In the previous section, we saw the steps to draw Lewis dot structures of 'polyatomic molecules'. In this section, we will see the steps to draw Lewis dot structures of 'polyatomic ions'
Some examples of polyatomic ions are: NH4+, CO32-
The following 6 steps will help us to make a Lewis dot structure of such ions
Step 1: Finding the number of dots
• In a Lewis dot structure, we see a large number of dots
• So the first step will be to determine the answer to this question:
How many dots will be present in the final structure?
■ The answer is simple:
• All ‘available valence electrons’ will be present in the final structure
♦ But anions will have some extra number of electrons. So final number will increase
♦ Also, cations have a deficiency of electrons. So final number will decrease
■ But, instead of calculating the 'final number', it is convenient to keep the numbers separate, as items (a) and (b) shown below:
(a) Total number of ‘available valence electrons’
(b) Number of electrons to be added/subtracted
• Final number = (a) ± (b)
■ Let us see an example. We will write it in steps:
(i) Consider the ion CO32-
• C has the electronic configuration 1s22s22p2
♦ So it has 4 valence electrons
• O has the electronic configuration 1s22s22p4
♦ So it has 6 valence electrons
(ii) So the total number of ‘available valence electrons’ = [4 + (6 × 3)] = 22
(iii) But there are two extra electrons (indicated by the charge of -2)
(iv) So we can write the numbers separately:
(a) Total number of ‘available valence electrons’ = 22
(b) Number of electrons to be added = 2
(v) Final number = [(a) ± (b)] = [(a) + (b)] = [22 + 2] = 24
Thus, there will be 24 dots in the final structure
■ Let us see another example. We will write it in steps:
(i) Consider the ion NH4+
• N has the electronic configuration 1s22s22p3
♦ So it has 5 valence electrons
• H has the electronic configuration 1s1
♦ So it has 1 valence electron
(ii) So the total number of ‘available valence electrons’ = [5 + (4 × 1)] = 9
(iii) But there is a deficiency of one electron (indicated by the charge of +1)
(iv) So we can write the numbers separately:
(a) Total number of ‘available valence electrons’ = 9
(b) Number of electrons to be subtracted = 1
(v) Final number = [(a) ± (b)] = [(a) - (b)] = [9 - 1] = 8
Thus, there will be 8 dots in the final structure
Step 2: The skeletal structure
(We have seen this in the previous section. However, we will write the steps again)
• Draw the skeletal structure of the ion
♦ The least electronegative atom will be the central atom
♦ The other atoms will be distributed around the central atom
We have seen examples in the previous section
Step 3: Preliminary single bonds
(We have seen this in the previous section. However, we will write the steps again)
• Connect the central atom to all the surrounding atoms with a single bond ('─')
• There will be at least one bond between atoms. So we can confidently put a ('─') at all possible connections
• The single bonds that we put in this step can be called preliminary bond
♦ This is because, in later steps, these single bonds may have to be changed to double or triple bonds
Step 4: Preliminary distribution of electrons
• Distribute all the 'available valence electrons' around the atoms
✰ Note that, it is the 'available valence electrons' that we distribute in this step
✰ We do not consider the extra or deficient electrons in this step
♦ Distribute the electrons in such a way that, the outer atoms get octet (duplet in case of H atoms) first
♦ Then give the remaining electrons to the central atom
• The distribution that we do in this step can be called preliminary distribution
♦ This is because, in later steps, this distribution may have to be changed
Step 5: Check for duplet and octet
• Check whether all H atoms (if present), have duplet
• Check whether all other atoms have octet
• If the required duplets and octets are not obtained, change the single bonds to double or triple bonds as necessary
■ In this step, we will see an interesting result:
♦ In the case of anions:
✰ The complete octet/duplet will be obtained only when we put the extra electrons
♦ In the case of cations:
✰ The complete octet/duplet will be obtained only when we take away the deficient electrons
• By the end of this step, the final structure should emerge
♦ All H atoms (if present), should be having duplet
♦ All other atoms should be having octet
Step 6: Check the number of dots
The total number of dots in the final structure must be same as the number obtained in step 1
Example 1: CO32-
Step 1: Finding the number of dots
• Number of valence electrons of C = 4
• Number of valence electrons of O = 6
• So total number of valence electrons = [4+(3 × 6)] = 22
• Two extra electrons are also present
■ We will write the number as two items:
(a) Total number of ‘available valence electrons’ = 22
(b) Number of electrons to be added = 2
• Final number = [(a) ± (b)] = [(a) + (b)] = [22 + 2] = 24
Step 2: The skeletal structure
• C is less electronegative than O
(Electronegativity of C is 2.55 and that of O is 3.44)
♦ So C is the central atom
♦ The three O atoms will be distributed around the C atom
• The skeletal structure is shown in fig.4.19(a) below:
Step 3: Preliminary single bonds
• The four atoms are joined by a '─' as shown in fig.4.19(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' = 22
• 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 C 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 = (22-16) = 6
♦ These 6 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 6 electrons
♦ So this atom needs 2 more electrons
• The C atom has got only 6 electrons
♦ So this atom also needs 2 more electrons
(i) Rearrangement: Change the preliminary single bond
♦ Change the top single bond to double bond as shown in the fig.d
(ii) Two electrons from the top O is used for making the new bond
♦ Now the left and right side O atoms have octet
♦ The C atom also has octet
♦ But the top O atom has got only 6 electrons
(iii) All the 22 electrons are used up. Still, complete octet is not achieved
• So, we will need external electrons
(iv) Get two external electrons from any suitable source
• Give them to the top O atom
• This is shown in fig.e
• Now all atoms have octet
(v) But the two external electrons will create a charge of -2
• So we put the structure inside square brackets and put a -2 at the top right corner
Step 6: Check the number of dots
♦ Total number of dots in fig.e = 24
♦ Number calculated in step 1 = 24
■ What 'source' would give two electrons so that, all atoms in CO32- have octet?
