In the previous section 3.10, we completed a discussion on the basics of oxidation states. In this section, we will see some basics about valence
1. The electrons present in the outermost main-shell of an atom are called valence electrons of that atom
2. Suppose that, we are given an element
• We need to count the ‘number of valence electrons’ present in the atom of that element
3. Why do we want that number?
Answer: That number determines a special property called valence of that element
4. To find the valence, we use the following rule:
(i) If the number of valence electrons is less than 4
Valence of that element = Number of valence electrons
(ii) If the number of valence electrons is greater than 4
Valence of that element = 8 - Number of valence electrons
(iii) If the number of valence electrons is equal to 4
Valence of that element = 4
Note: The above rules may not be always applicable to transition elements. This is because, in some transition elements, the electrons from inner main-shells also take part in chemical reactions. We will see such cases in later chapters. But the rules are applicable to representative elements (s-block and p-block elements)
To predict the formulas of compounds
• To make the prediction, all we need to do is this:
Interchange the valence
Let us see an example:
Solved example 3.21
Using the Periodic Table, predict the formulas of compounds which might be formed by the following pairs of elements; (a) silicon and bromine (b) aluminium and sulphur.
Solution:
Part (a)
1. Valence of silicon:
• Number of valence electrons = 4
• So valence = 4
2. Valence of bromine:
• Number of valence electrons = 7
• So valence = (8-7) = 1
3. Interchange the valence:
• The number of atoms of Si = valence of Br = 1
• The number of atoms of Br = valence of Si = 4
4. Which element is to be written first:
• The element with lower electronegativity should be written first
• Br is on the right side of Si
♦ That means Br has greater electronegativity
• So we must write Si first
5. Thus the formula is SiBr4
Part (b)
1. Valence of aluminium:
• Number of valence electrons = 3
• So valence = 3
2. Valence of sulphur:
• Number of valence electrons = 6
• So valence = (8-6) = 2
3. Interchange the valence:
• The number of atoms of Al = valence of S = 2
• The number of atoms of S = valence of Al = 3
4. Which element is to be written first:
• The element with lower electronegativity should be written first
• S is on the right side of Al
♦ That means S has greater electronegativity
• So we must write Al first
5. Thus the formula is Al2S3
Solved example 3.22
Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements
(a) Lithium and oxygen (b) Magnesium and nitrogen
(c) Aluminium and iodine (d) Silicon and oxygen
(e) Phosphorus and fluorine (f) Element 71 and fluorine
Solution:
Part (a)
1. Valence of lithium:
• Number of valence electrons = 1
• So valence = 1
2. Valence of oxygen:
• Number of valence electrons = 6
• So valence = (8-6) = 2
3. Interchange the valence:
• The number of atoms of Li = valence of O = 2
• The number of atoms of O = valence of Li = 1
4. Which element is to be written first:
• The element with lower electronegativity should be written first
• O is on the right side of Si
♦ That means O has greater electronegativity
• So we must write Li first
5. Thus the formula is Li2O
Part (b)
1. Valence of magnesium:
• Number of valence electrons = 2
• So valence = 2
2. Valence of nitrogen:
• Number of valence electrons = 5
• So valence = (8-5) = 3
3. Interchange the valence:
• The number of atoms of Mg = valence of N = 3
• The number of atoms of N = valence of Mg = 2
4. Which element is to be written first:
• The element with lower electronegativity should be written first
• N is on the right side of Mg
♦ That means N has greater electronegativity
• So we must write Mg first
5. Thus the formula is Mg3N2
Part (c)
1. Valence of aluminium:
• Number of valence electrons = 3
• So valence = 3
2. Valence of iodine:
• Number of valence electrons = 7
• So valence = (8-7) = 1
3. Interchange the valence:
• The number of atoms of Al = valence of I = 1
• The number of atoms of I = valence of Al = 3
4. Which element is to be written first:
• The element with lower electronegativity should be written first
• I is on the right side of Al
♦ That means I has greater electronegativity
• So we must write Al first
5. Thus the formula is AlI3
Part (d)
1. Valence of silicon:
• Number of valence electrons = 4
• So valence = 4
2. Valence of oxygen:
• Number of valence electrons = 6
• So valence = (8-6) = 2
3. Interchange the valence:
• The number of atoms of Si = valence of O = 2
• The number of atoms of O = valence of Si = 4
4. Which element is to be written first:
• The element with lower electronegativity should be written first
• O is on the right side of Si
♦ That means O has greater electronegativity
• So we must write Si first
5. Thus the formula is Si2O4
This is same as SiO2
Part (e)
1. Valence of phosphorus:
• Number of valence electrons = 5
• So valence = (8-5) = 3
2. Valence of fluorine:
• Number of valence electrons = 7
• So valence = (8-7) = 1
3. Interchange the valence:
• The number of atoms of P = valence of F = 1
• The number of atoms of F = valence of P = 3
4. Which element is to be written first:
• The element with lower electronegativity should be written first
• F is on the right side of P
♦ That means F has greater electronegativity
• So we must write P first
5. Thus the formula is PF3
Part (f)
1. Element 71 is lutetium
• Valence of lutetium:
• Number of valence electrons = 3
• So valence = 3
2. Valence of fluorine:
• Number of valence electrons = 7
• So valence = (8-7) = 1
3. Interchange the valence:
• The number of atoms of Lu = valence of F = 1
• The number of atoms of F = valence of P = 3
4. Which element is to be written first:
• The element with lower electronegativity should be written first
• F is on the right side of Lu
♦ That means F has greater electronegativity
• So we must write Lu first
5. Thus the formula is LuF3
1. The electrons present in the outermost main-shell of an atom are called valence electrons of that atom
2. Suppose that, we are given an element
• We need to count the ‘number of valence electrons’ present in the atom of that element
