidation number of sulphur in H₂SO₄
Solution:
• From the table, oxidation number of H = +1 
• Oxidation number of O = -2
• Let the oxidation number of S = x 
• Sum of oxidation numbers in any molecule is equal to zero. So we can write:
• [2 × (+1)] + [1 × x] + [4 × (-2)] = 0 ⇒ 2 + x - 8 = 0 ⇒ x = 8 - 2 ⇒ x = +6
• So the oxidation number of sulphur in H₂SO₄ is +6

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Now we will see some solved examples:
Solved example 3.7
Find the oxidation number of Manganese (Mn) in (i) KMnO₄ , (ii) MnO₂ , (iii) Mn₂O₃ , (iv) Mn₂O₇
Solution: (i) KMnO₄ :
• From the table, oxidation number of K = +1 
• Oxidation number of O = -2
• Let the oxidation number of Mn = x 
• Sum of oxidation numbers in any molecule is equal to zero. So we can write:
• [1 × (+1)] + [1 × x] + [4 × (-2)] = 0 ⇒ 1 + x - 8 = 0 ⇒ x = 8 - 1 ⇒ x = +7
• So the oxidation number of Mn in KMnO₄ is +7
(ii) MnO₂ 
• From the table, oxidation number of O = -2
• Let the oxidation number of Mn = x 
• Sum of oxidation numbers in any molecule is equal to zero. So we can write:
• [1 × x] + [2 × (-2)] = 0 ⇒ x - 4 = 0 ⇒ x = 4

• So the oxidation number of Mn in MnO₂ is +4
(iii) Mn₂O₃
• From the table, oxidation number of O = -2
• Let the oxidation number of Mn = x 
• Sum of oxidation numbers in any molecule is equal to zero. So we can write:
• [2 × x] + [3 × (-2)] = 0 ⇒ 2x - 6 = 0 ⇒ 2x = 6 ⇒ x = 3
• So the oxidation number of Mn in Mn₂O₃ is +3
(iii3) Mn₂O₇
• From the table, oxidation number of O = -2
• Let the oxidation number of Mn = x 
• Sum of oxidation numbers in any molecule is equal to zero. So we can write:
• [2 × x] + [7 × (-2)] = 0 ⇒ 2x - 14 = 0 ⇒ 2x = 14 ⇒ x = 7
• So the oxidation number of Mn in Mn₂O₃ is +7

Solved example 3.8
Oxidation number of oxygen is -2. Find the oxidation number of other atoms in the following compounds:
(i) H₂O , (ii) H₂CO₃ , (iii) HNO₃ , (iv) H₃PO₄
Solution:
(i) H₂O
• Given oxidation number of O = -2 
• Let the oxidation number of H = x 
• Sum of oxidation numbers in any molecule is equal to zero. So we can write:
• [2 × x] + [1 × (-2)] = 0 ⇒ 2x - 2 = 0 ⇒ 2x = 2 ⇒ x = 1
• So the oxidation number of H in H₂O is +1
When we solve this problem, we get the oxidation number of H as +1. We can use this value for solving the other problems
(ii) H₂CO₃
• Given oxidation number of O = -2 
• Oxidation number of H = +1
• Let the oxidation number of C = x 
• Sum of oxidation numbers in any molecule is equal to zero. So we can write:
• [2 × (+1)] + [1 × x] + [3 × (-2)] = 0 ⇒ 2 + x - 6 = 0 ⇒ x = 4
• So the oxidation number of C in H₂CO₃ is +4
(iii) HNO₃
• Given oxidation number of O = -2 
• Oxidation number of H = +1
• Let the oxidation number of N = x 
• Sum of oxidation numbers in any molecule is equal to zero. So we can write:
• [1 × (+1)] + [1 × x] + [3 × (-2)] = 0 ⇒ 1 + x - 6 = 0 ⇒ x = 5
• So the oxidation number of N in HNO₃ is +5
(iv) H₃PO₄
• Given oxidation number of O = -2 
• Oxidation number of H = +1
• Let the oxidation number of P = x 
• Sum of oxidation numbers in any molecule is equal to zero. So we can write:
• [3 × (+1)] + [1 × x] + [4 × (-2)] = 0 ⇒ 3 + x - 8 = 0 ⇒ x = 5
• So the oxidation number of P in H₃PO₄ is +5

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We have completed the discussion on oxidation numbers. In the next chapter, we will learn about the Periodic table. 

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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 3.5 - Chemical formula from Valency

In the previous section, we completed the discussion on valency. In this section, we will see an application of Valency.

