• In the previous section 4.39, we saw the details of the F2 and Ne2 molecules. In this section we will see hydrogen bonding. Later in this section, we will see some solved examples also
• Basic details about hydrogen bonding can be written in 6 steps:
1. Consider a HF (Hydrogen fluoride) molecule
• We have already seen it’s Lewis dot structure in a previous section. It is shown again in fig.4.223(a) below:
• We see that, two electrons (a pair) are shared between the H and F atoms
2. The F atom has a very high electronegativity
• So the shared pair of electrons will be attracted towards the F atom. Thus:
♦ The F atom will acquire a small partial negative charge
✰ This partial negative charge is denoted by 𝜹-
♦ The H atom will acquire a small partial positive charge
✰ This partial positive charge is denoted by 𝜹+
(𝜹 is the Greek small letter 'Delta')
• The partial charges are shown in fig.4.223(b)
3. Consider a small sample of HF
• It will contain a large number of HF molecules
• Consider any one HF molecule
♦ The H atom in ‘that HF molecule’ will attract the F atom in ‘another HF molecule'
♦ This is because,
✰ The H atom in the first HF is positively charged
✰ The F atom in the second HF is negatively charged
4. This 'attraction between H and F' is indicated by a dashed line as shown in fig.4.223 (c)
• Because of this attraction, the two HF molecules cannot easily separate away from each other
• We can say that, a bond is formed between the two HF molecules
■ This bond is called hydrogen bond
• Note that, in the fig.c, individual 'HF molecules' are outlined
♦ This is to give emphasis to the fact that, hydrogen bonding occurs 'between molecules'
5. Due to the hydrogen bonding, a large number of molecules will come together and form a group
• This is because,
♦ The left H will form a hydrogen bond with the F of a third HF molecule
♦ The right F will form a hydrogen bond with the H of a fourth HF molecule
♦ This is shown in fig.4.224 below
♦ This process continues, and a group will be formed
6. Figs.4.223 and 224 show the 'occurrence of hydrogen bond' when a covalent bond exists between H and F
• The hydrogen bond can occur in the following two cases also:
♦ When a covalent bond exists between H and O
♦ When a covalent bond exists between H and N
• This is because:
♦ Like F, the O can also attract the shared pair of electrons to itself
✰ Like F, the O is also highly electronegative
♦ Like F, the N can also attract the shared pair of electrons to itself
✰ Like F, the N is also highly electronegative
(i) Consider the cases where H atom enters into a covalent bond with F, N or O
(ii) In all those three cases, a bond is formed between 'the H in one molecule' to the F/N/O in another molecule
(iii) This bond is called hydrogen bond
• Basic details about hydrogen bonding can be written in 6 steps:
1. Consider a HF (Hydrogen fluoride) molecule
• We have already seen it’s Lewis dot structure in a previous section. It is shown again in fig.4.223(a) below:
 |
| Fig.4.223 |
2. The F atom has a very high electronegativity
• So the shared pair of electrons will be attracted towards the F atom. Thus:
♦ The F atom will acquire a small partial negative charge
✰ This partial negative charge is denoted by 𝜹-
♦ The H atom will acquire a small partial positive charge
✰ This partial positive charge is denoted by 𝜹+
(𝜹 is the Greek small letter 'Delta')
• The partial charges are shown in fig.4.223(b)
3. Consider a small sample of HF
• It will contain a large number of HF molecules
• Consider any one HF molecule
♦ The H atom in ‘that HF molecule’ will attract the F atom in ‘another HF molecule'
♦ This is because,
✰ The H atom in the first HF is positively charged
✰ The F atom in the second HF is negatively charged
4. This 'attraction between H and F' is indicated by a dashed line as shown in fig.4.223 (c)
• Because of this attraction, the two HF molecules cannot easily separate away from each other
• We can say that, a bond is formed between the two HF molecules
■ This bond is called hydrogen bond
• Note that, in the fig.c, individual 'HF molecules' are outlined
♦ This is to give emphasis to the fact that, hydrogen bonding occurs 'between molecules'
5. Due to the hydrogen bonding, a large number of molecules will come together and form a group
• This is because,
♦ The left H will form a hydrogen bond with the F of a third HF molecule