The answer can be written using an example. We will write it in steps:
(i) Consider the Ca atom
• It can donate two electrons and become Ca2+ ion
• The Ca2+ ion is stable because, it has octet
(ii) The two electrons donated by Ca can be used to make all the atoms in CO32- octet
• When all the atoms have octet, the 'CO32- ion as a whole' becomes stable
(iii) So we have two stable ions:
• The positive ion: Ca2+
• The negative ion: CO32-
(iv) An electrostatic force of attraction comes into effect between the two oppositely charged ions
• So the two ions begin to act together as a single unit
• We will not be able to separate the two ions easily from each other
(v) As a result, we get calcium carbonate (CaCO3)
• This is similar to the formation of NaCl from Na+ and Cl- ions
♦ The only difference is that, in our present case, one of the ions is a 'polyatomic ion'
Example 2: NO2-
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+(2 × 6)] = 17
• One extra electron is also present
■ We will write the number as two items:
(a) Total number of ‘available valence electrons’ = 17
(b) Number of electrons to be added = 1
• Final number = [(a) ± (b)] = [(a) + (b)] = [17 + 1] = 18
Step 2: The skeletal structure
• N is less electronegative than O
(Electronegativity of N is 3.04 and that of O is 3.44)
♦ So N is the central atom
♦ The two O atoms will be distributed around the N atom
• The skeletal structure is shown in fig.4.20(a) below:
Step 3: Preliminary single bonds
• The three atoms are joined by a '─' as shown in fig.4.20(b) above
Step 4: Preliminary distribution of electrons
(Remember that, the 'available valence electrons' are distributed in this step)
• First make the two outer O atoms octet
• Then give the remaining electrons to the N 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
• The two 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 = (17-16) = 1
♦ This 1 electron is given to the N atom
• There are no more electrons to distribute
Step 5: Check for octet
• The two O atoms have got 8 electrons each
• The N atom has got only 5 electrons
♦ So this atom needs 3 more electrons
(i) Rearrangement: Change the preliminary single bond
♦ Change the left side single bond to double bond as shown in the fig.d
(ii) Two electrons from the left side O is used for making the new bond
♦ Now the left and right side O atoms have octet
♦ But the N atom has got only 7 electrons
(iii) All the 17 electrons are used up. Still, complete octet is not achieved
• So, we will need one external electron
(iv) Get one external electron from any suitable source
• Give it to the N atom
• This is shown in fig.e
• Now all atoms have octet
(v) But the one external electron will create a charge of -1
• So we put the structure inside square brackets and put a '-' at the top right corner
Step 6: Check the number of dots
♦ Total number of dots in fig.e = 18
♦ Number calculated in step 1 = 18
■ What 'source' would give one electron so that, all atoms in NO2- have octet?
The answer can be written using an example. We will write it in steps:
(i) Consider the Na atom
• It can donate one electron and become Na+ ion
• The Na+ ion is stable because, it has octet
(ii) The electron donated by Na can be used to make all the atoms in NO2- octet
• When all the atoms have octet, the 'NO2- ion as a whole' becomes stable
(iii) So we have two stable ions:
• The positive ion: Na+
• The negative ion: NO2-
(iv) An electrostatic force of attraction comes into effect between the two oppositely charged ions
• So the two ions begin to act together as a single unit
• We will not be able to separate the two ions easily from each other
(v) As a result, we get sodium nitrite (NaNO2)
• This is similar to the formation of NaCl from Na+ and Cl- ions
♦ The only difference is that, in our present case, one of the ions is a 'polyatomic ion'
Some examples of polyatomic ions are: NH4+, CO32-
The following 6 steps will help us to make a Lewis dot structure of such ions
Step 1: Finding the number of dots
• In a Lewis dot structure, we see a large number of dots
• So the first step will be to determine the answer to this question:
How many dots will be present in the final structure?
