In the previous section 3.9, we completed a discussion on the basics of seven physical properties:
(i) Atomic radius (ii) Ionic radius (iii) Ionization energy (iv) Electron gain enthalpy (v) Electronegativity (vi) Metallic character (vii) Non-metallic character
• We also saw the basics about their periodic trends
• In this section, we will see some chemical properties
• The first chemical property that we are going to see is: Valence
(Nowadays, another chemical property called oxidation state is used more frequently instead of 'valence'. We can use either 'valence' or 'oxidation state' in the calculations. Both will give the same results. But 'oxidation state' will give more information)
• After seeing the basics of these properties, we will discuss their periodicity
■ Before beginning our present discussion, it is important that, we revise the lessons on these topics:
♦ Ionic and covalent bonds
♦ Polar nature
♦ Valency of elements
♦ Chemical formula from valency
♦ Oxidation and reduction
♦ Oxidation number ('Oxidation number' and 'oxidation state' are often used interchangeably)
• We have already seen those topics in our previous classes
• The reader may refer his/her old note books and text books. Or the following link can be used:
Chemical bonding
• We will write it in steps:
1. From the table 3.6 of the previous section, we get:
♦ The electronegativity of O is 3.5
♦ The electronegativity of F is 4.0
2. The difference in electronegativities is (4.0-3.5) = 0.5
• This difference is less than 1.7
• So OF2 will be covalent
3. Let us draw the electron dot diagram:
• We see that, on the right side, all the three atoms have the required 8 electrons in the outermost shell
4. Let us write the details about 'electron sharing':
• Consider the left F atom
♦ It shares 2 electrons (a single pair) with the O atom
♦ It is a single bond
• Consider the right F atom
♦ It shares 2 electrons (a single pair) with the O atom
♦ So it is a single bond
5. Oxidation states of F atoms
A. F atom on the left:
(i) F has greater electronegativity
• So the F atom on the left will attract the shared pair of electrons to itself
• This F atom thus has the pair close to it
(ii) But only 'one electron of the pair' can be considered to be 'gained'
• The other electron already belonged to F
(iii) So the F atom on the left has gained an electron
• 'Gaining electron' is reduction
• It is indicated by '-ve oxidation state'
• Since only one electron is gained, the oxidation state is '-1'
■ So we can write:
The F atom on the left has an oxidation state of -1
B. The same steps in 5 (A) can be written for the F atom on the right. For better clarity, we will write them again
F atom on the right:
(i) F has greater electronegativity
• So the F atom on the right will attract the shared pair of electrons to itself
• This F atom thus has the pair close to it
(ii) But only 'one electron of the pair' can be considered to be 'gained'
• The other electron already belonged to F
(iii) So the F atom on the right has gained an electron
• 'Gaining electron' is reduction
• It is indicated by '-ve oxidation state'
• Since only one electron is gained, the oxidation state is '-1'
■ So we can write:
The F atom on the right has an oxidation state of -1
6. Oxidation state of O atom
A. Left side of O atom
(i) F has greater electronegativity
• So the F atom on the left will attract the shared pair of electrons to itself
• The O atom thus loses the pair
(ii) But only 'one electron of the pair' can be considered to be 'lost'
• The other electron never belonged to O
(iii) So on the left side, the O atom has lost an electron
• 'Losing electron' is oxidation
• It is indicated by '+ve oxidation state'
• Since only one electron is gained, the oxidation state is '+1'
■ So we can write:
On the left side, the O atom has an oxidation state of +1
B. The same steps in 6 (A) can be written for the right side of O atom. For better clarity, we will write them again
Right side of O atom:
(i) F has greater electronegativity
• So the F atom on the right will attract the shared pair of electrons to itself
• The O atom thus loses the pair
(ii) But only 'one electron of the pair' can be considered to be 'lost'
• The other electron never belonged to O
(iii) So on the right side, the O atom has lost an electron
• 'Losing electron' is oxidation
• It is indicated by '+ve oxidation state'
• Since only one electron is gained, the oxidation state is '+1'
■ So we can write:
• On the right side, the O atom has an oxidation state of +1