♦ The right F will form a hydrogen bond with the H of a fourth HF molecule
♦ This is shown in fig.4.224 below
♦ This process continues, and a group will be formed
 |
| Fig.4.224 |
• The hydrogen bond can occur in the following two cases also:
♦ When a covalent bond exists between H and O
♦ When a covalent bond exists between H and N
• This is because:
♦ Like F, the O can also attract the shared pair of electrons to itself
✰ Like F, the O is also highly electronegative
♦ Like F, the N can also attract the shared pair of electrons to itself
✰ Like F, the N is also highly electronegative
Let us see an example in which O atom is involved. It can be written in 5 steps:
1. Consider a H2O (Water) molecule
• We have already seen it’s Lewis dot structure in a previous section. It is shown again in fig.4.225(a) below:
• We see that:
♦ Two electrons (a pair) are shared between the left H and the O atom
♦ Two electrons (a pair) are shared between the right H and the O atom
2. The O atom has a very high electronegativity
• So the shared pair of electrons will be attracted towards the O atom. Thus:
♦ The O atom will acquire a small partial negative charge
✰ This partial negative charge is denoted by 𝜹-
♦ The left H atom will acquire a small partial positive charge
✰ This partial positive charge is denoted by 𝜹+
♦ The right H atom will acquire a small partial positive charge
✰ This partial positive charge is denoted by 𝜹+
• The partial charges are shown in fig.4.223(b)
3. Consider a small sample of H2O
• It will contain a large number of H2O molecules
• Consider any one H2O molecule
♦ The H atom in ‘that H2O molecule’ will attract the O atom in ‘another H2O molecule'
♦ This is because,
✰ The H atom in the first H2O is positively charged
✰ The O atom in the second H2O is negatively charged
4. This 'attraction between H and O' is indicated by a dashed line as shown in fig.4.223 (c)
• Because of this attraction, the two H2O molecules cannot easily separate away from each other
• We can say that, a hydrogen bond is formed between the two H2O molecules
• Note that, in the fig.c, individual 'H2O molecules' are outlined
♦ This is to give emphasis to the fact that, hydrogen bonding occurs 'between molecules'
5. Due to the hydrogen bonding, a large number of H2O molecules will come together and form a group
♦ This is shown in fig.4.226 below:
1. Consider a H2O (Water) molecule
• We have already seen it’s Lewis dot structure in a previous section. It is shown again in fig.4.225(a) below:
 |
| Fig.4.225 |
♦ Two electrons (a pair) are shared between the left H and the O atom
♦ Two electrons (a pair) are shared between the right H and the O atom
2. The O atom has a very high electronegativity
• So the shared pair of electrons will be attracted towards the O atom. Thus:
♦ The O atom will acquire a small partial negative charge
✰ This partial negative charge is denoted by 𝜹-
♦ The left H atom will acquire a small partial positive charge
✰ This partial positive charge is denoted by 𝜹+
♦ The right H atom will acquire a small partial positive charge
✰ This partial positive charge is denoted by 𝜹+
• The partial charges are shown in fig.4.223(b)
3. Consider a small sample of H2O
• It will contain a large number of H2O molecules
• Consider any one H2O molecule
♦ The H atom in ‘that H2O molecule’ will attract the O atom in ‘another H2O molecule'
♦ This is because,
✰ The H atom in the first H2O is positively charged
✰ The O atom in the second H2O is negatively charged
4. This 'attraction between H and O' is indicated by a dashed line as shown in fig.4.223 (c)
• Because of this attraction, the two H2O molecules cannot easily separate away from each other
• We can say that, a hydrogen bond is formed between the two H2O molecules
• Note that, in the fig.c, individual 'H2O molecules' are outlined
♦ This is to give emphasis to the fact that, hydrogen bonding occurs 'between molecules'
5. Due to the hydrogen bonding, a large number of H2O molecules will come together and form a group
♦ This is shown in fig.4.226 below:
 |
| Fig.4.226 |
So now we can write the definition of hydrogen bond. It can be written in 3 steps:
(ii) In all those three cases, a bond is formed between 'the H in one molecule' to the F/N/O in another molecule
(iii) This bond is called hydrogen bond
• Consider the dashed lines in the above figs.