From Valency to Chemical formula

Given below are the chemical formulae of some compounds:
• Sodium chloride NaCl • Magnesium chloride MgCl₂ • Aluminium chloride AlCl₃ •  Carbontetrachloride CCl₄  
There are four compounds. Chlorine is present in all four of them. But the number of chlorine atoms present is different in all the four compounds. Why is this so?

We will be able to understand the reason if we analyse the reaction taking place in each of them.
Consider NaCl: 
• Sodium has an electronic configuration 2,8,1. So it needs to lose one electron to attain octet.
• Chlorine has an electronic configuration 2,8,7. So it needs to gain one electron to attain octet.
• So the electron which is lost by one sodium atom is readily accepted by one chlorine atom. 
• The reaction is complete. We have seen it's details here. It is shown again below:

• So one chlorine atom is sufficient to accept the one electron given away by sodium
■ We can bring in 'valency' into the above discussion:
• Sodium has one electron in the outer most shell. So it's valency is 1. 
    ♦ We find that Valency of sodium is equal to the number of chlorine atoms in NaCl 
• Again, Chlorine has 7 electrons in the outermost shell. So it's valency is 8-7 = 1.
    ♦ We find that Valency of chlorine is equal to the number of sodium atoms in NaCl 
• There seems to be an interchange of valency numbers:
    ♦ Valency of Sodium is equal to the number of chlorine atoms, and in return, valency of chlorine is equal to the number of sodium atoms

Now consider MgCl₂  
• Magnesium has an electronic configuration 2,8,2. So it needs to lose two electrons to attain octet.
• A single chlorine atom will not be able to accommodate the two electons given away by magnesium. So one more chlorine atom comes in
• The two chlorine atoms can together complete the reaction, by each of them accepting 'one of the two electrons' from magnesium. We have seen it's details here. It is shown again below: 

• So we see that there will be two chlorine atoms in magnesium chloride
■ We can bring in 'valency' into the above discussion:
• Magnesium has two electrons in the outer most shell. So it's valency is 2. 
    ♦ We find that Valency of Magnesium is equal to the number of chlorine atoms in MgCl₂ 
• Again, Chlorine has 7 electrons in the outermost shell. So it's valency is 8-7 = 1.
    ♦ We find that Valency of chlorine is equal to the number of Magnesium atoms in MgCl₂
• There seems to be an interchange of valency numbers:
    ♦ Valency of Magnesium is equal to the number of chlorine atoms, and in return, valency of chlorine is equal to the number of Magnesium atoms

Now consider AlCl₃
• Aluminium has an electronic configuration 2,8,3. So it needs to lose three electrons to attain octet.
• A single chlorine atom will not be able to accommodate the three electons given away by Aluminium. So two more chlorine atoms comes in
• The three chlorine atoms can together complete the reaction, by each of them accepting 'one of the three electrons' from Aluminium
• So we see that there will be three chlorine atoms in Aluminium chloride
■ We can bring in 'valency' into the above discussion:
• Aluminium has three electrons in the outer most shell. So it's valency is 3. 
    ♦ We find that Valency of Aluminium is equal to the number of chlorine atoms in AlCl₃ 
• Again, Chlorine has 7 electrons in the outermost shell. So it's valency is 8-7 = 1.
    ♦ We find that Valency of chlorine is equal to the number of Aluminium atoms in AlCl₃
• There seems to be an interchange of valency numbers:
   ♦ Valency of Aluminium is equal to the number of chlorine atoms, and in return, valency of chlorine is equal to the number of Aluminium atoms

Now consider CCl₄
• Carbon has an electronic configuration 2,4. So it needs four more electrons to attain octet.
• Chlorine is also in need for electron. So they form a covalent bond. We have seen it's details here. It is shown again below:

• A single carbon atom needs to form covalent bonds with 4 chlorine atoms. Then only there will be octet. 
• So we find that there will be four chlorine atoms in CCl₄ 
■ We can bring in 'valency' into the above discussion:
• Carbon has four electrons in the outer most shell. So it's valency is 4. 
    ♦ We find that Valency of Carbon is equal to the number of chlorine atoms in CCl₄ 
• Again, Chlorine has 7 electrons in the outermost shell. So it's valency is 8-7 = 1.
    ♦ We find that Valency of chlorine is equal to the number of Carbon atoms in CCl₄
• There seems to be an interchange of valency numbers:
    ♦ Valency of Carbon is equal to the number of chlorine atoms, and in return, valency of chlorine is equal to the number of Carbon atoms

