Chapter 8.3 - Alkynes Homologous series

In the previous section, we saw the properties of the alkene homologous series. In this section we will see one more homologous series.

In the previous section we started with C₂H₄, and saw that it was the first member of alkene series. In this section we will start with another compound C₂H₂. It is a molecule of 'ethyne'. We want to know how the atoms are arranged in the molecule. Let us find out:
Step 1: In C₂H₂, there are two carbon atoms. The first step is to connect those two, using a single bond. It is shown in fig.8.13(a) below:

Fig.8.13

Step 2: • Consider the carbon atom on the left in fig.(a). It is in single bond with the right side carbon atom. So it has acquired one electron through this single bond. Three more electrons are required. 
• There are not enough hydrogen atoms to supply these 3 electrons. There are only a total of two hydrogen atoms. 
• So we will equally share the 2 hydrogen atoms and see what happens. This step is shown in fig.8.13(b)
Step 3: In the fig.(b), the two hydrogen atoms have attained octet. But both the carbon atoms still need 3 electrons each. This can be supplied by changing the single bond between the carbon atoms to a triple bond. This is shown in fig.(c). Now all the atoms have attained octet. 
■ With the above three steps, we get the final arrangement of atoms in a molecule of C₂H₂. The electron dot diagram is shown in the fig.8.14 below:

Fig.8.14

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Now we will consider another example: C₃H₄. We will write the required steps as in the previous example.

Step 1: In C₃H₄, there are 3 carbon atoms. The first step is to connect the three using a single bond. It is shown in fig.8.15(a) below:

Fig.8.15

Step 2: • Consider the left most carbon atom. It is in single bond with the middle carbon atom. So it has acquired one electron through this single bond. Three more electrons are required.
• If we bond it to three hydrogen atoms, the middle carbon atom and the right side carbon atom will be left with only one hydrogen. Because there are only 4 hydrogen atoms in total. We cannot share 1 hydrogen atoms among 2 carbon atoms. So we can not give 3 hydrogen atoms to the left side carbon.
• Also, we cannot share the total 4 hydrogen atoms among the total 3 carbon atoms equally. 
• So we will give importance to symmetry and see what happens. That is., we will give 2 electrons to the middle carbon atom. And one hydrogen atom each to the carbon atoms on left and right. This step is shown in fig.8.15(b)
Step 3: In fig.(b), the requirements of all the hydrogen atoms are satisfied. There is also symmetry. But consider the left most carbon atom. It needs two more electrons. So change the single bond between it and the middle carbon atom to a triple bond. This step is shown in fig.c.
Step 4: • In fig.c, all the hydrogen atoms has octet
• The left most carbon atom has octet
• But the middle carbon atom has two excess electrons. [3 single bonds and 1 triple bond]
• Also the right most carbon atom is in shortage of two electrons
• So remove two hydrogen from the middle carbon
• Bond this removed hydrogen atoms to the right most carbon atom.
• This step is shown in fig.d. Now all atoms have octet

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So we determined the arrangement of atoms in the following two molecules:

(i) Molecule with 2 carbon atoms and 2 hydrogen atoms
(ii) Molecule with 3 carbon atoms and 4 hydrogen atoms
• A pattern is beginning to emerge. The next two cases will be:
(iii) Molecule with 4 carbon atoms and 6 hydrogen atoms
(iv) Molecule with 5 carbon atoms and 8 hydrogen atoms
• Note that number of carbon atoms is increasing by 1. Also, number of hydrogen atoms is increasing by 2
• It is like a series. Let us draw the arrangement for the first six members of the series:

