We were learning about resonance structures in section 4.9, We saw the examples of ozone molecule, carbon dioxide molecule and carbonate ion. In this section, we will see two more examples
1. Let us first draw the Lewis dot structure of SO3 (Sulfur trioxide)
(We have seen the steps in an earlier section 4.2)
Step 1: Finding the number of dots
• Number of valence electrons of C = 6
• Number of valence electrons of O = 6
• So total number of valence electrons = [6+(3 × 6)] = 24
Step 2: The skeletal structure
• The skeletal structure is shown in fig.4.56(a) below:
Step 3: Preliminary single bonds
• The four atoms are joined by '─' as shown in fig.4.55(b) above
Step 4: Preliminary distribution of electrons
• First make the three outer O atoms octet
• This is shown in fig.c
• In the fig.c, we see that:
♦ The valence electrons of the O atoms are shown in green color
♦ The valence electrons of the C atom are shown in red color
• All the O atoms have 8 electrons (including the one red dot in the single bonds)
♦ So the three O atoms use up (3 × 8) = 24 electrons
♦ The number of remaining electrons = (24-24) = 0
• There are no more electrons to distribute
Step 5: Check for octet
• All the O atoms have got 8 electrons each
• The S atom has got only 6 electrons
♦ So this atom needs 2 more electrons
■ Rearrangement: Change the preliminary single bond
♦ Take a lone pair from the top O atom
♦ Using those electrons, change the top single bond to double bond as shown in the fig.d
• Now all atoms have octet
• The structure in fig.4.56(d) is stable
2. Another possible rearrangement
• We already know how to obtain the structure in fig.4.56(d) above
♦ We obtained it by working from fig.4.56(c)
• That same fig.4.56(c) is shown again in fig.4.57(c) below:
• Earlier, we took a pair from the top O atom
♦ This time, we take a pair from the left O atom and make a double bond
♦ This is shown in fig.4.57(d')
• The structure in fig.4.57(d') is stable
3. Yet another possible rearrangement
• We already know how to obtain the structure in fig.4.56(d) above
♦ We obtained it by working from fig.4.56(c)
• That same fig.4.56(c) is shown again in fig.4.58(c) below:
• Earlier, we took a pair from the top O atom
♦ This time, we take a pair from the right O atom and make a double bond
♦ This is shown in fig.4.58(d'')
• The structure in fig.4.58(d'') is stable
4. So we have three possible structures of SO3
♦ The structure in fig.4.56(d)
♦ The structure in fig.4.57(d')
♦ The structure in fig.4.58(d'')
• They are shown together in fig.4.59 below:
5. Now the next question arises:
■ In reality, which is the correct form in which SO3 exists? Fig.4.59 (d), (d') or (d'')?
• Let us try to find the answer:
(i) A S-O single bond will have a certain length
(ii) A S=O double bond will have a different length
(iii) With this information, we examine the bond lengths in an actual SO3 molecule
• Surprisingly, the actual values are different from (i) and (ii)
• In fact there are no 'values'. There is only one value
• The distance between S and O atoms in all the three pairs are the same
6. The structures in figs (d) (d') and (d'') are called canonical structures of SO3
• They are also called resonance structures of SO3
7. Resonance structures are indicated by giving double headed arrows between them
1. Let us first draw the Lewis dot structure of NO32- (nitrate ion)
(We have seen the steps in an earlier section 4.3)
Step 1: Finding the number of dots
• Number of valence electrons of N = 5
• Number of valence electrons of O = 6
• So total number of valence electrons = [5+(3 × 6)] = 23
• One extra electron is also present
■ We will write the number as two items:
(a) Total number of ‘available valence electrons’ = 23
(b) Number of electrons to be added = 1
• Final number = [(a) ± (b)] = [(a) + (b)] = [23 + 1] = 24
Step 2: The skeletal structure
• The skeletal structure is shown in fig.4.60(a) below:
Step 3: Preliminary single bonds
• The four atoms are joined by '─' as shown in fig.4.60(b) above
Step 4: Preliminary distribution of electrons
(Remember that, the 'available valence electrons' are distributed in this step)
• First make the three outer O atoms octet
♦ For that (3×8) = 24 electrons will be required
♦ But the number of 'available valence electrons' = 23
• So first, we will make the left and right O atoms octet
• Then give the remaining electrons to the top O atom
• This is shown in fig.c
• In the fig.c, we see that:
♦ The valence electrons of the O atoms are shown in green color
♦ The valence electrons of the N atom are shown in red color
• Left and right side O atoms have 8 electrons (including the one red dot in the single bonds)
♦ So the two O atoms use up (2 × 8) = 16 electrons
♦ The number of remaining electrons = (23-16) = 7
♦ These 7 electrons are given to the top O atom
• There are no more electrons to distribute
Step 5: Check for octet
• The left and right side O atoms have got 8 electrons each
• The top O atom has got only 7 electrons
♦ So this atom needs 1 more electron
• The N atom has got only 6 electrons
♦ So this atom also needs 2 more electrons
■ Rearrangement: Change the preliminary single bond
♦ Take a lone pair from the top O atom