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In all the four cases that we saw above, there is an inter change of valency number and the number of atoms present. Let us write a summary: 
■ In the chemical formula NaCl, 
• The number of atoms of Na present is 1. This '1' is the valency of Cl 
• The number of atoms of Cl present is 1. This '1' is the valency of Na
■ In the chemical formula MgCl₂, 
• The number of atoms of Mg present is 1. This '1' is the valency of Cl 
• The number of atoms of Cl present is 2. This '2' is the valency of Mg
■ In the chemical formula AlCl₃, 
• The number of atoms of Al present is 1. This '1' is the valency of Cl 
• The number of atoms of Cl present is 3. This '3' is the valency of Al
■ In the chemical formula CCl₄, 
• The number of atoms of C present is 1. This '1' is the valency of Cl 
• The number of atoms of Cl present is 4. This '4' is the valency of C

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If there is indeed such an interchange, we get an easy method to write the chemical formulae of any compound. Let us see two more examples:

Consider MgO  
• Magnesium has an electronic configuration 2,8,2. So it needs to lose two electrons to attain octet.
• Oxygen has an electronic configuration 2,6. So it needs to gain two electrons to attain octet.
• So the electrons which are lost by one magnesium atom is readily accepted by one oxygen atom.
• So we see that there will be one magnesium atom, and one oxygen atom in magnesium oxide
We can bring in 'valency' into the above discussion:
• Magnesium has two electrons in the outer most shell. So it's valency is 2. 
• Oxygen has 6 electrons in the outermost shell. So it's valency is 8-6 = 2
• Let us interchange the valency numbers and write the chemical formula. (Here, interchanging does not make any difference because, both valencies are '2'. But we will follow the procedure, and assume that they are interchanged)
• We get the chemical formula as: Mg₂O₂
    ♦ In the above chemical equation, the suffix '2' of magnesium is the valency of Oxygen
    ♦ The suffix '2' of oxygen is the valency of magnesium
• Now divide the 'interchanged valencies' by the common factor. '2' and '2' have the common factor '2'. So we get 2 /2 = 1
• The chemical formula becomes: Mg₁O₁. If the suffix is '1', it is not usually written
• So the final chemical formula is MgO

Consider CO₂
• Carbon has an electronic configuration 2,4. So it needs four more electrons to attain octet.
• Oxygen has an electronic configuration 2,6. So it needs to gain two more electrons to attain octet.
• Both are in need of electrons. So they form a covalent bond as shown in fig.3.20 below:

Fig.3.20

• A single carbon atom needs to combine with 2 oxygen atoms. Then only there will be octet.
• In each bond, 2 pairs (ie., 4 electrons) are shared
• Carbon has a valency of 4, and oxygen has a valency of 2
• Interchange these numbers. So the chemical formula becomes: C₂O₄
• Divide the 'interchanged valencies' by the common factor. '2' and '4' have the common factor '2'. So we get 2 /2 = 1, and 4/2 = 2
• The chemical formula becomes: C₁O₂. If the suffix is '1', it is not usually written
• So the final chemical formula is CO₂ 

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So we find that, just by knowing the valencies of the combining elements, we can write the chemical formula of the compound. The procedure is as follows:

1. Write the symbol of the element with the lower electronegativity first

2. Interchange the valency of each element, and write as suffix

3. Divide each suffix with the common factor. (If there is a common factor)

4. If the suffix of any element is 1, it need not be written

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We will now see some solved examples
Solved example 3.6
Some elements and their valency are given below. 
• Chlorine(Cl), Valency = 1;  • Lithium(Li)  Valency = 1;   • Oxygen(O)  Valency =2;   • Zinc(Zn)  Valency = 2; • Calcium(Ca)  Valency = 2
Write the chemical formulae of the compounds that are formed when they react with each other.
Solution:
In the given list, some elements are metals, and the rest are non-metals. 
• Metals usually do not combine with other metals. 
• Non-metallic elements, some times combine with other non-metallic elements to form compounds
• The most probable cases are those in which metals combine with non-metals
• In the given list, Li, Zn, and Ca are metals. Cl and O are the non-metals

So the possible combinations are:
(i) Li with Cl, (ii) Li with O, (iii) Zn with Cl, (iv) Zn with O, (v) Ca with Cl, (vi) Ca with O, (vii) Cl with O
Let us consider each:
(i) Li with Cl:
• Valency of Li = 1, Electronegativity of Li = 0.98 
• Valency of Cl = 1, Electronegativity of Cl = 3.16
• Li has the lower Electronegativity. So we write it first: LiCl
• Both valencies are the same. So interchanging gives the same result.We get Li₁Cl₁
• Suffix '1' is not written. So we get the chemical formula as: LiCl 

