In the previous section 4.22, we saw the formation of the bond between two H atoms. Based on that, we will continue our discussion on VB theory
1. We have seen that, the H2 molecule is formed when two H atoms come close to each other such that, the distance between their nuclei is 74 pm
2. We know that in hydrogen, the nucleus is situated at the center of the 1s orbital
• That is.,
♦ The single electron of H creates a cloud
♦ This cloud is spherical in shape
♦ We call this sphere as the ‘1s orbital’ of H
♦ The nucleus of H is situated at the center of the 1s sphere
3. The electron inside the 1s orbital can be represented using an arrow
• This is shown in fig.4.129(a) below
♦ The 1s orbital is shown in orange color
✰ It is given a bit of transparency so that, the nucleus is also visible
✰ The nucleus is shown as a small red sphere
4. When the distance is 74 pm, the two 1s orbitals penetrates into each other
• This is shown in fig.4.129(b) above
• This inter penetration is called overlapping of orbitals
• When the overlapping occurs, the density of the cloud will be greater in the region between the two nuclei
• Both the electrons will be present in that overlapping region
5. Thus both the electrons will be attracted to both the nuclei
• This results in the bonding between the two H atoms
• Let us see how CH4 (methane) is formed. It can be written in steps:
1. We know the Lewis dot structure of CH4. It is shown in the fig.4.130(a) below:
• The central atom is C. It forms single covalent bonds with four H atoms
2. The electronic configuration of C is 1s22s22p2
• We want the electrons in the valence shell. They are the electrons which take part in the chemical reactions. So we want 2s2 and 2p2
• They are shown in fig.4.130(b)
3. The fig.b shows the energy levels also
• We see that:
♦ The two electrons in the 2p orbitals have a greater energy
♦ The two electrons in the 2s orbitals have a lesser energy
4. We have four H atoms ready to combine with the C atom
• We know that, an H atom has only one orbital: the 1s orbital
• Let us connect the various orbitals of C with the 1s orbital of H atoms:
♦ The 1s orbital of the first H atom overlap with the px orbital of C
♦ The 1s orbital of the second H atom overlap with the py orbital of C
5. Now we have a problem. It can be written in 3 steps:
(i) The 2s orbital is completely filled. Because, there are two electrons in it
(Recall Pauli's exclusion principle)
(ii) So the 1s of the third H atom cannot overlap with the 2s of C
(iii) Also we see no place for the fourth H atom at all
6. This problem can be solved in the following way. It involves 2 steps:
(i) Give enough energy so that one electron in the 2s orbital of C gets excited and jumps to the 2pz orbital
(ii) Now there are four half filled orbitals. This is shown in fig.c
♦ The 1s orbital of the first H atom overlap with the px orbital of C
♦ The 1s orbital of the second H atom overlap with the py orbital of C
♦ The 1s orbital of the third H atom overlap with the pz orbital of C
♦ The 1s orbital of the fourth H atom overlap with the 2s orbital of C
7. Based on the information in (6) above, let us try to make a model. It can be written in 3 steps:
(i) Fig.4.131(a) below shows the three 2p orbitals of C
• Each of them contains one electron
(ii) The point of intersection of the three axes is the position of the 'nucleus of the C atom'
• We know that the center of the 2s sphere will be at the center of the atom
• So the 2s and 2p orbitals can be shown together as in fig.b
• The 2s also contains one electron
(iii) Now we attach the 1s orbital of a H atom to each of the half filled orbitals of C
• This is shown in fig.c
• The four H atoms are numbered as Hi, Hii, Hiii, Hiv
• Note that, 'electron pairs' are shown at the overlapping regions of the various orbitals
• They are discussed below:
Drawback 1
• This can be written in 3 steps:
1. From the fig.4.131(c), we directly get the following bond angles:
∠HiCHii = 90o, ∠HiCHiii = 90o, ∠HiiCHiii = 90o
2. These angles are readily available because, the p orbitals lie along the three coordinate axes
3. But the above 90o values are not acceptable because, we know that in CH4, all the bond angles are equal to 109.5o
Drawback 2