3. Why do we want that number?
Answer: That number determines a special property called valence of that element
4. To find the valence, we use the following rule:
(i) If the number of valence electrons is less than 4
Valence of that element = Number of valence electrons
(ii) If the number of valence electrons is greater than 4
Valence of that element = 8 - Number of valence electrons
(iii) If the number of valence electrons is equal to 4
Valence of that element = 4
Note: The above rules may not be always applicable to transition elements. This is because, in some transition elements, the electrons from inner main-shells also take part in chemical reactions. We will see such cases in later chapters. But the rules are applicable to representative elements (s-block and p-block elements)
■ In our previous classes, we have seen an interesting application of valence:
• To make the prediction, all we need to do is this:
Interchange the valence
Let us see an example:
Solved example 3.21
Using the Periodic Table, predict the formulas of compounds which might be formed by the following pairs of elements; (a) silicon and bromine (b) aluminium and sulphur.
Solution:
Part (a)
1. Valence of silicon:
• Number of valence electrons = 4
• So valence = 4
2. Valence of bromine:
• Number of valence electrons = 7
• So valence = (8-7) = 1
3. Interchange the valence:
• The number of atoms of Si = valence of Br = 1
• The number of atoms of Br = valence of Si = 4
4. Which element is to be written first:
• The element with lower electronegativity should be written first
• Br is on the right side of Si
♦ That means Br has greater electronegativity
• So we must write Si first
5. Thus the formula is SiBr4
Part (b)
1. Valence of aluminium:
• Number of valence electrons = 3
• So valence = 3
2. Valence of sulphur:
• Number of valence electrons = 6
• So valence = (8-6) = 2
3. Interchange the valence:
• The number of atoms of Al = valence of S = 2
• The number of atoms of S = valence of Al = 3
4. Which element is to be written first:
• The element with lower electronegativity should be written first
• S is on the right side of Al
♦ That means S has greater electronegativity
• So we must write Al first
5. Thus the formula is Al2S3
Solved example 3.22
Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements
(a) Lithium and oxygen (b) Magnesium and nitrogen
(c) Aluminium and iodine (d) Silicon and oxygen
(e) Phosphorus and fluorine (f) Element 71 and fluorine
Solution:
Part (a)
1. Valence of lithium:
• Number of valence electrons = 1
• So valence = 1
2. Valence of oxygen:
• Number of valence electrons = 6
• So valence = (8-6) = 2
3. Interchange the valence:
• The number of atoms of Li = valence of O = 2
• The number of atoms of O = valence of Li = 1
4. Which element is to be written first:
• The element with lower electronegativity should be written first
• O is on the right side of Si
♦ That means O has greater electronegativity
• So we must write Li first
5. Thus the formula is Li2O
Part (b)
1. Valence of magnesium:
• Number of valence electrons = 2
• So valence = 2
2. Valence of nitrogen:
• Number of valence electrons = 5
• So valence = (8-5) = 3
3. Interchange the valence:
• The number of atoms of Mg = valence of N = 3
• The number of atoms of N = valence of Mg = 2
4. Which element is to be written first:
• The element with lower electronegativity should be written first
• N is on the right side of Mg
♦ That means N has greater electronegativity
• So we must write Mg first
5. Thus the formula is Mg3N2
Part (c)
1. Valence of aluminium:
• Number of valence electrons = 3
• So valence = 3
2. Valence of iodine:
• Number of valence electrons = 7
• So valence = (8-7) = 1
3. Interchange the valence:
• The number of atoms of Al = valence of I = 1
• The number of atoms of I = valence of Al = 3
4. Which element is to be written first:
• The element with lower electronegativity should be written first
• I is on the right side of Al
♦ That means I has greater electronegativity
• So we must write Al first
5. Thus the formula is AlI3
Part (d)
1. Valence of silicon:
• Number of valence electrons = 4
• So valence = 4
2. Valence of oxygen:
• Number of valence electrons = 6
• So valence = (8-6) = 2
3. Interchange the valence:
• The number of atoms of Si = valence of O = 2
• The number of atoms of O = valence of Si = 4
4. Which element is to be written first:
• The element with lower electronegativity should be written first
• O is on the right side of Si
♦ That means O has greater electronegativity