• So in total, the O atom has lost two electrons and thus it's oxidation state is +2
7. Thus in total, we can write:
• Each of the F atoms has an oxidation state of -1
• The O atom has an oxidation state of +2
■ In short form, we represent this as: $\mathbf\small{\rm{O^{+2}F_2^{-1}}}$
• We have seen an easier method also. In this method, we can apply any of the eight simple rules:
(i) The oxidation state of a free element is always 0
(ii) The oxidation state of a mono-atomic ion equals the charge of the ion
(iii) The oxidation state of H is +1
♦ H usually donates it's one and only electron
• But it is -1 in when combined with less electronegative elements
♦ H being more electronegative, pulls one electron and attains the stable 1s2 configuration
(iv) The oxidation state of a Group 1 element in a compound is +1
♦ The general outermost electronic configuration for Group 1 elements is ns1
♦ The Group 1 elements loses this electron
(v) The oxidation state of a Group 2 element in a compound is +2
♦ The general outermost electronic configuration for Group 2 elements is ns2
♦ The Group 2 elements loses these two electrons
(vi) The oxidation state of a Group 17 element in a binary compound (a compound which has exactly two different elements) is -1
♦ The general outermost electronic configuration for Group 17 elements is ns2np5
♦ They need one more electron to become ns2np6
(vii) The sum of the oxidation states of all of the atoms in a neutral compound is 0
(viii) The sum of the oxidation states in a polyatomic ion is equal to the charge of the ion
♦ We need to find the oxidation state of O. It is the only unknown
♦ Oxidation state of F is obvious from rule (vi)
1. Let the oxidation state of O be 'x'
2. From rule (vi), the oxidation state of F = -1
3. So the oxidation states in OF2 can be represented as: $\mathbf\small{\rm{O^{x}F_2^{-1}}}$
4. Applying rule 7, we get: [x+(2 × -1)] = 0
⇒ [x - 2] = 0
⇒ x = 2
4. So the final representation is: $\mathbf\small{\rm{O^{+2}F_2^{-1}}}$
• We will write it in steps:
1. From the table 3.7 of the previous section, we get:
♦ The electronegativity of O is 3.5
♦ The electronegativity of Na is 0.9
2. The difference in electronegativities is (3.5 - 0.9) = 2.6
• This difference is greater than 1.7
• So Na2O will be ionic
3. Let us draw the electron dot diagram:
• We see that, on the right side, all the three atoms have the required 8 electrons in the outermost shell
♦ 1s22s22p63s1 of Na has become 1s22s22p6
♦ 1s22s22p4 of O has become 1s22s22p6
4. Let us write the details about 'electron transfer':
• Consider the left Na atom
♦ It donates 1 electron to the O atom
♦ It then becomes Na+ ion
• Consider the right Na atom
♦ It donates 1 electron to the O atom
♦ It then becomes Na+ ion
5. Oxidation states of Na atoms
A. Na atom on the left:
• This Na atom has lost an electron
• 'Losing electron' is oxidation
• It is indicated by '+ve oxidation state'
• Since only one electron is lost, the oxidation state is '+1'
■ So we can write:
The Na atom on the left has an oxidation state of +1
B. The same steps in 5 (A) can be written for the Na atom on the right. For better clarity, we will write them again
Na atom on the right:
• This Na atom has lost an electron
• 'Losing electron' is oxidation
• It is indicated by '+ve oxidation state'
• Since only one electron is lost, the oxidation state is '+1'
■ So we can write:
The Na atom on the right has an oxidation state of +1
6. Oxidation state of O atom
A. Left side of O atom
• On this side, the O atom has gained an electron
• 'Gaining electron' is reduction
• It is indicated by '-ve oxidation state'
• Since only one electron is gained on the left side, the oxidation state is '-1'
■ So we can write:
On the left side, the O atom has an oxidation state of -1
B. The same steps in 6 (A) can be written for the right side of O atom. For better clarity, we will write them again
Right side of O atom:
• On this side, the O atom has gained an electron
• 'Gaining electron' is reduction
• It is indicated by '-ve oxidation state'
• Since only one electron is gained on the right side, the oxidation state is '-1'
■ So we can write:
• On the right side, the O atom has an oxidation state of -1
• So in total, the O atom has gained two electrons and thus it's oxidation state is -2
7. Thus in total, we can write:
• Each of the Na atoms has an oxidation state of +1
• The O atom has an oxidation state of -2
■ In short form, we represent this as: $\mathbf\small{\rm{Na_2^{+1}O^{-2}}}$
♦ We need to find the oxidation state of O. It is the only unknown