(ii) Li with O :
• Valency of Li = 1, Electronegativity of Li = 0.98 
• Valency of O = 2, Electronegativity of O = 3.44
• Li has the lower Electronegativity. So we write it first: LiO
• Now we rewrite the above with the interchanged valencies: We get Li₂O₁
• Suffix '1' is not written. So we get the chemical formula as: Li₂O

(iii) Zn with Cl:
• Valency of Zn = 2, Electronegativity of Zn = 1.65 
• Valency of Cl = 1, Electronegativity of Cl = 3.16
• Zn has the lower Electronegativity. So we write it first: ZnCl
• Now we rewrite the above with the interchanged valencies: We get Zn₁Cl2
• Suffix '1' is not written. So we get the chemical formula as: ZnCl2

(iv) Zn with O:
• Valency of Zn = 2, Electronegativity of Zn = 1.65 
• Valency of O = 2, Electronegativity of O = 3.44
• Zn has the lower Electronegativity. So we write it first: ZnO
• Both valencies are the same. So interchanging gives the same result.We get: Zn₂O₂
• Now divide the 'interchanged valencies' by the common factor. '2' and '2' have the common factor '2'. So we get 2 /2 = 1
• The chemical formula becomes: Zn₁O₁. If the suffix is '1', it is not usually written
• So the final chemical formula is ZnO

(v) Ca with Cl:
• Valency of Ca = 2, Electronegativity of Ca = 1.00 
• Valency of Cl = 1, Electronegativity of Cl = 3.16
• Ca has the lower Electronegativity. So we write it first: CaCl
• Now we rewrite the above with the interchanged valencies: We get Ca₁Cl2
• Suffix '1' is not written. So we get the chemical formula as: CaCl2

(vi) Ca with O:
• Valency of Ca = 2, Electronegativity of Ca = 1.00 
• Valency of O = 2, Electronegativity of O = 3.44
• Ca has the lower Electronegativity. So we write it first: CaO
• Both valencies are the same. So interchanging gives the same result.We get: Ca₂O₂
• Now divide the 'interchanged valencies' by the common factor. '2' and '2' have the common factor '2'. So we get 2 /2 = 1
• The chemical formula becomes: Ca₁O₁. If the suffix is '1', it is not usually written
• So the final chemical formula is CaO

(vii) O with Cl:
• Valency of O = 2, Electronegativity of O = 3.44 
• Valency of Cl = 1, Electronegativity of Cl = 3.16
• Cl has the lower Electronegativity. So we write it first: ClO
• Now we rewrite the above with the interchanged valencies: We get Cl₂O1
• Suffix '1' is not written. So we get the chemical formula as: Cl₂O

Solved example 3.7
Some elements and their valency are given below. 
• Barium (B), Valency = 2; Chlorine(Cl), Valency = 1; • Zinc(Zn)  Valency = 2   • Oxygen(O)  Valency =2; (i) Write the chemical formula of Barium chloride
(ii) Write the chemical formula of Zinc oxide
(iii) Chemical formula of calcium oxide is CaO. What is the valency of calcium?

Solution:
(i) Ba with Cl:
• Valency of Ba = 2
• Valency of Cl = 1
• Basic form of barium chloride is: BaCl
• Now we rewrite the above with the interchanged valencies: We get Ba₁Cl2
• Suffix '1' is not written. So we get the chemical formula as: BaCl2
(ii) Zn with O:
• Valency of Zn = 2
• Valency of O = 2
• Basic form of zinc oxide is: ZnO
• Both valencies are the same. So interchanging gives the same result.We get: Zn₂O₂
• Now divide the 'interchanged valencies' by the common factor. '2' and '2' have the common factor '2'. So we get 2 /2 = 1
• The chemical formula becomes: Za₁O₁. If the suffix is '1', it is not usually written
• So the final chemical formula is ZnO
(iii) The final chemical equation is given as CaO. Here we have to work in a sort of reverse order
• Let valency of Ca = x
• Valency of O = 2
• Basic form can be written as: CaO
• Now we rewrite the above with the interchanged valencies: We get Ca₂Ox
• The final form is given to us as CaO. This means:
    ♦ 2 when divided by the 'common factor of 2 and x', gives 1
    ♦ x when divided by the 'common factor of 2 and x', gives 1
This is possible only if x is equal to 2
• So we can write: valency of Ca = x = 2

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We saw how to determine the chemical formula from the valencies. In the next section, we will see Oxidation and Reduction.

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