• So we must write Si first
5. Thus the formula is Si2O4
This is same as SiO2
Part (e)
1. Valence of phosphorus:
• Number of valence electrons = 5
• So valence = (8-5) = 3
2. Valence of fluorine:
• Number of valence electrons = 7
• So valence = (8-7) = 1
3. Interchange the valence:
• The number of atoms of P = valence of F = 1
• The number of atoms of F = valence of P = 3
4. Which element is to be written first:
• The element with lower electronegativity should be written first
• F is on the right side of P
♦ That means F has greater electronegativity
• So we must write P first
5. Thus the formula is PF3
Part (f)
1. Element 71 is lutetium
• Valence of lutetium:
• Number of valence electrons = 3
• So valence = 3
2. Valence of fluorine:
• Number of valence electrons = 7
• So valence = (8-7) = 1
3. Interchange the valence:
• The number of atoms of Lu = valence of F = 1
• The number of atoms of F = valence of P = 3
4. Which element is to be written first:
• The element with lower electronegativity should be written first
• F is on the right side of Lu
♦ That means F has greater electronegativity
• So we must write Lu first
5. Thus the formula is LuF3
• We have seen the basics of valence. We will now see it's periodic trends
• First we will consider the periodicity along groups
1. Imagine that, we are moving from top to bottom along the Group 1
A. Hydrides
• Consider the elements Li, Na and K
♦ Li will combine with H to give LiH (Lithium hydride)
♦ Na will combine with H to give NaH (Sodium hydride)
♦ K will combine with H to give KH (Potassium hydride)
• Consider LiH
♦ The number of H atoms will give the valence of Li
♦ It is clear that, the valence of Li is 1
• Consider NaH
♦ The number of H atoms will give the valence of Na
♦ It is clear that, the valence of Na is 1
• Consider KH
♦ The number of H atoms will give the valence of K
♦ It is clear that, the valence of K is 1
• In all the 3 compounds, we see that, the valence (of the element which combines with H) is 1
B. Oxides
• Similar to the above hydrides, we have oxides also:
♦ Li2O (Lithium oxide)
♦ Na2O (Sodium oxide)
♦ K2O (Potassium oxide)
• The number of O atoms will give the valence (of the element which combines with O)
♦ We see that the number is '1' in all the three compounds
■ So in general, we can write:
The Group 1 elements show a valence of 1
2. Imagine that, we are moving from top to bottom along the Group 2
A. Hydrides
• Consider the elements Be, Mg and Ca
♦ Be will combine with H to give BeH2 (Beryllium hydride)
♦ Mg will combine with H to give MgH2 (Magnesium hydride)
♦ Ca will combine with H to give CaH2 (Calcium hydride)
• Consider BeH2
♦ The number of H atoms will give the valence of Be
♦ It is clear that, the valence of Be is 2
• Consider MgH2
♦ The number of H atoms will give the valence of Mg
♦ It is clear that, the valence of Mg is 2
• Consider CaH2
♦ The number of H atoms will give the valence of Ca
♦ It is clear that, the valence of Ca is 2
• In all the 3 compounds, we see that, the valence (of the element which combines with H) is 2
B. Oxides
• Similar to the above hydrides, we have oxides also:
♦ MgO (Magnesium oxide)
♦ CaO (Calcium oxide)
♦ SrO (Strontium oxide)
♦ BaO (Barium oxide)
• Here, the number of O atoms will not give the valence. But we can draw the electron dot diagram. We will see that, each of the Group 2 elements donate the 2 outermost electrons to form the respective oxides
■ So in general, we can write:
The Group 2 elements show a valence of 2
3. Imagine that, we are moving from top to bottom along the Group 13
(We skip the transition elements which are Groups 3 to 12. This is because, they show variable valence. We shall see those details in later chapters)
A. Hydrides
• Consider the elements B and Al
♦ B will combine with H to give BH3 (Boron hydride)
✰ Note that, BH3 is the empirical formula. It is not the actual formula of the molecule
♦ Al will combine with H to give AlH3 (Aluminium hydride)
• Consider BH3
♦ The number of H atoms will give the valence of B
♦ It is clear that, the valence of B is 3
• Consider AlH3
♦ The number of H atoms will give the valence of Al
♦ It is clear that, the valence of Al is 3
• In both the compounds, we see that, the valence (of the element which combines with H) is 3
B. Oxides
• Similar to the above hydrides, we have oxides also:
♦ B2O3 (Boric oxide)
♦ Al2O3 (Aluminium oxide)
♦ Ga2O3 (Gallium oxide)
♦ In2O3 (Indium oxide)
• The number of O atoms will give the valence (of the element which combines with O)
♦ We see that the number is '3' in all the four compounds