■ The answer is simple:
• All ‘available valence electrons’ will be present in the final structure
♦ But anions will have some extra number of electrons. So final number will increase
♦ Also, cations have a deficiency of electrons. So final number will decrease
■ But, instead of calculating the 'final number', it is convenient to keep the numbers separate, as items (a) and (b) shown below:
(a) Total number of ‘available valence electrons’
(b) Number of electrons to be added/subtracted
• Final number = (a) ± (b)
■ Let us see an example. We will write it in steps:
(i) Consider the ion CO32-
• C has the electronic configuration 1s22s22p2
♦ So it has 4 valence electrons
• O has the electronic configuration 1s22s22p4
♦ So it has 6 valence electrons
(ii) So the total number of ‘available valence electrons’ = [4 + (6 × 3)] = 22
(iii) But there are two extra electrons (indicated by the charge of -2)
(iv) So we can write the numbers separately:
(a) Total number of ‘available valence electrons’ = 22
(b) Number of electrons to be added = 2
(v) Final number = [(a) ± (b)] = [(a) + (b)] = [22 + 2] = 24
Thus, there will be 24 dots in the final structure
■ Let us see another example. We will write it in steps:
(i) Consider the ion NH4+
• N has the electronic configuration 1s22s22p3
♦ So it has 5 valence electrons
• H has the electronic configuration 1s1
♦ So it has 1 valence electron
(ii) So the total number of ‘available valence electrons’ = [5 + (4 × 1)] = 9
(iii) But there is a deficiency of one electron (indicated by the charge of +1)
(iv) So we can write the numbers separately:
(a) Total number of ‘available valence electrons’ = 9
(b) Number of electrons to be subtracted = 1
(v) Final number = [(a) ± (b)] = [(a) - (b)] = [9 - 1] = 8
Thus, there will be 8 dots in the final structure
Step 2: The skeletal structure
(We have seen this in the previous section. However, we will write the steps again)
• Draw the skeletal structure of the ion
♦ The least electronegative atom will be the central atom
♦ The other atoms will be distributed around the central atom
We have seen examples in the previous section
Step 3: Preliminary single bonds
(We have seen this in the previous section. However, we will write the steps again)
• Connect the central atom to all the surrounding atoms with a single bond ('─')
• There will be at least one bond between atoms. So we can confidently put a ('─') at all possible connections
• The single bonds that we put in this step can be called preliminary bond
♦ This is because, in later steps, these single bonds may have to be changed to double or triple bonds
Step 4: Preliminary distribution of electrons
• Distribute all the 'available valence electrons' around the atoms
✰ Note that, it is the 'available valence electrons' that we distribute in this step
✰ We do not consider the extra or deficient electrons in this step
♦ Distribute the electrons in such a way that, the outer atoms get octet (duplet in case of H atoms) first
♦ Then give the remaining electrons to the central atom
• The distribution that we do in this step can be called preliminary distribution
♦ This is because, in later steps, this distribution may have to be changed
Step 5: Check for duplet and octet
• Check whether all H atoms (if present), have duplet
• Check whether all other atoms have octet
• If the required duplets and octets are not obtained, change the single bonds to double or triple bonds as necessary
■ In this step, we will see an interesting result:
♦ In the case of anions:
✰ The complete octet/duplet will be obtained only when we put the extra electrons
♦ In the case of cations:
✰ The complete octet/duplet will be obtained only when we take away the deficient electrons
• By the end of this step, the final structure should emerge
♦ All H atoms (if present), should be having duplet
♦ All other atoms should be having octet
Step 6: Check the number of dots
The total number of dots in the final structure must be same as the number obtained in step 1
Let us now apply the above rules to some ions:
Step 1: Finding the number of dots
• Number of valence electrons of C = 4
• Number of valence electrons of O = 6
• So total number of valence electrons = [4+(3 × 6)] = 22
• Two extra electrons are also present
■ We will write the number as two items:
(a) Total number of ‘available valence electrons’ = 22
(b) Number of electrons to be added = 2
• Final number = [(a) ± (b)] = [(a) + (b)] = [22 + 2] = 24
Step 2: The skeletal structure
• C is less electronegative than O
(Electronegativity of C is 2.55 and that of O is 3.44)
♦ So C is the central atom
♦ The three O atoms will be distributed around the C atom
• The skeletal structure is shown in fig.4.19(a) below:
 |
| Fig.4.19 |
• The four atoms are joined by a '─' as shown in fig.4.19(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' = 22
• 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 C 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 = (22-16) = 6
