In the previous section, we completed the discussion on valency. In this section, we will see an application of Valency.
• 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
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
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:
• 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
(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
We saw how to determine the chemical formula from the valencies. In the next section, we will see Oxidation and Reduction.
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:
■ 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:
• 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
■ 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
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 |
• 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
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
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
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
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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