♦ Using those electrons, change the top single bond to double bond as shown in the fig.d
♦ Now the left and right side O atoms have octet
♦ The N atom also has octet
♦ But the top O atom has got only 7 electrons
• All the 23 electrons are used up. Still, complete octet is not achieved
• So, we will need an external electron
• Get one external electron from any suitable source
• Give it to the top O atom
• This is shown in fig.e
• Now all atoms have octet
(v) But the external electron will create a charge of -1
• So we put the structure inside square brackets and put a -1 at the top right corner
• The structure in fig.4.60(e) is stable
2. Another possible rearrangement
• We already know how to obtain the structure in fig.4.60(e) above
♦ We obtained it by working from fig.4.60(b)
• That same fig.4.60(b) is shown again in fig.4.61(b) below:
• Earlier, we made the left and right O atoms octet
♦ This time, we make the top and right O atoms octet. This is shown in fig.4.61(c')
• Next we take two electrons from the left O atom and make a double bond
♦ This is shown in fig.4.61(d')
• Finally, we add the extra electron to the left side O atom to attain octet
♦ This is shown in fig.4.61(e')
• The structure in fig.4.61(e') is stable
3. Yet another possible rearrangement
• We already know how to obtain the structure in fig.4.60(e) above
♦ We obtained it by working from fig.4.60(b)
• That same fig.4.60(b) is shown again in fig.4.62(b) below:
• Earlier, we made the left and right O atoms octet
♦ This time, we make the top and left O atoms octet. This is shown in fig.4.62(c'')
• Next we take two electrons from the right O atom and make a double bond
♦ This is shown in fig.4.62(d'')
• Finally, we add the extra electron to the right side O atom to attain octet
♦ This is shown in fig.4.62(e'')
• The structure in fig.4.62(e'') is stable
4. So we have three possible structures of NO3-
♦ The structure in fig.4.60(e)
♦ The structure in fig.4.61(e')
♦ The structure in fig.4.62(e'')
• They are shown together in fig.4.63 below:
5. Now the next question arises:
■ In reality, which is the correct form in which NO3- exists? Fig.4.63 (e), (e') or (e'')?
• Let us try to find the answer:
(i) A N-O single bond will have a certain length
(ii) A N=O double bond will have a different length
(iii) With this information, we examine the bond lengths in an actual NO3- ion
• Surprisingly, the actual values are different from (i) and (ii)
• In fact there are no 'values'. There is only one value
• The distance between N and O atoms in all the three pairs are the same
6. The structures in figs (e) (e') and (e'') are called canonical structures of NO3-
• They are also called resonance structures of NO3-
7. Resonance structures are indicated by giving double headed arrows between them
Resonance structures of Sulfur trioxide
1. Let us first draw the Lewis dot structure of SO3 (Sulfur trioxide)
(We have seen the steps in an earlier section 4.2)
Step 1: Finding the number of dots
• Number of valence electrons of C = 6
• Number of valence electrons of O = 6
• So total number of valence electrons = [6+(3 × 6)] = 24
Step 2: The skeletal structure
• The skeletal structure is shown in fig.4.56(a) below:
 |
| Fig.4.56 |
• The four atoms are joined by '─' as shown in fig.4.55(b) above
Step 4: Preliminary distribution of electrons
• First make the three outer O atoms octet
• This is shown in fig.c
• In the fig.c, we see that:
♦ The valence electrons of the O atoms are shown in green color
♦ The valence electrons of the C atom are shown in red color
• All the O atoms have 8 electrons (including the one red dot in the single bonds)
♦ So the three O atoms use up (3 × 8) = 24 electrons
♦ The number of remaining electrons = (24-24) = 0
• There are no more electrons to distribute
Step 5: Check for octet
• All the O atoms have got 8 electrons each
• The S atom has got only 6 electrons
♦ So this atom needs 2 more electrons
■ Rearrangement: Change the preliminary single bond
♦ Take a lone pair from the top O atom
♦ Using those electrons, change the top single bond to double bond as shown in the fig.d
• Now all atoms have octet
• The structure in fig.4.56(d) is stable
2. Another possible rearrangement
• We already know how to obtain the structure in fig.4.56(d) above
♦ We obtained it by working from fig.4.56(c)
• That same fig.4.56(c) is shown again in fig.4.57(c) below:
 |
| Fig.4.57 |
♦ This time, we take a pair from the left O atom and make a double bond
♦ This is shown in fig.4.57(d')
• The structure in fig.4.57(d') is stable
3. Yet another possible rearrangement
• We already know how to obtain the structure in fig.4.56(d) above
♦ We obtained it by working from fig.4.56(c)
• That same fig.4.56(c) is shown again in fig.4.58(c) below:
 |
| Fig.4.58 |
♦ This time, we take a pair from the right O atom and make a double bond
♦ This is shown in fig.4.58(d'')
• The structure in fig.4.58(d'') is stable
4. So we have three possible structures of SO3
♦ The structure in fig.4.56(d)
♦ The structure in fig.4.57(d')
♦ The structure in fig.4.58(d'')
• They are shown together in fig.4.59 below:
 |
| Fig.4.59 |
■ In reality, which is the correct form in which SO3 exists? Fig.4.59 (d), (d') or (d'')?