With the above table, the analysis about the series becomes easy. Let us learn the properties of the series:
Property 1:
• The 1^st member of the series has 2 carbon atoms
And 2 hydrogen atoms
• The 2^nd member of the series has 3 carbon atoms
And 4 hydrogen atoms
• The 3^rd member of the series has 4 carbon atoms
And 6 hydrogen atoms
So on...
■ So there exists a definite relation between the two quantities below:
• No. of carbon atoms
• No. of hydrogen atoms
■ If the no. of carbon atoms is 'n', then the no. of hydrogen atoms will be (2n-2)
So we can represent the members of the series by a general formula: C_nH_(2n-2)
Property 2:
• Consider the second row. Consider the item in the 'Condensed formula' column in this row. We have: CH = C – CH₃
• Now consider the third row, same column. We have: CH = C – CH₂ – CH₃
The difference is a CH₂ group
• Now consider the fourth row, same column. We have:   CH = C – CH₂ – CH₂ – CH₃
Here also, the difference between the previous third row is a CH₂ group
• We will find the same difference through out the series. That is:
■ Successive members differ by a CH₂ group

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In the previous sections, we saw that a series of compounds having the above two properties is called a homologous series. So this series is also a homologous series. The name given to this series is: Alkynes.

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In the next section, we will learn about hydrocarbons. 

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Chapter 8.2 - Alkene Homologous series

In the previous section, we saw the properties of a homologous series. We also saw the Alkane series. In this section we will see another homologous series.

In the previous section we started with CH₄, and saw that it was the first member of a series. In this section we will start with another compound C₂H₄. It is a molecule of 'ethene'. We want to know how the atoms are arranged in the molecule. Let us find out:
Step 1: In C₂H₄, there are two carbon atoms. The first step is to connect those two, using a single bond. It is shown in fig.8.10(a) below:

Fig.8.10

Step 2: • Consider the carbon atom on the left in fig.(a). It is in single bond with the right side carbon atom. So it has acquired one electron through this single bond. Three more electrons are required. 
• If we bond it to three hydrogen atoms, the right side carbon atom will be left with only one hydrogen. Because there are only 4 hydrogen atoms in total. So we can not give 3 hydrogen atoms to the left side carbon.
• So we will equally share the 4 hydrogen atoms and see what happens. This step is shown in fig.8.10(b)
Step 3: In the fig.(b), all the hydrogen atoms have attained octet. But both the carbon atoms still need one electron each. This can be supplied by changing the single bond between the carbon atoms to a double bond. This is shown in fig.(c). Now all the atoms have attained octet. 
■ With the above three steps, we get the final arrangement of atoms in a molecule of C₂H₄. The electron dot diagram is shown in the fig.8.11 below:

Fig.8.11

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Now we will consider another example: C₃H₆. We will write the required steps as in the previous example.

Step 1: In C₃H₆, there are 3 carbon atoms. The first step is to connect the three using a single bond. It is shown in fig.8.12(a) below:

Fig.8.12

Step 2: • Consider the left most carbon atom. It is in single bond with the middle carbon atom. So it has acquired one electron through this single bond. Three more electrons are required.
• If we bond it to three hydrogen atoms, the middle carbon atom and the right side carbon atom will be left with only three hydrogen. Because there are only 6 hydrogen atoms in total. We cannot share 3 hydrogen atoms among 2 carbon atoms. So we can not give 3 hydrogen atoms to the left side carbon.
• So we will equally share the 6 hydrogen atoms and see what happens. This step is shown in fig.8.12(b)
Step 3: In fig.(b), the requirements of all the hydrogen atoms are satisfied. But consider the left most carbon atom. It needs one more electron. So change the single bond between it and the middle carbon atom to a double bond. This step is shown in fig.c.
Step 4: • In fig.c, all the hydrogen atoms has octet
• The left most carbon atom has octet
• But the middle carbon atom has an excess electron. [3 single bonds and 1 double bond]
• Also the right most carbon atom is in shortage of one electron
• So remove one hydrogen from the middle carbon
• Bond this removed hydrogen atom to the right most carbon atom.
• This step is shown in fig.d. Now all atoms have octet

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So we determined the arrangement of atoms in the following two molecules:

(i) Molecule with 2 carbon atoms and 4 hydrogen atoms
(ii) Molecule with 3 carbon atoms and 6 hydrogen atoms
• A pattern is beginning to emerge. The next two cases will be:
(iii) Molecule with 4 carbon atoms and 8 hydrogen atoms
(iv) Molecule with 5 carbon atoms and 10 hydrogen atoms
• Note that number of carbon atoms is increasing by 1. Also, number of hydrogen atoms is increasing by 2
• It is like a series. Let us tabulate the arrangement for the first six members of the series:

With the above table, the analysis about the series becomes easy. Let us learn the properties of the series:
Property 1:
• The 1^st member of the series has 2 carbon atoms
And 4 hydrogen atoms
• The 2^nd member of the series has 3 carbon atoms
And 6 hydrogen atoms
• The 3^rd member of the series has 4 carbon atoms
And 8 hydrogen atoms
So on...
■ So there exists a definite relation between the two quantities below:
• No. of carbon atoms
• No. of hydrogen atoms
■ If the no. of carbon atoms is 'n', then the no. of hydrogen atoms will be 2n
So we can represent the members of the series by a general formula: C_nH_2n
Property 2:
• Consider the second row. Consider the item in the 'Condensed formula' column in this row. We have: CH₂ = CH – CH₃
• Now consider the third row, same column. We have: CH₂ = CH – CH₂ – CH₃
The difference is a CH₂ group
• Now consider the fourth row, same column. We have:   CH₂ = CH – CH₂  – CH₂ – CH₃
Here also, the difference between the previous third row is a CH₂ group
• We will find the same difference through out the series. That is:
■ Successive members differ by a CH₂ group

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In the previous section, we saw that a series of compounds having the above two properties is called a homologous series. So this series is also a homologous series. The name given to this series is: Alkenes.

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In the next section, we will see one more homologous series. 

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Chapter 8.1 - Properties and definition of Homologous Series

In the previous section, we saw the reasons for formation of very large number of carbon compounds. In this section we will see the arrangement of atoms in Carbon compounds. Based on that we will discuss about Homologous series.

Consider the molecule C₂H₆. It is a molecule of 'ethane'. We want to know how the atoms are arranged in the molecule. Let us find out:
Step 1: In C₂H₆, there are two carbon atoms. The first step is to connect those two, using a single bond. It is shown in fig.8.6(a) below:

Fig.8.6

Step 2: • Consider the carbon atom on the left. It is in single bond with the right side carbon atom. So it has acquired one electron through this single bond. Three more electrons are required. 
• So bond it to three hydrogen atoms. A bond between a carbon atom and a hydrogen atom is possible only through a single bond. This is because hydrogen has only one electron to share.
• Thus the left side carbon atom is bonded to three hydrogen atoms through single bonds. With this step, the left side carbon atom attains octet. The three hydrogen atoms also attain octet. This step is shown in fig.8.6(b)
Step 3: The same situation we saw in step 2, exists with the right side carbon atom. But we will write the steps again:
• Consider the carbon atom on the right. It is in single bond with the left side carbon atom. So it has acquired one electron through this single bond. Three more electrons are required. 
• So bond it to the remaining three hydrogen atoms. As mentioned above, a bond between a carbon atom and a hydrogen atom is possible only through a single bond. This is because hydrogen has only one electron to share.
• Thus the right side carbon atom is bonded to three hydrogen atoms through single bonds. With this step, the right side carbon atom attains octet. The three hydrogen atoms also attain octet. This step is shown in fig.c
■ With the above three steps, we get the final arrangement of atoms in a molecule of C₂H₆. The electron dot diagram is shown in the fig.8.7 below:

Fig.8.7

Note that the 'pair of electrons between the two carbon atoms' are both green in colour. This is different from all other pairs. The reader may write the reason for this 'difference in colour' in his/her own notebooks.

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Now we will consider another example: C₃H₈. We will write the required steps as in the previous example.