• This can be written in 3 steps:
1. The 90o angles obtained above, involves Hi, Hii and Hiii only
• We want the angles related to Hiv also
2. The angles related to Hiv can be calculated using 3 steps:
(i) Imagine a line joining the centers of the 'Hiv orange' and 'C cyan' spheres
(ii) Write the angles which this line makes with CHi, CHii and CHiii
(iii) Thus we will get ∠HiCHiv, ∠HiiCHiv and ∠HiiiCHiv
3. But there is a problem. It can be written in steps:
(i) The 'Hiv orange' and 'C cyan' are both spheres
♦ They can overlap from any direction
♦ There are infinite number of possible directions
♦ Three of them are shown in fig.4.132 below:
(ii) So there are infinite number of possible values for the angles related to Hiv
(iii) This is not acceptable because, we know that, all bond angles in CH4 are equal to 109.5
Drawback 3
• This can be written in 3 steps:
1. Let us consider the bond lengths
• We can easily see that, lengths of the following three bond are equal:
♦ Bond between Hi and C
♦ Bond between Hii and C
♦ Bond between Hiii and C
• But the bond length between Hiv and C will be different
2. That means:
♦ Three out of the four bonds in CH4 are of the same length
♦ The fourth bond is of a different length
3. This is not acceptable because experiments show that, all bond lengths in CH4 are equal
Drawback 4
• This can be written in 5 steps:
1. Let us consider the bond energies
• We have:
Energy of the electron in 2px
= Energy of the electron in 2py
= Energy of the electron in 2pz
≠ Energy of the electron in 2s
2. So we can write:
♦ The electrons of Hi, Hii and Hiii will be combining with electrons of the same energy
♦ The electrons of Hiv will be combining with an electron of a different energy
3. That means:
• Energies of the following three bond are equal:
♦ Bond between Hi and C
♦ Bond between Hii and C
♦ Bond between Hiii and C
• But the energy of the bond between Hiv and C will be different
4. That means:
♦ Three out of the four bonds in CH4 have the same energy
♦ The fourth bond has a different energy
5. This is not acceptable because experiments show that, all bond lengths in CH4 have the same energy
Orbital overlap concept
1. We have seen that, the H2 molecule is formed when two H atoms come close to each other such that, the distance between their nuclei is 74 pm
2. We know that in hydrogen, the nucleus is situated at the center of the 1s orbital
• That is.,
♦ The single electron of H creates a cloud
♦ This cloud is spherical in shape
♦ We call this sphere as the ‘1s orbital’ of H
♦ The nucleus of H is situated at the center of the 1s sphere
3. The electron inside the 1s orbital can be represented using an arrow
• This is shown in fig.4.129(a) below
♦ The 1s orbital is shown in orange color
✰ It is given a bit of transparency so that, the nucleus is also visible
✰ The nucleus is shown as a small red sphere
4. When the distance is 74 pm, the two 1s orbitals penetrates into each other
• This is shown in fig.4.129(b) above
• This inter penetration is called overlapping of orbitals
• When the overlapping occurs, the density of the cloud will be greater in the region between the two nuclei
• Both the electrons will be present in that overlapping region
5. Thus both the electrons will be attracted to both the nuclei
• This results in the bonding between the two H atoms
• In this way, different orbitals can overlap to form bonds
1. We know the Lewis dot structure of CH4. It is shown in the fig.4.130(a) below:
• The central atom is C. It forms single covalent bonds with four H atoms
 |
| Fig.4.130 |
• We want the electrons in the valence shell. They are the electrons which take part in the chemical reactions. So we want 2s2 and 2p2
• They are shown in fig.4.130(b)
3. The fig.b shows the energy levels also
• We see that:
♦ The two electrons in the 2p orbitals have a greater energy
♦ The two electrons in the 2s orbitals have a lesser energy
4. We have four H atoms ready to combine with the C atom
• We know that, an H atom has only one orbital: the 1s orbital
• Let us connect the various orbitals of C with the 1s orbital of H atoms:
♦ The 1s orbital of the first H atom overlap with the px orbital of C
♦ The 1s orbital of the second H atom overlap with the py orbital of C