■ So in general, we can write:
The Group 13 elements show a valence of 3
• First we will consider the periodicity along groups
1. Imagine that, we are moving from top to bottom along the Group 1
A. Hydrides
• Consider the elements Li, Na and K
♦ Li will combine with H to give LiH (Lithium hydride)
♦ Na will combine with H to give NaH (Sodium hydride)
♦ K will combine with H to give KH (Potassium hydride)
• Consider LiH
♦ The number of H atoms will give the valence of Li
♦ It is clear that, the valence of Li is 1
• Consider NaH
♦ The number of H atoms will give the valence of Na
♦ It is clear that, the valence of Na is 1
• Consider KH
♦ The number of H atoms will give the valence of K
♦ It is clear that, the valence of K is 1
• In all the 3 compounds, we see that, the valence (of the element which combines with H) is 1
B. Oxides
• Similar to the above hydrides, we have oxides also:
♦ Li2O (Lithium oxide)
♦ Na2O (Sodium oxide)
♦ K2O (Potassium oxide)
• The number of O atoms will give the valence (of the element which combines with O)
♦ We see that the number is '1' in all the three compounds
■ So in general, we can write:
The Group 1 elements show a valence of 1
2. Imagine that, we are moving from top to bottom along the Group 2
A. Hydrides
• Consider the elements Be, Mg and Ca
♦ Be will combine with H to give BeH2 (Beryllium hydride)
♦ Mg will combine with H to give MgH2 (Magnesium hydride)
♦ Ca will combine with H to give CaH2 (Calcium hydride)
• Consider BeH2
♦ The number of H atoms will give the valence of Be
♦ It is clear that, the valence of Be is 2
• Consider MgH2
♦ The number of H atoms will give the valence of Mg
♦ It is clear that, the valence of Mg is 2
• Consider CaH2
♦ The number of H atoms will give the valence of Ca
♦ It is clear that, the valence of Ca is 2
• In all the 3 compounds, we see that, the valence (of the element which combines with H) is 2
B. Oxides
• Similar to the above hydrides, we have oxides also:
♦ MgO (Magnesium oxide)
♦ CaO (Calcium oxide)
♦ SrO (Strontium oxide)
♦ BaO (Barium oxide)
• Here, the number of O atoms will not give the valence. But we can draw the electron dot diagram. We will see that, each of the Group 2 elements donate the 2 outermost electrons to form the respective oxides
■ So in general, we can write:
The Group 2 elements show a valence of 2
3. Imagine that, we are moving from top to bottom along the Group 13
(We skip the transition elements which are Groups 3 to 12. This is because, they show variable valence. We shall see those details in later chapters)
A. Hydrides
• Consider the elements B and Al
♦ B will combine with H to give BH3 (Boron hydride)
✰ Note that, BH3 is the empirical formula. It is not the actual formula of the molecule
♦ Al will combine with H to give AlH3 (Aluminium hydride)
• Consider BH3
♦ The number of H atoms will give the valence of B
♦ It is clear that, the valence of B is 3
• Consider AlH3
♦ The number of H atoms will give the valence of Al
♦ It is clear that, the valence of Al is 3
• In both the compounds, we see that, the valence (of the element which combines with H) is 3
B. Oxides
• Similar to the above hydrides, we have oxides also:
♦ B2O3 (Boric oxide)
♦ Al2O3 (Aluminium oxide)
♦ Ga2O3 (Gallium oxide)
♦ In2O3 (Indium oxide)
• The number of O atoms will give the valence (of the element which combines with O)
♦ We see that the number is '3' in all the four compounds
■ So in general, we can write:
The Group 13 elements show a valence of 3
• Continuing like this, by citing suitable compounds, we can demonstrate that:
♦ The Group 14 elements show a valence of 4
♦ The Group 15 elements show a valence of 3 OR [(8-3) = 5]
♦ The Group 16 elements show a valence of 2
♦ The Group 17 elements show a valence of 1
■ So we can write a general trend in the 'periodicity of valence' when we move along a group:
When we move from top to bottom along a group, the valence does not change. All elements of a group will have the same valence
♦ The Group 14 elements show a valence of 4
♦ The Group 15 elements show a valence of 3 OR [(8-3) = 5]
♦ The Group 16 elements show a valence of 2
♦ The Group 17 elements show a valence of 1
■ So we can write a general trend in the 'periodicity of valence' when we move along a group:
When we move from top to bottom along a group, the valence does not change. All elements of a group will have the same valence
• In the next section, we will see Anomalous properties of 2nd period elements
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