Step 1: In C₃H₈, there are 3 carbon atoms. The first step is to connect the three using single bonds. It is shown in fig.8.8(a) below:

Fig.8.8

Step 2: • Consider the left most carbon atom. It is in single bond with the middle carbon atom. So it has acquired one electron through this single bond. Three more electrons are required.
• So bond it to three hydrogen atoms. A bond between a carbon atom and a hydrogen atom is possible only through a single bond. This is because hydrogen has only one electron to share.
• Thus the left most carbon atom is bonded to three hydrogen atoms through single bonds. With this step, the left most carbon atom attains octet. The three bonded hydrogen atoms also attain octet. This step is shown in fig.b
Step 3: Consider the middle carbon atom. It is already bonded to two carbon atoms. They are two single bonds. So it has already acquired two electrons (one from each bond). Two more electrons are required. 
• So this carbon atom is bonded to two hydrogen atoms. One at top and the other at bottom. With this step, the middle carbon atom attains octet. The two hydrogen atoms bonded to it also attains octet. This step is shown in fig.c.
Step 4: Now the only remaining carbon atom is the right most one. It is in the same situation as in the left most carbon atom. But we will write the steps again:
• Consider the right most carbon atom. It is in single bond with the middle carbon atom. So it has acquired one electron through this single bond. Three more electrons are required. 
• For that, bond it to the three remaining hydrogen atoms. With this step, the right most carbon atom attains octet. The three hydrogen atoms which remained also attain octet. This step is shown in fig.d
■ With the above four steps, we get the final arrangement of atoms in a molecule of C₃H₈. The electron dot diagram is shown in the fig.8.9 below:

Fig.8.9

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So we determined the arrangement of atoms in the following two molecules:

(i) Molecule with 2 carbon atoms and 6 hydrogen atoms
(ii) Molecule with 3 carbon atoms and 8 hydrogen atoms
• A pattern is beginning to emerge. The next two cases will be:
(iii) Molecule with 4 carbon atoms and 10 hydrogen atoms
(iv) Molecule with 5 carbon atoms and 12 hydrogen atoms
• Note that number of carbon atoms is increasing by 1. Also, number of hydrogen atoms is increasing by 2
• It is like a series. Let us tabulate the arrangement for the first six members of the series:

With the above table, the analysis about the series becomes easy. Let us learn the properties of the series:
Property 1:
• The 1^st member of the series has 1 carbon atom
And 4 hydrogen atoms
• The 2^nd member of the series has 2 carbon atoms
And 6 hydrogen atoms
• The 3^rd member of the series has 3 carbon atoms
And 8 hydrogen atoms
So on...
■ So there exists a definite relation between the two quantities below:
• No. of carbon atoms
• No. of hydrogen atoms
■ If the no. of carbon atoms is 'n', then the no. Of hydrogen atoms will be (2n+2)
So we can represent the members of the series by a general formula: C_nH_(2n+2)
Property 2:
• Consider the second row. Consider the item in the 'Condensed formula' column in this row. We have: CH₃ – CH₃
A Condensed formula is a system of writing molecules in the same line as other text. It shows all atoms, but omits the vertical bonds. It also omits most or all the horizontal single bonds. 
• Now consider the third row, same column. We have: CH₃ – CH₂ –  CH₃
The difference from the previous second row is a CH₂ group
• Now consider the fourth row, same column. We have:  CH₃ – CH₂ – CH₂ –  CH₃ 
Here also, the difference between the previous third row is a CH₂ group
• We will find the same difference through out the series. That is:
■ Successive members differ by a CH₂ group

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A series of compounds having the above two properties is called a homologous series. We can write the official definition:

■ A Homologous Series is a group of chemical compounds satisfying the following conditions:
• All members of the series can be represented by a general formula
• Successive members differ by a CH₂ group

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• Members of a homologous series show similarity in chemical properties

• There is a regular gradation in their physical properties. That is., as we move down the series, the melting point, boiling point etc., increases.

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So now we know what a 'homologous series' is. The series that we saw just above is given a special name: Alkanes. In the next section, we will see another homologous series. 

