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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Chapter 3.2 - Covalent bond

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

Covalent bonding

We have seen that the atoms attain octet by donating or accepting electrons. When the donation and acceptance takes place, cations and anions are formed. These ions will then be held together by the electrostatic force of attraction. And thus a molecule is formed. We also learned that, the above type of bonding between atoms is called Ionic bond.

But this type of bonding is not always possible. Consider the case of fluorine. It has a configuration 2,7. It has 7 electrons in the outer most shell. So it needs one more electron to attain octet. What if the atom available nearby is also fluorine? Both the atoms will be 'in need of an electron'. Neither one of them will be able to donate an electron. In such cases, sharing of electrons takes place. One pair of electrons is shared between two fluorine atoms. This is shown in the fig.3.11 below:

Fig.3.11

• Note that, one pair (that is., two electrons) is shared. The two electrons in the pair, belongs to both the atoms. So altogether, each atom will have 8 electrons in the outer most shell. 
• As the pair belongs to both the atoms, the two atoms will not be able to move away from each other
• That is., the two fluorine atoms will have to always stick together. Thus a bond is formed between the two fluorine atoms

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The chemical bond formed as a result of the sharing of electrons between combining atoms is called covalent bond.

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Since one pair of electrons is shared between two fluorine atoms, it is called a single bond. A single bond is represented by a small line between the symbols of the combining atoms. So the single bond between fluorine atoms can be represented as: F-F 

Another example: Consider chlorine with atomic number 17. It has a configuration 2,8,7. It has 7 electrons in the outer most shell. So it is in need of an electron to attain octet. It is a case similar to fluorine: ‘need of a single electron’. If we have two chlorine atoms, neither one of them will be able to donate an electron. So the only way to attain octet is to form a covalent bond as shown in fig.3.12 below:

Fig.3.12

The details about this bond can be written as:
• One pair (that is., two electrons) is shared. The two electrons in the pair, belongs to both the chlorine atoms. So altogether, each atom will have 8 electrons in the outer most shell. 
• As the pair belongs to both the atoms, the two atoms will not be able to move away from each other
• That is., the two chlorine atoms will have to always stick together. Thus a bond is formed between the two chlorine atoms
• Since one pair of electrons is shared between two chlorine atoms, it is a single bond. It is represented as: Cl-Cl

Another example: Consider oxygen with atomic number 8. It has a configuration 2,6. It has 6 electrons in the outer most shell. So it is in need of two electrons to attain octet. If we have two oxygen atoms, neither one of them will be able to donate two electrons. So the only way to attain octet is to form a covalent bond as shown in fig.3.13 below:

Fig.3.13

The details about this bond can be written as:
• Two pairs (that is., four electrons) are shared. The four electrons in the pairs, belongs to both the oxygen atoms. So altogether, each atom will have 8 electrons in the outer most shell. 
• As the pairs belongs to both the atoms, the two atoms will not be able to move away from each other
• That is., the two oxygen atoms will have to always stick together. Thus a bond is formed between the two oxygen atoms
• Since two pairs of electrons are shared between two oxygen atoms, it is called a double bond. A double bond is represented by two small lines between the symbols of the combining atoms. So the double bond between oxygen atoms can be represented as: O=O

Another example: Consider nitrogen with atomic number 7. It has a configuration 2,5. It has 5 electrons in the outer most shell. So it is in need of three electrons to attain octet. If we have two nitrogen atoms, neither one of them will be able to donate three electrons. So the only way to attain octet is to form a covalent bond as shown in fig.3.14 below:

Fig.3.14

The details about this bond can be written as:
• Three pairs (that is., six electrons) are shared. The six electrons in the pairs, belongs to both the nitrogen atoms. So altogether, each atom will have 8 electrons in the outer most shell. 
• As the pairs belongs to both the atoms, the two atoms will not be able to move away from each other
• That is., the two nitrogen atoms will have to always stick together. Thus a bond is formed between the two nitrogen atoms
• Since three pairs of electrons are shared between two nitrogen atoms, it is called a triple bond. A triple bond is represented by three small lines between the symbols of the combining atoms. So the triple bond between nitrogen atoms can be represented as: N≡N

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So we have seen that a covalent bond may be a single, double or triple bond. It is related to the ‘number of pairs’ of electrons shared. This will be more clear from the following table:

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The covalent bonds that we have seen so far, are formed between atoms of the same element. In the above cases, two atoms of the same element combined to form a molecule. Thus, two individual atoms, who would be unstable by themselves, attained stability by forming a covalent bond between them. It may be noted that, since two atoms are present, it is a diatomic molecule. We have see the basics about mono, di and poly atomic molecules here.