• All those dashed lines are drawn 'between molecules'
• This indicates that, the 'bonds indicated by those dashed lines' act between molecules
■ Such bonds are called Inter-molecular hydrogen bonds
■ There is another type called Intra-molecular hydrogen bond
It can be explained using an example in 3 steps:
1. Consider a molecule of O-nitrophenol. It is shown in fig.4.227(a) below
• It contains a NO2 group and a OH group
2. These groups are expanded in fig.4.227(b)
• Consider the NO2 group
♦ O being more electronegative than N, pulls the shared electrons. Thus:
✰ O becomes $\mathbf\small{\rm{O}^{\delta-}}$
✰ N becomes $\mathbf\small{\rm{N}^{\delta+}}$
• Consider the OH group
♦ O being more electronegative than H, pulls the shared electrons. Thus:
✰ O becomes $\mathbf\small{\rm{O}^{\delta-}}$
✰ H becomes $\mathbf\small{\rm{H}^{\delta+}}$
3. Now, an electrostatic force of attraction comes into play between the $\mathbf\small{\rm{O}^{\delta-}}$ and $\mathbf\small{\rm{H}^{\delta+}}$
• This is shown in fig.4.227 (c)
• This hydrogen bonding is between two atoms in the same molecule
• It is called a intra-molecular hydrogen bond
Solved example 4.9
Describe the change in hybridization (if any) of the Al atom in the following reaction:
$\mathbf\small{\rm{AlCl_3+Cl^-\longrightarrow AlCl_4^-}}$
Solution:
1. We have to find the hybridization in both AlCl3 and AlCl4-
• And then see if there is any difference
2. First we will draw the Lewis dot structure of the entire reaction
• It is shown in fig.4.228 below:
• In the above fig., we can see the Lewis structure of AlCl3, Cl- and AlCl4-
3. We see that, in AlCl3, the Al has not attained octet. The reason can be explained in 2 steps:
(i) The electronic configuration of Al is [Ne]3s23p1
• So Al needs 'five more' electrons
(ii) But in AlCl3, the Al atom attains only 'three more' electrons
4. So the Al atom tries to attain 'two more' electrons
• If AlCl3 combines with a Cl atom, only 'one more' electron can be attained
• So AlCl3 combines with the Cl- ion
■ As a result, AlC4- is formed
• In AlCl4-, the Al atom has attained octet. This is clear from the Lewis dot structure of AlCl4-
(We have seen the method to draw Lewis dot structures of polyatomic ions in section 4.3)
5. Next we will see the application of the 'concept of hybridization' for the above reaction
(i) Consider the electronic configuration of Al in the ground state: [Ne]3s23p1
(ii) In the excited state, one electron from the 3s orbital jumps to a 3p orbital
♦ Thus the configuration becomes: [Ne]3s13px13py1
(iii) The 3s orbital mix together with the 3px and 3py
♦ Thus we get three 'sp2 hybridized orbitals'
♦ One Cl atom attaches to each of those three hybrid orbitals. Thus we get AlCl3
♦ We have seen this type of molecule formation in BCl3. See fig.4.167 of section 4.29
6. Next we will see the hybridization in AlCl4-
(i) Consider the electronic configuration of Al in the ground state: [Ne]3s23p1
(ii) In the excited state, one electron from the 3s orbital jumps to a 3p orbital
♦ Thus the configuration becomes: [Ne]3s13px13py1
(iii) The 3s orbital mix together with the 3px, 3py and 3pz
♦ Thus we get four 'sp2 hybridized orbitals'
♦ (Note that, the 3pz also takes part in the hybridization, even though it is empty)
♦ One Cl atom attaches to each of those four hybrid orbitals. Thus we get AlCl4-
7. So we can write:
■ In AlCl3, the Al atom undergoes sp2 hybridization
■ In AlCl4-, the Al atom undergoes sp3 hybridization
Solved example 4.10
Draw diagrams showing the formation of a double bond and a triple bond between
carbon atoms in C2H4 and C2H2 molecules
Solution:
• Details about C2H4 are given in section 4.26
• Details about C2H2 are given in section 4.27
Solved example 4.11
What is the total number of sigma and pi bonds in the following molecules?