5. Now we have a problem. It can be written in 3 steps:
(i) The 2s orbital is completely filled. Because, there are two electrons in it
(Recall Pauli's exclusion principle)
(ii) So the 1s of the third H atom cannot overlap with the 2s of C
(iii) Also we see no place for the fourth H atom at all
6. This problem can be solved in the following way. It involves 2 steps:
(i) Give enough energy so that one electron in the 2s orbital of C gets excited and jumps to the 2pz orbital
(ii) Now there are four half filled orbitals. This is shown in fig.c
♦ The 1s orbital of the first H atom overlap with the px orbital of C
♦ The 1s orbital of the second H atom overlap with the py orbital of C
♦ The 1s orbital of the third H atom overlap with the pz orbital of C
♦ The 1s orbital of the fourth H atom overlap with the 2s orbital of C
7. Based on the information in (6) above, let us try to make a model. It can be written in 3 steps:
(i) Fig.4.131(a) below shows the three 2p orbitals of C
• Each of them contains one electron
 |
| Fig.4.131 |
• We know that the center of the 2s sphere will be at the center of the atom
• So the 2s and 2p orbitals can be shown together as in fig.b
• The 2s also contains one electron
(iii) Now we attach the 1s orbital of a H atom to each of the half filled orbitals of C
• This is shown in fig.c
• The four H atoms are numbered as Hi, Hii, Hiii, Hiv
• Note that, 'electron pairs' are shown at the overlapping regions of the various orbitals
• The fig.4.131(c) appears to be a satisfactory model of CH4. But there are four drawbacks
Drawback 1
• This can be written in 3 steps:
1. From the fig.4.131(c), we directly get the following bond angles:
∠HiCHii = 90o, ∠HiCHiii = 90o, ∠HiiCHiii = 90o
2. These angles are readily available because, the p orbitals lie along the three coordinate axes
3. But the above 90o values are not acceptable because, we know that in CH4, all the bond angles are equal to 109.5o
Drawback 2
• This can be written in 3 steps:
• We want the angles related to Hiv also
2. The angles related to Hiv can be calculated using 3 steps:
(i) Imagine a line joining the centers of the 'Hiv orange' and 'C cyan' spheres
(ii) Write the angles which this line makes with CHi, CHii and CHiii
(iii) Thus we will get ∠HiCHiv, ∠HiiCHiv and ∠HiiiCHiv
3. But there is a problem. It can be written in steps:
(i) The 'Hiv orange' and 'C cyan' are both spheres
♦ They can overlap from any direction
♦ There are infinite number of possible directions
♦ Three of them are shown in fig.4.132 below:
 |
| Fig.4.132 |
(iii) This is not acceptable because, we know that, all bond angles in CH4 are equal to 109.5
Drawback 3
• This can be written in 3 steps:
1. Let us consider the bond lengths
• We can easily see that, lengths of the following three bond are equal:
♦ Bond between Hi and C
♦ Bond between Hii and C
♦ Bond between Hiii and C
• But the bond length between Hiv and C will be different
2. That means:
♦ Three out of the four bonds in CH4 are of the same length
♦ The fourth bond is of a different length
3. This is not acceptable because experiments show that, all bond lengths in CH4 are equal
Drawback 4
• This can be written in 5 steps:
1. Let us consider the bond energies
• We have:
Energy of the electron in 2px
= Energy of the electron in 2py
= Energy of the electron in 2pz
≠ Energy of the electron in 2s
2. So we can write:
♦ The electrons of Hi, Hii and Hiii will be combining with electrons of the same energy
♦ The electrons of Hiv will be combining with an electron of a different energy
3. That means:
• Energies of the following three bond are equal:
♦ Bond between Hi and C
♦ Bond between Hii and C
♦ Bond between Hiii and C
• But the energy of the bond between Hiv and C will be different
4. That means:
♦ Three out of the four bonds in CH4 have the same energy
♦ The fourth bond has a different energy
5. This is not acceptable because experiments show that, all bond lengths in CH4 have the same energy
■ Because of the above four drawbacks, the model shown in fig.4.131(c) is not acceptable
• In the next section, we will see how scientists solved this problem
• In the next section, we will see how scientists solved this problem
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