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We will now see a molecule formed from two different atoms. Consider the formation of  hydrogen chloride.

• Hydrogen has an atomic number 1. It’s electronic configuration is also 1. It has 1 electron in the outer most shell. It needs one more electron to attain octet. 
• Chlorine has an atomic number 17. It’s electronic configuration is 2,8,7. It has 7 electrons in the outer most shell. It also needs one more electron to attain octet.
• We see that both hydrogen and chlorine are in need for a single electron. They cannot donate any electrons. So the only solution is to form a covalent bond as shown in fig.3.15 below:

Fig.3.15

The details about this bond can be written as:
• One pair (that is., two electrons) is shared. The two electrons in the pair, belongs to both the hydrogen atom and the chlorine atom.
• The hydrogen atom now has the required two electrons for octet. 
• The chlorine atom now has the required 8 electrons for octet
• As the pair belongs to both the atoms, the two atoms will not be able to move away from each other
• That is., the hydrogen and chlorine atoms will have to always stick together. Thus a bond is formed between the two atoms
• Since one pair of electrons is shared, it is a single bond. It is represented as: H-Cl

Another example: Consider the formation of carbontetrachloride
• Carbon has an atomic number 6. It’s electronic configuration is 2,4. It has 4 electrons in the outer most shell. It needs four more electrons to attain octet. 
• Chlorine has an atomic number 17. It’s electronic configuration is 2,8,7. It has 7 electrons in the outer most shell. It needs one more electron to attain octet.
• We see that both carbon and chlorine are in ‘need for electrons’. They cannot donate any electrons. So the only solution is to form a covalent bond as shown in fig.3.16 below:

Fig.3.16

• The details about this bond can be written as:
• One carbon atom combines with 4 chlorine atoms. Each of the four chlorine atoms share one pair of electrons with a carbon atom
• In total, four pairs (that is., eight electrons) are shared. The eight electrons in the pairs, belongs to both the carbon atom and the chlorine atom.
• The carbon atom now has the required 8 electrons for octet. 
• The chlorine atom now has the required 8 electrons for octet
• As the pair belongs to both the atoms, the two atoms will not be able to move away from each other
• That is., the carbon and chlorine atoms will have to always stick together. Thus a bond is formed between the five atoms 

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We will now see some solved examples

Solved example 3.4

Illustrate the chemical bond in the following covalent compounds using electron dot diagram.

(i) CH₄,  (ii) HF,  (iii) H₂O

Solution:

• (i) CH₄: Carbon has an atomic number 6. It’s electronic configuration is 2,4. It has 4 electrons in the outer most shell. It needs four more electrons to attain octet. 
• Hydrogen has an atomic number 1. It’s electronic configuration is also 1. It has 1 electron in the outer most shell. It needs one more electron to attain octet.
• We see that both carbon and hydrogen are in ‘need for electrons’. They cannot donate any electrons. So the only solution is to form a covalent bond as shown in fig.3.17 below:

Fig.3.17

• The details about this bond can be written as:
• One carbon atom combines with 4 hydrogen atoms. Each of the four hydrogen atoms share one pair of electrons with a carbon atom
• In total, four pairs (that is., eight electrons) are shared. The eight electrons in the pairs, belongs to both the carbon atom and the hydrogen atom.
• The carbon atom now has the required 4 electrons for octet. 
• The hydrogen atom now has the required 2 electrons for octet
• As the pair belongs to both the atoms, the two atoms will not be able to move away from each other
• That is., the carbon and hydrogen atoms will have to always stick together. Thus a bond is formed between the five atoms

• (ii) HF: Hydrogen has an atomic number 1. It’s electronic configuration is also 1. It has 1 electron in the outer most shell. It needs one more electron to attain octet. 
• Fluorine has an atomic number 9. It’s electronic configuration is 2,7. It has 7 electrons in the outer most shell. It also needs one more electron to attain octet.
• We see that both hydrogen and fluorine are in need for a single electron. They cannot donate any electrons. So the only solution is to form a covalent bond as shown in  fig.3.18 below:

Fig.3.18

The details about this bond can be written as:
• One pair (that is., two electrons) is shared. The two electrons in the pair, belongs to both the hydrogen atom and the fluorine atom.
• The hydrogen atom now has the required two electrons for octet. 
• The fluorine atom now has the required 8 electrons for octet
• As the pair belongs to both the atoms, the two atoms will not be able to move away from each other
• That is., the hydrogen and fluorine atoms will have to always stick together. Thus a bond is formed between the two atoms
• Since one pair of electrons is shared, it is a single bond. It is represented as: H-F