♦ Oxidation state of Na is obvious from rule (iv)
1. Let the oxidation state of O be 'x'
2. From rule (iv), the oxidation state of Na = +1
3. So the oxidation states in Na2O can be represented as: $\mathbf\small{\rm{Na_2^{+1}O^{x}}}$
4. Applying rule (vii), we get: [(2 × 1) + x] = 0
⇒ [2 + x] = 0
⇒ x = -2
4. So the final representation is: $\mathbf\small{\rm{Na_2^{+1}O^{-2}}}$
$\mathbf\small{\rm{O^{+2}F_2^{-1}}}$ and $\mathbf\small{\rm{Na_2^{+1}O^{-2}}}$
• We can easily see the relation between the two items:
(i) The ‘oxidation state of an element’ when it is part of a compound
(ii) The ‘charge that the element acquires’ when it is in that compound
■ Let us elaborate with examples:
• In OF2:
♦ O atom loses two electrons and thus acquires a positive charge of +2
✰ The oxidation state of O in OF2 is also +2
♦ Each of the F atoms gains one electron and thus acquires a negative charge of -1
✰ The oxidation state of ‘each F’ in OF2 is also -1
• In Na2O:
♦ O atom gains two electrons and thus acquires a negative charge of -2
✰ The oxidation state of O in Na2O is also -2
♦ Each of the Na atoms loses one electron and thus acquires a positive charge of +1
✰ The oxidation state of ‘each Na’ in Na2O is also +1
1. We define oxidation state for a particular element
• This particular element,
♦ may exhibit one oxidation state in a compound
♦ may exhibit another oxidation state in another compound
• For example, oxidation state of O is
♦ +2 in OF2
♦ -2 in Na2O
2. So when we define the oxidation state of an element, we must specify the compound (in which the element is present) also
3. In that compound, all the elements will be having various values of electronegativities
• Elements with higher electronegativities will be able to pull electrons to themselves
• So we can write:
♦ Elements with higher electronegativities will acquire negative charges
♦ Elements with lower electronegativities will acquire positive charges
4. Now we concentrate on the 'element under consideration'
■ The charge acquired by that element ‘based on the electronegativity consideration’ is called the oxidation state of that element
(i) Atomic radius (ii) Ionic radius (iii) Ionization energy (iv) Electron gain enthalpy (v) Electronegativity (vi) Metallic character (vii) Non-metallic character
• We also saw the basics about their periodic trends
• In this section, we will see some chemical properties
• The first chemical property that we are going to see is: Valence
(Nowadays, another chemical property called oxidation state is used more frequently instead of 'valence'. We can use either 'valence' or 'oxidation state' in the calculations. Both will give the same results. But 'oxidation state' will give more information)
• After seeing the basics of these properties, we will discuss their periodicity
■ Before beginning our present discussion, it is important that, we revise the lessons on these topics:
♦ Ionic and covalent bonds
♦ Polar nature
♦ Valency of elements
♦ Chemical formula from valency
♦ Oxidation and reduction
♦ Oxidation number ('Oxidation number' and 'oxidation state' are often used interchangeably)
• We have already seen those topics in our previous classes
• The reader may refer his/her old note books and text books. Or the following link can be used:
Chemical bonding
• Based on those previous lessons, we are now going to see the details of the compound: OF2
1. From the table 3.6 of the previous section, we get:
♦ The electronegativity of O is 3.5
♦ The electronegativity of F is 4.0
2. The difference in electronegativities is (4.0-3.5) = 0.5
• This difference is less than 1.7
• So OF2 will be covalent
3. Let us draw the electron dot diagram:
 |
| Fig.3.19 |
4. Let us write the details about 'electron sharing':
• Consider the left F atom
♦ It shares 2 electrons (a single pair) with the O atom
♦ It is a single bond
• Consider the right F atom
♦ It shares 2 electrons (a single pair) with the O atom
♦ So it is a single bond
5. Oxidation states of F atoms
A. F atom on the left:
(i) F has greater electronegativity
• So the F atom on the left will attract the shared pair of electrons to itself
• This F atom thus has the pair close to it
(ii) But only 'one electron of the pair' can be considered to be 'gained'
• The other electron already belonged to F
(iii) So the F atom on the left has gained an electron
• 'Gaining electron' is reduction
• It is indicated by '-ve oxidation state'
• Since only one electron is gained, the oxidation state is '-1'
■ So we can write:
The F atom on the left has an oxidation state of -1