♦ These 6 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 6 electrons
♦ So this atom needs 2 more electrons
• The C atom has got only 6 electrons
♦ So this atom also needs 2 more electrons
(i) Rearrangement: Change the preliminary single bond
♦ Change the top single bond to double bond as shown in the fig.d
(ii) Two electrons from the top O is used for making the new bond
♦ Now the left and right side O atoms have octet
♦ The C atom also has octet
♦ But the top O atom has got only 6 electrons
(iii) All the 22 electrons are used up. Still, complete octet is not achieved
• So, we will need external electrons
(iv) Get two external electrons from any suitable source
• Give them to the top O atom
• This is shown in fig.e
• Now all atoms have octet
(v) But the two external electrons will create a charge of -2
• So we put the structure inside square brackets and put a -2 at the top right corner
Step 6: Check the number of dots
♦ Total number of dots in fig.e = 24
♦ Number calculated in step 1 = 24
Now a question arises:
The answer can be written using an example. We will write it in steps:
(i) Consider the Ca atom
• It can donate two electrons and become Ca2+ ion
• The Ca2+ ion is stable because, it has octet
(ii) The two electrons donated by Ca can be used to make all the atoms in CO32- octet
• When all the atoms have octet, the 'CO32- ion as a whole' becomes stable
(iii) So we have two stable ions:
• The positive ion: Ca2+
• The negative ion: CO32-
(iv) An electrostatic force of attraction comes into effect between the two oppositely charged ions
• So the two ions begin to act together as a single unit
• We will not be able to separate the two ions easily from each other
(v) As a result, we get calcium carbonate (CaCO3)
• This is similar to the formation of NaCl from Na+ and Cl- ions
♦ The only difference is that, in our present case, one of the ions is a 'polyatomic ion'
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+(2 × 6)] = 17
• One extra electron is also present
■ We will write the number as two items:
(a) Total number of ‘available valence electrons’ = 17
(b) Number of electrons to be added = 1
• Final number = [(a) ± (b)] = [(a) + (b)] = [17 + 1] = 18
Step 2: The skeletal structure
• N is less electronegative than O
(Electronegativity of N is 3.04 and that of O is 3.44)
♦ So N is the central atom
♦ The two O atoms will be distributed around the N atom
• The skeletal structure is shown in fig.4.20(a) below:
 |
| Fig.4.20 |
• The three atoms are joined by a '─' as shown in fig.4.20(b) above
Step 4: Preliminary distribution of electrons
(Remember that, the 'available valence electrons' are distributed in this step)
• First make the two outer O atoms octet
• Then give the remaining electrons to the N 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
• The two 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 = (17-16) = 1
♦ This 1 electron is given to the N atom
• There are no more electrons to distribute
Step 5: Check for octet
• The two O atoms have got 8 electrons each
• The N atom has got only 5 electrons
♦ So this atom needs 3 more electrons
(i) Rearrangement: Change the preliminary single bond
♦ Change the left side single bond to double bond as shown in the fig.d
(ii) Two electrons from the left side O is used for making the new bond
♦ Now the left and right side O atoms have octet
♦ But the N atom has got only 7 electrons
(iii) All the 17 electrons are used up. Still, complete octet is not achieved
• So, we will need one external electron
(iv) Get one external electron from any suitable source
• Give it to the N atom
• This is shown in fig.e
• Now all atoms have octet
(v) But the one external electron will create a charge of -1
• So we put the structure inside square brackets and put a '-' at the top right corner
Step 6: Check the number of dots
♦ Total number of dots in fig.e = 18
♦ Number calculated in step 1 = 18
Now a question arises:
The answer can be written using an example. We will write it in steps:
(i) Consider the Na atom
• It can donate one electron and become Na+ ion
• The Na+ ion is stable because, it has octet
(ii) The electron donated by Na can be used to make all the atoms in NO2- octet
• When all the atoms have octet, the 'NO2- ion as a whole' becomes stable
(iii) So we have two stable ions:
• The positive ion: Na+
• The negative ion: NO2-
(iv) An electrostatic force of attraction comes into effect between the two oppositely charged ions
• So the two ions begin to act together as a single unit
• We will not be able to separate the two ions easily from each other
(v) As a result, we get sodium nitrite (NaNO2)
• This is similar to the formation of NaCl from Na+ and Cl- ions
♦ The only difference is that, in our present case, one of the ions is a 'polyatomic ion'
• The above discussion will enable us to 'draw Lewis dot structures' of some polyatomic ions
• In the next section, we will see Formal charge
• In the next section, we will see Formal charge
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