(a) C2H2 (b) C2H4
Solution:
Part (a):
1. From the previous solved example 4.10, we get the details about the structure of C2H2
• We see that there are a total of three bonds:
(i) Two numbers of C-H bonds
♦ Each of those two, is a single bond
♦ Also each of those two, are sigma bonds
(ii) One number C≡C bond
♦ This is a triple bond
♦ Within this triple bond,
✰ One is a sigma bond
✰ The other two are pi bonds
2. So we get:
♦ Total number of sigma bonds = 3
♦ Total number of pi bonds = 2
Part (b):
1. From the previous solved example 4.10, we get the details about the structure of C2H4
• We see that there are a total of five bonds:
(i) Four numbers of C-H bonds
♦ Each of those five, is a single bonds
♦ Also each of those five, are sigma bonds
(ii) One number C=C bond
♦ This is a double bond
♦ Within this double bond
✰ One is a sigma bond
✰ The other is a pi bond
2. So we get:
♦ Total number of sigma bonds = 5
♦ Total number of pi bonds = 1
Solved example 4.12
Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why? (a) 1s and 1s (b) 1s and 2px ; (c) 2py and 2py (d) 1s and 2s
Solution:
Part (a):
1. x-axis is the inter-nuclear axis
♦ So the x-axis passes through the centers of both the 1s orbitals
2.Then the two 1s spheres can over lap in a end-to-end manner
♦ So 1s and 1s will form a sigma bond
Part (b):
1. x-axis is the inter-nuclear axis
♦ So the x-axis passes through the center of the 1s orbital
♦ The 2px will lie along the same x-axis
2. Then the 1s sphere can over lap in a end-to-end manner with the px
♦ So 1s and 2px will form a sigma bond
Part (c):
1. x-axis is the inter-nuclear axis
♦ The two 2py orbitals will be perpendicular to the x-axis
2. Then the two 2py orbitals can over lap only in a side-wise manner
♦ So 2py and 2py will form only a pi bond
Part (d):
1. x-axis is the inter-nuclear axis
♦ So the x-axis passes through the centers of both the 1s and 2s orbitals
2. Then the two spheres can over lap in a end-to-end manner
♦ So 1s and 2s will form a sigma bond
■ So the answer is: (c) 2py and 2py
Solved example 4.13
Which hybrid orbitals are used by carbon atoms in the following molecules?