B. The same steps in 5 (A) can be written for the F atom on the right. For better clarity, we will write them again
F atom on the right:
(i) F has greater electronegativity
• So the F atom on the right will attract the shared pair of electrons to itself
• This F atom thus has the pair close to it
(ii) But only 'one electron of the pair' can be considered to be 'gained'
• The other electron already belonged to F
(iii) So the F atom on the right has gained an electron
• 'Gaining electron' is reduction
• It is indicated by '-ve oxidation state'
• Since only one electron is gained, the oxidation state is '-1'
■ So we can write:
The F atom on the right has an oxidation state of -1
6. Oxidation state of O atom
A. Left side of O atom
(i) F has greater electronegativity
• So the F atom on the left will attract the shared pair of electrons to itself
• The O atom thus loses the pair
(ii) But only 'one electron of the pair' can be considered to be 'lost'
• The other electron never belonged to O
(iii) So on the left side, the O atom has lost an electron
• 'Losing electron' is oxidation
• It is indicated by '+ve oxidation state'
• Since only one electron is gained, the oxidation state is '+1'
■ So we can write:
On the left side, the O atom has an oxidation state of +1
B. The same steps in 6 (A) can be written for the right side of O atom. For better clarity, we will write them again
Right side of O atom:
(i) F has greater electronegativity
• So the F atom on the right will attract the shared pair of electrons to itself
• The O atom thus loses the pair
(ii) But only 'one electron of the pair' can be considered to be 'lost'
• The other electron never belonged to O
(iii) So on the right side, the O atom has lost an electron
• 'Losing electron' is oxidation
• It is indicated by '+ve oxidation state'
• Since only one electron is gained, the oxidation state is '+1'
■ So we can write:
• On the right side, the O atom has an oxidation state of +1
• So in total, the O atom has lost two electrons and thus it's oxidation state is +2
7. Thus in total, we can write:
• Each of the F atoms has an oxidation state of -1
• The O atom has an oxidation state of +2
■ In short form, we represent this as: $\mathbf\small{\rm{O^{+2}F_2^{-1}}}$
• We have seen the above method for finding the oxidation states in our earlier classes
(i) The oxidation state of a free element is always 0
(ii) The oxidation state of a mono-atomic ion equals the charge of the ion
(iii) The oxidation state of H is +1
♦ H usually donates it's one and only electron
• But it is -1 in when combined with less electronegative elements
♦ H being more electronegative, pulls one electron and attains the stable 1s2 configuration
(iv) The oxidation state of a Group 1 element in a compound is +1
♦ The general outermost electronic configuration for Group 1 elements is ns1
♦ The Group 1 elements loses this electron
(v) The oxidation state of a Group 2 element in a compound is +2
♦ The general outermost electronic configuration for Group 2 elements is ns2
♦ The Group 2 elements loses these two electrons
(vi) The oxidation state of a Group 17 element in a binary compound (a compound which has exactly two different elements) is -1
♦ The general outermost electronic configuration for Group 17 elements is ns2np5
♦ They need one more electron to become ns2np6
(vii) The sum of the oxidation states of all of the atoms in a neutral compound is 0
(viii) The sum of the oxidation states in a polyatomic ion is equal to the charge of the ion
• Let us apply the rules to OF2:
♦ Oxidation state of F is obvious from rule (vi)
1. Let the oxidation state of O be 'x'
2. From rule (vi), the oxidation state of F = -1
3. So the oxidation states in OF2 can be represented as: $\mathbf\small{\rm{O^{x}F_2^{-1}}}$
4. Applying rule 7, we get: [x+(2 × -1)] = 0
⇒ [x - 2] = 0
⇒ x = 2
4. So the final representation is: $\mathbf\small{\rm{O^{+2}F_2^{-1}}}$
Next we are going to see the same details about another compound: Na2O
1. From the table 3.7 of the previous section, we get:
♦ The electronegativity of O is 3.5
♦ The electronegativity of Na is 0.9
2. The difference in electronegativities is (3.5 - 0.9) = 2.6
• This difference is greater than 1.7
• So Na2O will be ionic
3. Let us draw the electron dot diagram:
 |
| Fig.3.20 |
♦ 1s22s22p63s1 of Na has become 1s22s22p6
♦ 1s22s22p4 of O has become 1s22s22p6
4. Let us write the details about 'electron transfer':
• Consider the left Na atom
♦ It donates 1 electron to the O atom
♦ It then becomes Na+ ion