• (iii) H₂O: Oxygen has an atomic number 8. It’s electronic configuration is 2,6. It has 6 electrons in the outer most shell. It needs two more electrons to attain octet. 
• Hydrogen has an atomic number 1. It’s electronic configuration is also 1. It has 1 electron in the outer most shell. It needs one more electron to attain octet.
• We see that both oxygen and hydrogen are in ‘need for electrons’. They cannot donate any electrons. So the only solution is to form a covalent bond as shown in  fig.3.19 below:

Fig.3.19

• The details about this bond can be written as:
• One oxygen atom combines with 2 hydrogen atoms. Each of the 2 hydrogen atoms share one pair of electrons with a oxygen atom
• In total, two pairs (that is., four electrons) are shared. The four electrons in the pairs, belongs to both the oxygen atom and the hydrogen atom.
• The oxygen atom now has the required 8 electrons for octet. 
• The hydrogen atom now has the required 2 electrons for octet
• As the pair belongs to both the atoms, the two atoms will not be able to move away from each other
• That is., the oxygen and hydrogen atoms will have to always stick together. Thus a bond is formed between the three atoms.

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In the next section, we will learn about electronegativity.

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Chapter 3.1 - Cations and Anions

In the previous section, we saw ionic bond and the electron dot diagram.  In this section, we will see some solved examples. We will also learn about Cations and Anions

Solved example 3.1
Some elements (symbols are not real) with their atomic number are given below:
₉P, ₁₇Q, ₁₀R, ₁₂S 
(i) Write the electronic configuration of each element
(ii) Which element is the most stable one? Why?
(iii) Which element donates electrons in chemical reaction?
(iv) Write the chemical formula of the compound formed by combining element S with element P
Solution:
(i) Electronic configuration of the elements are given below (details here):
₉P : 2,7 ; ₁₇Q : 2, 8, 7 ; ₁₀R : 2, 8 ; ₁₂S : 2, 8, 2
(ii) ₁₀R is the most stable element. Because, it has 8 electrons in it's outer most shell
(iii) • ₉P will have to donate 7 electrons to attain octet. It is easier to accept 1 electron, than to donate 7
• ₁₇Q will have to donate 7 electrons to attain octet. It is easier to accept 1 electron, than to donate 7
• ₁₀R has already has octet. It will not accept or donate any electrons
• ₁₂S can donate 2 electrons to attain octet. It is easier to donate 2, than to accept 6
• So the answer is ₁₂S


(iv) Combination of S and P: 
• ₁₂S has a configuration 2, 8, 2. So it has 2 electrons in the outer most shell
• It needs 6 more electrons to attain octet. It is easier to lose the 2 electrons than to obtain 6 electrons
• ₉P has a configuration 2, 7. So it has 7 electrons in the outer most shell
• It needs 1 more electron to attain octet. 
• So 2 atoms of P will combine with one atom of S. So that:
    ♦ One P will accept 'one of the 2 electrons' donated by S
    ♦ The other P will accept the 'other of the 2 electrons' donated by S
• With the above information, we can draw the electron dot diagram as shown in fig.3.7 below: 

Fig.3.7

• The outer most electrons are shown as dots around the symbol of the atom 
• By counting the number of dots, we can see that, S has lost two electrons, and the two P atoms have gained those two electrons. (one electron for each P)
• The initial and final configurations are also given. S and P have attained octet
• S has become a positive ion. This is due to the loss of electrons. Two electrons are lost. So the charge is 2+. So we write '2+' on the top right
• Each P has become a negative ion. This is due to the gain of electron. One electrons is gained by each P. So the charge is 1- for each. So we write '-' on the top right
• The 'oppositely charged' ions are held together by electrostatic force of attraction. Thus a molecule SP₂ is formed

Solved example 3.2
Draw the electron dot diagram for the ionic bonding in the following compounds:
(i) Sodium fluoride (ii) Magnesium fluoride
(Hint: Atomic numbers are: ₁₁Na, ₉F, ₁₂Mg)

Solution:
(i) Formation of Sodium fluoride: 
• ₁₁Na has a configuration 2, 8, 1. So it has 1 electron in the outer most shell
• It needs 7 more electrons to attain octet. It is easier to lose the 1 electron than to obtain 7 electrons
• ₉F has a configuration 2, 7. So it has 7 electrons in the outer most shell
• It needs 1 more electron to attain octet. 
• So 1 atom of Na will combine with one atom of F. So that, F will accept the 1 electron donated by Na
• With the above information, we can draw the electron dot diagram as shown in fig.3.8 below:

Fig.3.8

• The outer most electrons are shown as dots around the symbol of the atom 
• By counting the number of dots, we can see that, Na has lost one electrons, and F have gained that electron
• The initial and final configurations are also given. Na and F have attained octet
• Na has become a positive ion. This is due to the loss of electron. One electron is  lost. So the charge is 1+. So we write '+' on the top right
• F has become a negative ion. This is due to the gain of electron. One electrons is gained by P. So the charge is 1-. So we write '-' on the top right
• The 'oppositely charged' ions are held together by electrostatic force of attraction. Thus a molecule NaF is formed

(i) Formation of Magnesium fluoride: 
• ₁₂Mg has a configuration 2, 8, 2. So it has 2 electrons in the outer most shell
• It needs 6 more electrons to attain octet. It is easier to lose the 2 electrons than to obtain 6 electrons
• ₉F has a configuration 2, 7. So it has 7 electrons in the outer most shell
• It needs 1 more electron to attain octet. 
• So 2 atoms of F will combine with one atom of Mg. So that:
    ♦ One F will accept 'one of the 2 electrons' donated by Mg
    ♦ The other F will accept the 'other of the 2 electrons' donated by Mg
• With the above information, we can draw the electron dot diagram as shown in fig.3.9 below:

Fig.3.9

• The outer most electrons are shown as dots around the symbol of the atom 
• By counting the number of dots, we can see that, Mg has lost two electrons, and the two F atoms have gained those two electrons. (one electron for each F)
• The initial and final configurations are also given. Mg and F have attained octet
• Mg has become a positive ion. This is due to the loss of electrons. Two electrons are lost. So the charge is 2+. So we write '2+' on the top right
• Each F has become a negative ion. This is due to the gain of electron. One electrons is gained by each F. So the charge is 1- for each. So we write '-' on the top right
• The 'oppositely charged' ions are held together by electrostatic force of attraction. Thus a molecule MgF₂ is formed

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Cations and Anions

Let us consider the formation of NaCl that we saw in the previous section. It's electron dot diagram that we saw in fig.3.4 is shown again below:

As a result of the donation and acceptance of electron, 'sodium ion' and 'chlorine ion' is formed. The formation of these ions can be represented using simple equations:
• Na → Na⁺ + 1e^-
• Cl + 1e^- → Cl ^-
• The first equation indicates that: Sodium atom has donated an electron and became a positive ion.
• The second equation indicates that: Chlorine atom has accepted the electron, and became a negative ion.
■ Positive ions are called cations
■ Negative ions are called anions

Another example:
Let us consider the formation of MgO that we saw in fig.3.5 in the previous section. It is shown again below:

The equations can be written as:
• Mg → Mg²⁺ + 2e^-
• O + 2e^- → O^2-
Here Mg²⁺ is the cation and O^2- is the anion

One more example:
Consider the formation of Na₂O that we saw in fig.3.6 in the previous section. It is shown again below:

The equations can be written as:
• 2Na → 2Na¹⁺ + 2e^-
• O + 2e^- → O^2-
Here Na¹⁺ is the cation and O^2- is the anion

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Now we will see a solved example:
Solved example 3.3
Write all the details regarding the formation of Magnesium chloride MgCl₂
Solution:
(i) Formation of Magnesium chloride: 
• ₁₂Mg has a configuration 2, 8, 2. So it has 2 electrons in the outer most shell
• It needs 6 more electrons to attain octet. It is easier to lose the 2 electrons than to obtain 6 electrons
• ₁₇Cl has a configuration 2, 8, 7. So it has 7 electrons in the outer most shell
• It needs 1 more electron to attain octet. 
• So 2 atoms of Cl will combine with one atom of Mg. So that:
    ♦ One Cl will accept 'one of the 2 electrons' donated by Mg
    ♦ The other Cl will accept the 'other of the 2 electrons' donated by Mg
• With the above information, we can draw the electron dot diagram as shown in fig.3.10 below:

Fig.3.10

• The outer most electrons are shown as dots around the symbol of the atom 
• By counting the number of dots, we can see that, Mg has lost two electrons, and the two Cl atoms have gained those two electrons. (one electron for each Cl)
• The initial and final configurations are also given. Mg and Cl have attained octet
• Mg has become a positive ion. This is due to the loss of electrons. Two electrons are lost. So the charge is 2+. So we write '2+' on the top right
• 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. So we write '-' on the top right
• The 'oppositely charged' ions are held together by electrostatic force of attraction. Thus a molecule MgCl₂ is formed

The ion equations can be written as:
• Mg → Mg²⁺ + 2e^-
• 2Cl + 2e^- → 2Cl^-
Here Mg²⁺ is the cation and Cl^- is the anion

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So we have completed the discussion on Ionic bonds. In the next section, we will see Covalent bonds.

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