(a) CH3–CH3; (b) CH3–CH=CH2; (c) CH3-CH2-OH; (d) CH3-CHO (e) CH3COOH
Solution:
• In this problem, we are given the condensed formulas. The reader may draw the structural formula of each. Based on the structural formula, we can write the answers:
Part (a): CH3–CH3
• The structural formula is shown in fig.4.229(a) below:
1. The first C atom has four single bonds around it. So this C atom is sp3 hybridized
2. The second C atom has four single bonds around it. So this C atom is also sp3 hybridized
Part (b): CH3–CH=CH2
• The structural formula is shown in fig.4.229(b) above
1. The first C atom has four single bonds around it. So this C atom is sp3 hybridized
2. The second C atom has three bonds around it:
(i) Single bond with the left C atom
♦ This is a sigma bond
(ii) Single bond with the H atom
♦ This is a sigma bond
(iii) Double bond with the right C atom
♦ This is a sigma bond and a pi bond
• The pi bond mentioned in 2(iii) is possible only if the 2pz is left unhybridized
♦ So only 2s, 2px and 2py will participate in the hybridization
♦ So it is a sp2 hybridization
3. The third C atom has three bonds around it:
(i) Single bond with a H atom
♦ This is a sigma bond
(ii) Single bond with the other H atom
♦ This is a sigma bond
(iii) Double bond with the middle C atom
♦ This is a sigma bond and a pi bond
• The pi bond mentioned in 3(iii) is possible only if the 2pz is left unhybridized
♦ So only 2s, 2px and 2py will participate in the hybridization
♦ So it is a sp2 hybridization
Part (c): CH3-CH2-OH
• The structural formula is shown in fig.4.229(c) below:
1. The first C atom has four single bonds around it. So this C atom is sp3 hybridized
2. The second C atom has four single bonds around it. So this C atom is sp3 hybridized
Part (d): CH3-CHO
• The structural formula is shown in fig.4.229(d) above
1. The first C atom has four single bonds around it. So this C atom is sp3 hybridized
2. The second C atom has three bonds around it:
(i) Single bond with the left C atom
♦ This is a sigma bond
(ii) Single bond with the H atom
♦ This is a sigma bond
(iii) Double bond with the O atom
♦ This is a sigma bond and a pi bond
• The pi bond mentioned in 2(iii) is possible only if the 2pz is left unhybridized
♦ So only 2s, 2px and 2py will participate in the hybridization
♦ So it is a sp2 hybridization
Part (e): CH3COOH
• The structural formula is shown in fig.4.229(e) above
1. The first C atom has four single bonds around it. So this C atom is sp3 hybridized
2. The second C atom has three bonds around it:
(i) Single bond with the left C atom
♦ This is a sigma bond
(ii) Single bond with the O atom
♦ This is a sigma bond
(iii) Double bond with the other O atom
♦ This is a sigma bond and a pi bond
• The pi bond mentioned in 2(iii) is possible only if the 2pz is left unhybridized
♦ So only 2s, 2px and 2py will participate in the hybridization
♦ So it is a sp2 hybridization
• All those dashed lines are drawn 'between molecules'
• This indicates that, the 'bonds indicated by those dashed lines' act between molecules
■ Such bonds are called Inter-molecular hydrogen bonds
■ There is another type called Intra-molecular hydrogen bond
It can be explained using an example in 3 steps:
1. Consider a molecule of O-nitrophenol. It is shown in fig.4.227(a) below
• It contains a NO2 group and a OH group
 |
| Fig.4.227 |
• Consider the NO2 group
♦ O being more electronegative than N, pulls the shared electrons. Thus:
✰ O becomes $\mathbf\small{\rm{O}^{\delta-}}$
✰ N becomes $\mathbf\small{\rm{N}^{\delta+}}$
• Consider the OH group
♦ O being more electronegative than H, pulls the shared electrons. Thus:
✰ O becomes $\mathbf\small{\rm{O}^{\delta-}}$
✰ H becomes $\mathbf\small{\rm{H}^{\delta+}}$
3. Now, an electrostatic force of attraction comes into play between the $\mathbf\small{\rm{O}^{\delta-}}$ and $\mathbf\small{\rm{H}^{\delta+}}$
• This is shown in fig.4.227 (c)
• This hydrogen bonding is between two atoms in the same molecule
• It is called a intra-molecular hydrogen bond
So we have completed a discussion on the various topics in this chapter. We will see some solved examples
Solved example 4.9
Describe the change in hybridization (if any) of the Al atom in the following reaction:
$\mathbf\small{\rm{AlCl_3+Cl^-\longrightarrow AlCl_4^-}}$
Solution:
1. We have to find the hybridization in both AlCl3 and AlCl4-
• And then see if there is any difference