• Let us try to find the answer:
(i) A S-O single bond will have a certain length
(ii) A S=O double bond will have a different length
(iii) With this information, we examine the bond lengths in an actual SO3 molecule
• Surprisingly, the actual values are different from (i) and (ii)
• In fact there are no 'values'. There is only one value
• The distance between S and O atoms in all the three pairs are the same
6. The structures in figs (d) (d') and (d'') are called canonical structures of SO3
• They are also called resonance structures of SO3
7. Resonance structures are indicated by giving double headed arrows between them
Resonance structures of nitrate ion
(We have seen the steps in an earlier section 4.3)
Step 1: Finding the number of dots
• Number of valence electrons of N = 5
• Number of valence electrons of O = 6
• So total number of valence electrons = [5+(3 × 6)] = 23
• One extra electron is also present
■ We will write the number as two items:
(a) Total number of ‘available valence electrons’ = 23
(b) Number of electrons to be added = 1
• Final number = [(a) ± (b)] = [(a) + (b)] = [23 + 1] = 24
Step 2: The skeletal structure
• The skeletal structure is shown in fig.4.60(a) below:
 |
| Fig.4.60 |
• The four atoms are joined by '─' as shown in fig.4.60(b) above
Step 4: Preliminary distribution of electrons
(Remember that, the 'available valence electrons' are distributed in this step)
• First make the three outer O atoms octet
♦ For that (3×8) = 24 electrons will be required
♦ But the number of 'available valence electrons' = 23
• So first, we will make the left and right O atoms octet
• Then give the remaining electrons to the top O atom
• This is shown in fig.c
• In the fig.c, we see that:
♦ The valence electrons of the O atoms are shown in green color
♦ The valence electrons of the N atom are shown in red color
• Left and right side O atoms have 8 electrons (including the one red dot in the single bonds)
♦ So the two O atoms use up (2 × 8) = 16 electrons
♦ The number of remaining electrons = (23-16) = 7
♦ These 7 electrons are given to the top O atom
• There are no more electrons to distribute
Step 5: Check for octet
• The left and right side O atoms have got 8 electrons each
• The top O atom has got only 7 electrons
♦ So this atom needs 1 more electron
• The N atom has got only 6 electrons
♦ So this atom also needs 2 more electrons
■ Rearrangement: Change the preliminary single bond
♦ Take a lone pair from the top O atom
♦ Using those electrons, change the top single bond to double bond as shown in the fig.d
♦ Now the left and right side O atoms have octet
♦ The N atom also has octet
♦ But the top O atom has got only 7 electrons
• All the 23 electrons are used up. Still, complete octet is not achieved
• So, we will need an external electron
• Get one external electron from any suitable source
• Give it to the top O atom
• This is shown in fig.e
• Now all atoms have octet
(v) But the external electron will create a charge of -1
• So we put the structure inside square brackets and put a -1 at the top right corner
• The structure in fig.4.60(e) is stable
2. Another possible rearrangement
• We already know how to obtain the structure in fig.4.60(e) above
♦ We obtained it by working from fig.4.60(b)
• That same fig.4.60(b) is shown again in fig.4.61(b) below:
 |
| Fig.4.61 |
♦ This time, we make the top and right O atoms octet. This is shown in fig.4.61(c')
• Next we take two electrons from the left O atom and make a double bond
♦ This is shown in fig.4.61(d')
• Finally, we add the extra electron to the left side O atom to attain octet
♦ This is shown in fig.4.61(e')
• The structure in fig.4.61(e') is stable
3. Yet another possible rearrangement
• We already know how to obtain the structure in fig.4.60(e) above
♦ We obtained it by working from fig.4.60(b)
• That same fig.4.60(b) is shown again in fig.4.62(b) below:
 |
| Fig.4.62 |
♦ This time, we make the top and left O atoms octet. This is shown in fig.4.62(c'')
• Next we take two electrons from the right O atom and make a double bond
♦ This is shown in fig.4.62(d'')
• Finally, we add the extra electron to the right side O atom to attain octet
♦ This is shown in fig.4.62(e'')
• The structure in fig.4.62(e'') is stable
4. So we have three possible structures of NO3-
♦ The structure in fig.4.60(e)
♦ The structure in fig.4.61(e')
♦ The structure in fig.4.62(e'')
• They are shown together in fig.4.63 below:
 |
| Fig.4.63 |
■ In reality, which is the correct form in which NO3- exists? Fig.4.63 (e), (e') or (e'')?
• Let us try to find the answer:
(i) A N-O single bond will have a certain length
(ii) A N=O double bond will have a different length
(iii) With this information, we examine the bond lengths in an actual NO3- ion
• Surprisingly, the actual values are different from (i) and (ii)
• In fact there are no 'values'. There is only one value
• The distance between N and O atoms in all the three pairs are the same
6. The structures in figs (e) (e') and (e'') are called canonical structures of NO3-
• They are also called resonance structures of NO3-
7. Resonance structures are indicated by giving double headed arrows between them
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