• Consider the right Na atom
♦ It donates 1 electron to the O atom
♦ It then becomes Na+ ion
5. Oxidation states of Na atoms
A. Na atom on the left:
• This Na atom has lost an electron
• 'Losing electron' is oxidation
• It is indicated by '+ve oxidation state'
• Since only one electron is lost, the oxidation state is '+1'
■ So we can write:
The Na atom on the left has an oxidation state of +1
B. The same steps in 5 (A) can be written for the Na atom on the right. For better clarity, we will write them again
Na atom on the right:
• This Na atom has lost an electron
• 'Losing electron' is oxidation
• It is indicated by '+ve oxidation state'
• Since only one electron is lost, the oxidation state is '+1'
■ So we can write:
The Na atom on the right has an oxidation state of +1
6. Oxidation state of O atom
A. Left side of O atom
• On this side, the O atom has gained an electron
• 'Gaining electron' is reduction
• It is indicated by '-ve oxidation state'
• Since only one electron is gained on the left side, the oxidation state is '-1'
■ So we can write:
On the left side, the O atom has an oxidation state of -1
B. The same steps in 6 (A) can be written for the right side of O atom. For better clarity, we will write them again
Right side of O atom:
• On this side, the O atom has gained an electron
• 'Gaining electron' is reduction
• It is indicated by '-ve oxidation state'
• Since only one electron is gained on the right side, the oxidation state is '-1'
■ So we can write:
• On the right side, the O atom has an oxidation state of -1
• So in total, the O atom has gained two electrons and thus it's oxidation state is -2
7. Thus in total, we can write:
• Each of the Na atoms has an oxidation state of +1
• The O atom has an oxidation state of -2
■ In short form, we represent this as: $\mathbf\small{\rm{Na_2^{+1}O^{-2}}}$
• Let us apply the easier method to Na2O:
♦ Oxidation state of Na is obvious from rule (iv)
1. Let the oxidation state of O be 'x'
2. From rule (iv), the oxidation state of Na = +1
3. So the oxidation states in Na2O can be represented as: $\mathbf\small{\rm{Na_2^{+1}O^{x}}}$
4. Applying rule (vii), we get: [(2 × 1) + x] = 0
⇒ [2 + x] = 0
⇒ x = -2
4. So the final representation is: $\mathbf\small{\rm{Na_2^{+1}O^{-2}}}$
• So we have two results:
• We can easily see the relation between the two items:
(i) The ‘oxidation state of an element’ when it is part of a compound
(ii) The ‘charge that the element acquires’ when it is in that compound
■ Let us elaborate with examples:
• In OF2:
♦ O atom loses two electrons and thus acquires a positive charge of +2
✰ The oxidation state of O in OF2 is also +2
♦ Each of the F atoms gains one electron and thus acquires a negative charge of -1
✰ The oxidation state of ‘each F’ in OF2 is also -1
• In Na2O:
♦ O atom gains two electrons and thus acquires a negative charge of -2
✰ The oxidation state of O in Na2O is also -2
♦ Each of the Na atoms loses one electron and thus acquires a positive charge of +1
✰ The oxidation state of ‘each Na’ in Na2O is also +1
Based on the above discussion, we can write a 'definition for oxidation state'. We can write it in 4 steps:
• This particular element,
♦ may exhibit one oxidation state in a compound
♦ may exhibit another oxidation state in another compound
• For example, oxidation state of O is
♦ +2 in OF2
♦ -2 in Na2O
2. So when we define the oxidation state of an element, we must specify the compound (in which the element is present) also
3. In that compound, all the elements will be having various values of electronegativities
• Elements with higher electronegativities will be able to pull electrons to themselves
• So we can write:
♦ Elements with higher electronegativities will acquire negative charges
♦ Elements with lower electronegativities will acquire positive charges
4. Now we concentrate on the 'element under consideration'
■ The charge acquired by that element ‘based on the electronegativity consideration’ is called the oxidation state of that element
■ We can write the definition in a brief form also:
The oxidation state of an element in a particular compound can be defined as the charge acquired by it’s atom on the basis of electronegativity considerations from other atoms in the molecule
The oxidation state of an element in a particular compound can be defined as the charge acquired by it’s atom on the basis of electronegativity considerations from other atoms in the molecule
• In the next section, we will see some details about valence
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