2. First we will draw the Lewis dot structure of the entire reaction
• It is shown in fig.4.228 below:
 |
| Fig.4.228 |
3. We see that, in AlCl3, the Al has not attained octet. The reason can be explained in 2 steps:
(i) The electronic configuration of Al is [Ne]3s23p1
• So Al needs 'five more' electrons
(ii) But in AlCl3, the Al atom attains only 'three more' electrons
4. So the Al atom tries to attain 'two more' electrons
• If AlCl3 combines with a Cl atom, only 'one more' electron can be attained
• So AlCl3 combines with the Cl- ion
■ As a result, AlC4- is formed
• In AlCl4-, the Al atom has attained octet. This is clear from the Lewis dot structure of AlCl4-
(We have seen the method to draw Lewis dot structures of polyatomic ions in section 4.3)
5. Next we will see the application of the 'concept of hybridization' for the above reaction
(i) Consider the electronic configuration of Al in the ground state: [Ne]3s23p1
(ii) In the excited state, one electron from the 3s orbital jumps to a 3p orbital
♦ Thus the configuration becomes: [Ne]3s13px13py1
(iii) The 3s orbital mix together with the 3px and 3py
♦ Thus we get three 'sp2 hybridized orbitals'
♦ One Cl atom attaches to each of those three hybrid orbitals. Thus we get AlCl3
♦ We have seen this type of molecule formation in BCl3. See fig.4.167 of section 4.29
6. Next we will see the hybridization in AlCl4-
(i) Consider the electronic configuration of Al in the ground state: [Ne]3s23p1
(ii) In the excited state, one electron from the 3s orbital jumps to a 3p orbital
♦ Thus the configuration becomes: [Ne]3s13px13py1
(iii) The 3s orbital mix together with the 3px, 3py and 3pz
♦ Thus we get four 'sp2 hybridized orbitals'
♦ (Note that, the 3pz also takes part in the hybridization, even though it is empty)
♦ One Cl atom attaches to each of those four hybrid orbitals. Thus we get AlCl4-
7. So we can write:
■ In AlCl3, the Al atom undergoes sp2 hybridization
■ In AlCl4-, the Al atom undergoes sp3 hybridization
Solved example 4.10
Draw diagrams showing the formation of a double bond and a triple bond between
carbon atoms in C2H4 and C2H2 molecules
Solution:
• Details about C2H4 are given in section 4.26
• Details about C2H2 are given in section 4.27
Solved example 4.11
What is the total number of sigma and pi bonds in the following molecules?
(a) C2H2 (b) C2H4
Solution:
Part (a):
1. From the previous solved example 4.10, we get the details about the structure of C2H2
• We see that there are a total of three bonds:
(i) Two numbers of C-H bonds
♦ Each of those two, is a single bond
♦ Also each of those two, are sigma bonds
(ii) One number C≡C bond
♦ This is a triple bond
♦ Within this triple bond,
✰ One is a sigma bond
✰ The other two are pi bonds
2. So we get:
♦ Total number of sigma bonds = 3
♦ Total number of pi bonds = 2
Part (b):
1. From the previous solved example 4.10, we get the details about the structure of C2H4
• We see that there are a total of five bonds:
(i) Four numbers of C-H bonds
♦ Each of those five, is a single bonds
♦ Also each of those five, are sigma bonds
(ii) One number C=C bond
♦ This is a double bond
♦ Within this double bond
✰ One is a sigma bond
✰ The other is a pi bond
2. So we get:
♦ Total number of sigma bonds = 5
♦ Total number of pi bonds = 1
Solved example 4.12
Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why? (a) 1s and 1s (b) 1s and 2px ; (c) 2py and 2py (d) 1s and 2s
Solution:
Part (a):
1. x-axis is the inter-nuclear axis
♦ So the x-axis passes through the centers of both the 1s orbitals
2.Then the two 1s spheres can over lap in a end-to-end manner
♦ So 1s and 1s will form a sigma bond
Part (b):
1. x-axis is the inter-nuclear axis
♦ So the x-axis passes through the center of the 1s orbital
♦ The 2px will lie along the same x-axis
2. Then the 1s sphere can over lap in a end-to-end manner with the px
♦ So 1s and 2px will form a sigma bond
Part (c):
1. x-axis is the inter-nuclear axis
♦ The two 2py orbitals will be perpendicular to the x-axis
2. Then the two 2py orbitals can over lap only in a side-wise manner
♦ So 2py and 2py will form only a pi bond
Part (d):
1. x-axis is the inter-nuclear axis
♦ So the x-axis passes through the centers of both the 1s and 2s orbitals
2. Then the two spheres can over lap in a end-to-end manner
♦ So 1s and 2s will form a sigma bond
■ So the answer is: (c) 2py and 2py
Solved example 4.13
Which hybrid orbitals are used by carbon atoms in the following molecules?