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Chapter 3 - Chemical bonding

In the previous section, we completed the discussion on the Bohr model electronic configuration of atoms. We also saw isotopes. In this section, we will learn about Chemical bonding.

We have seen how the electrons are arranged in the various shells. Each element has a unique arrangement of electrons. This arrangement is called the configuration. We have learned to write the configuration of a large number of elements.

Now we consider the number of electrons in the outer most shell of an atom. Note that, the outer most shell is what matters for our present discussion. We want the number of electrons in this outer most shell.

If the number of electrons in the outer most shell is '8', then it is called an Octet electron configuration. Neon and Argon are good examples. They are shown in the fig. 2.1 below:

Fig.3.1

Elements, whose atoms have octet electron configuration, generally will not take part in chemical reactions. This is because, it is a stable configuration. So what about those elements which do not have this octet?
They will try to attain the octet, so that, there will be stability. Let us see how this octet can be attained:

• We will take the example of Sodium. It has a configuration of 2, 8, 1. (see fig.3.2 below)
• The outer most shell has 1 electron. So it needs 7 more electrons to attain octet

• Another example is Magnesium. It has a configuration of 2, 8, 2. (see fig.3.2 below)
• The outer most shell has 2 electrons. So it needs 6 more electrons to attain octet

• Yet another example is Chlorine. It has a configuration of 2, 8, 7. (see fig.3.2 below) 
• The outer most shell has 7 electrons. So it needs 1 more electron to attain octet

There are many such elements which are in 'need of electrons' to attain octet. How do they get the required number of electrons? 
■ Consider the example of sodium and chlorine. 
• We have seen the configuration of sodium above
• It has 1 electron in the outer most shell. So it needs 7 more electrons to attain octet
• It is easier to 'lose the 1 electron', than to 'obtain 7 electrons'
• When the one electron in the M-shell is taken away, the M-shell as a whole will disappear
• The L-shell will then become the 'outer most shell'
• The L-shell already have 8 electrons. So when L-shell becomes the outer most shell, it is an octet


• Now consider chlorine. We have seen it's configuration above
• It has 7 electrons in it's outer most shell. So it needs one more electron to attain octet
[Another way to attain octet is to lose the 7 electrons in the outer most shell. But it is easier to accept one electron than to lose 7 electrons]
• So it will accept one electron from the sodium atom that we saw above

When this donation and acceptance takes place, some other important changes also occurs:
• Normally, an atom is neutral because, the number of protons is equal to the number of electrons
• But when one electron is donated, the number of protons become excess. It becomes excess by '1'
• So the sodium atom which was initially neutral, has now become 'charged'
• We call such charged particles as ions. When an atom has become charged, we can no longer call it an 'atom'. We call it an 'ion'
• So our 'sodium atom' has become a 'sodium ion'. And this sodium ion has a positive charge

• Similarly, the chlorine atom has accepted an electron
• Number of electrons have become excess by 1
• It has now become a chlorine ion. And this chlorine ion has a negative charge

• So we see that the atoms have become ions
• Sodium atom has become sodium ion, and chlorine atom has become chlorine ion
• An electrostatic force of attraction will develop between the two ions. This force develops because, they are 'oppositely charged'
• This force will hold the two ions together
• Thus a molecule consisting of one ion of sodium and one ion of chlorine will be formed
■ This molecule is the 'molecule of sodium chloride'. It is represented as NaCl

The transfer of electron between sodium and chlorine can be represented pictorially as shown in the fig. 3.3 below:

Fig.3.3

• Consider the left side of the arrow: On either sides of the '+' sign, we have bohr models of sodium and chlorine. This shows the initial state 
• Sodium has one electron in the outer most shell. On the right side of the arrow, this atom of sodium has become an 'ion'. The outer most electron is lost. So the bohr model is shown inside square brackets, and a '+' sign is given at top right. This indicates one positive charge
• On the left side of the arrow, the bohr model of the chlorine atom has 7 electrons in the outer most shell. On the right side, the number of electrons in the outer most shell becomes 8. The chlorine atom becomes a chlorine ion. The model is put inside square brackets, and a '-' sign is given at top right. This indicates one negative charge
• Each entity inside square brackets is an ion. These 'oppositely charged' ions are held together by electrostatic force of attraction. So the diagram gives us a clear picture of how sodium and chlorine attain octet. It also shows us the formation of sodium chloride

Another method to represent the above process, is by using Electron dot diagram. In this method, only those electrons in the outer most shell are shown. This is because, they are the only electrons taking part in the bond formation. The dot diagram showing the formation of sodium chloride is shown in fig.3.4 below:

Fig.3.4

• The outer most electrons are shown as dots around the symbol of the atom 
• By counting the number of dots, we can see that, sodium has lost one electron, and chlorine has gained one electron
• The initial and final configurations are also given. Sodium and chlorine has attained octet
• Sodium has become a positive ion. This is due to the loss of electron. One electron is lost. so the charge is 1+. For a charge of magnitude one, the '1' is not written. So we just write '+' on the top right
• Chlorine has become a negative ion. This is due to the gain of electron. One electron is gained. so the charge is 1-. For a charge of magnitude one, the '1' is not written. So we just write '-' on the top right
• The 'oppositely charged' ions are held together by electrostatic force of attraction. Thus a molecule NaCl of sodium chloride is formed

Formation of Magnesium oxide
• Magnesium has a configuration 2, 8, 2. (See here). So it has 2 electrons in the outer most shell
• It needs 6 more electrons to attain octet. It is easier to lose the 2 electrons than to obtain 6 electrons
• Oxygen has a configuration 2, 6. So it has 6 electrons in the outer most shell
• It needs 2 more electrons to attain octet. So it will accept the 2 electrons which are donated by magnesium
• With the above information, we can draw the electron dot diagram as shown in fig.3.5 below:

Fig.3.5

• The outer most electrons are shown as dots around the symbol of the atom 
• By counting the number of dots, we can see that, magnesium has lost two electrons, and oxygen has gained those two electrons
• The initial and final configurations are also given. Magnesium and oxygen has attained octet
• Magnesium has become a positive ion. This is due to the loss of electrons. Two electrons are lost. So the charge is 2+. So we write '2+' on the top right
• Chlorine has become a negative ion. This is due to the gain of electron. Two electrons are gained. So the charge is 2-. So we write '2-' on the top right
• The 'oppositely charged' ions are held together by electrostatic force of attraction. Thus a molecule MgO of Magnesium oxide is formed

Formation of sodium oxide
• Na has a configuration 2, 8, 1. So it has 1 electron in the outer most shell
• It needs 7 more electrons to attain octet. It is easier to lose the 1 electron than to obtain 7 electrons
• O has a configuration 2, 6. So it has 6 electrons in the outer most shell
• It needs 2 more electron to attain octet. 
• So 2 atoms of Na will combine with one atom of O. So that:
    ♦ One Na will donate one electron to O
    ♦ The other Na will also donate one electron to O
    ♦ So, in total, O gets two electrons
• With the above information, we can draw the electron dot diagram as shown in fig.3.6 below:

Fig.3.6

• The outer most electrons are shown as dots around the symbol of the atom 
• By counting the number of dots, we can see that, each Na has lost one electron, and the  O atom have gained those two electrons. (one electron from each Na)
• The initial and final configurations are also given. Na and O have attained octet
• Na has become a positive ion. This is due to the loss of electron. One electron is lost. So the charge is 1+. So we write '+' on the top right
• O has become a negative ion. This is due to the gain of electron. Two electrons are gained by O. So the charge is 2- . So we write '-' on the top right

• The 'oppositely charged' ions are held together by electrostatic force of attraction. Thus a molecule Na₂O is formed

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In the next section, we will see some solved examples. We will also learn about Cations and Anions.

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