(a) CH3–CH3; (b) CH3–CH=CH2; (c) CH3-CH2-OH; (d) CH3-CHO (e) CH3COOH
Solution:
• In this problem, we are given the condensed formulas. The reader may draw the structural formula of each. Based on the structural formula, we can write the answers:
Part (a): CH3–CH3
• The structural formula is shown in fig.4.229(a) below:
 |
| Fig.4.229 |
1. The first C atom has four single bonds around it. So this C atom is sp3 hybridized
2. The second C atom has four single bonds around it. So this C atom is also sp3 hybridized
Part (b): CH3–CH=CH2
• The structural formula is shown in fig.4.229(b) above
1. The first C atom has four single bonds around it. So this C atom is sp3 hybridized
2. The second C atom has three bonds around it:
(i) Single bond with the left C atom
♦ This is a sigma bond
(ii) Single bond with the H atom
♦ This is a sigma bond
(iii) Double bond with the right C atom
♦ This is a sigma bond and a pi bond
• The pi bond mentioned in 2(iii) is possible only if the 2pz is left unhybridized
♦ So only 2s, 2px and 2py will participate in the hybridization
♦ So it is a sp2 hybridization
3. The third C atom has three bonds around it:
(i) Single bond with a H atom
♦ This is a sigma bond
(ii) Single bond with the other H atom
♦ This is a sigma bond
(iii) Double bond with the middle C atom
♦ This is a sigma bond and a pi bond
• The pi bond mentioned in 3(iii) is possible only if the 2pz is left unhybridized
♦ So only 2s, 2px and 2py will participate in the hybridization
♦ So it is a sp2 hybridization
Part (c): CH3-CH2-OH
• The structural formula is shown in fig.4.229(c) below:
 |
| Fig.4.229 |
2. The second C atom has four single bonds around it. So this C atom is sp3 hybridized
Part (d): CH3-CHO
• The structural formula is shown in fig.4.229(d) above
1. The first C atom has four single bonds around it. So this C atom is sp3 hybridized
2. The second C atom has three bonds around it:
(i) Single bond with the left C atom
♦ This is a sigma bond
(ii) Single bond with the H atom
♦ This is a sigma bond
(iii) Double bond with the O atom
♦ This is a sigma bond and a pi bond
• The pi bond mentioned in 2(iii) is possible only if the 2pz is left unhybridized
♦ So only 2s, 2px and 2py will participate in the hybridization
♦ So it is a sp2 hybridization
Part (e): CH3COOH
• The structural formula is shown in fig.4.229(e) above
1. The first C atom has four single bonds around it. So this C atom is sp3 hybridized
2. The second C atom has three bonds around it:
(i) Single bond with the left C atom
♦ This is a sigma bond
(ii) Single bond with the O atom
♦ This is a sigma bond
(iii) Double bond with the other O atom
♦ This is a sigma bond and a pi bond
• The pi bond mentioned in 2(iii) is possible only if the 2pz is left unhybridized
♦ So only 2s, 2px and 2py will participate in the hybridization
♦ So it is a sp2 hybridization
• In the next chapter, we will see states of matter
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