Chapter 13.18 - Reason For The Stability of Benzene

In the previous section, we saw that, benzene has unusual stability. In this section, we will see the explanation for such a stability.

The explanation can be written in 12 steps:
1. Each of the six C atoms in benzene is sp² hybridized. So each of them will have three hybrid orbitals.
2. Consider any one C atom.
• It can use two hybrid orbitals to form 𝜎 bonds with two adjacent C atoms. Such 𝜎 bonds are shown in red color in fig.13.99 below:

Fig.13.99

• The remaining hybrid orbital can be used to form a 𝜎 bond with the s-orbital of a H atom. Such 𝜎 bonds are shown in green color in the fig.13.99.
• For example:
   ♦ The C₃ forms 𝜎 bonds with C₂ and C₄
   ♦ This C₃ atom forms 𝜎 bond with a H atom also.
• All the 𝜎 bonds lie on a hexagonal plane.
3. Recall that, any C atom which is sp² hybridized, will have an unhybridized p orbital. This p orbital will be perpendicular to the hexagonal plane. Those p orbitals are shown in magenta color in fig.13.99.
4. The p orbitals can overlap in lateral direction to form π bonds.
• Two p orbitals are required to form one π bond. We have six p orbitals. So there will be three π bonds.
◼ In fig.13.100(a) below,
   ♦ p orbitals of C₁ and C₂ overlaps laterally to form a π bond.
   ♦ p orbitals of C₃ and C₄ overlaps laterally to form a π bond.
   ♦ p orbitals of C₅ and C₆ overlaps laterally to form a π bond.

Fig.13.100

 
• This arrangement will result in the alternate single bonds and double bonds as shown in fig.13.100(b)
◼ In fig.13.101(a) below,
   ♦ p orbitals of C₂ and C₃ overlaps laterally to form a π bond.
   ♦ p orbitals of C₄ and C₅ overlaps laterally to form a π bond.
   ♦ p orbitals of C₆ and C₁ overlaps laterally to form a π bond.

Fig.13.101

   
• This arrangement will result in the alternate single bonds and double bonds as shown in fig.13.101(b).
5. Consider the two arrangements:
   ♦ Fig.13.100(a) shows one possible arrangement.
   ♦ Fig.13.101(b) shows the second possible arrangement.
• Both the arrangements are equally likely to occur.
• That means, at any instant, we will be seeing an electron cloud distributed over the entire area.
• This is shown in fig.13.102(a) below:

Fig.13.102

6. When the electrons belong to particular C atoms, we say that those electrons are localized.
   ♦ In fig.100 and 101, the electrons in the p orbitals are localized.
• But when the electrons do not belong to particular C atoms, we say that, those electrons are delocalized.
   ♦ In fig.102, the electrons in the p orbitals are delocalized.
7. The delocalization as in fig.102 will result in an electron cloud which is distributed over the whole area of the molecule.
• Fig.103 below shows three layers.
(i) The upper red layer is the cloud formed by all the upper lobes of the p orbitals.
(ii) The lower red layer is the cloud formed by all the lower lobes of the p orbitals.
(iii) The middle grey hexagonal layer is the plane in which the six C atoms are situated.

Fig.13.103

8. The three layers act together as a single unit.
• Such an unit will have complete symmetry.
• We will not be able to say that:
   ♦ some portions of the molecule have single bonds.
   ♦ some other portions have double bonds.
9 . When the electron clouds of the p orbitals are distributed over the entire area, they are better attracted by the nuclei of the C atoms.
• If the electron clouds are localized between particular C atoms, such a high amount of attraction will not be possible.
• So the delocalization gives extra stability to the benzene molecule.
10. Let us see some experimental results which can be used as proof for the structure shown in fig.103. It can be written in two steps:
(i) X-ray diffraction data shows that, benzene has a planar structure.
(ii) Imagine that, benzene indeed has alternate single bonds and double bonds. Then:
   ♦ The distance between some of the adjacent C atoms would be 154 pm
(The C-C single bond has a bond length of 154 pm)  
   ♦ The distance between the remaining of the adjacent C atoms would be 133 pm
(The C=C double bond has a bond length of 133 pm)  
• But X-ray data shows that distance between any two adjacent C atoms in benzene are same, which is 139 pm.
• Note that, 139 is a value which is intermediate between 133 and 154. This indicates that:
   ♦ There are no pure single bonds in benzene.
   ♦ There are no pure double bonds in benzene.
11. The absence of pure double bonds is the reason why benzene cannot undergo addition reactions in normal conditions. The reluctance to addition reactions puzzled scientists because, they expect all molecules containing double bonds to undergo addition reactions. Also, double or triple bonds are necessary to explain how six H atoms are able to satisfy the valencies of six C atoms.
12. It is not convenient to always draw the 3D views showing the details of p orbitals. So we adopt the method shown in fig.104 below:

Fig.13.104

• The fig.104 shows that, the actual benzene molecule is a hybrid of (A) and (B). The hybrid is shown in (C). Since the π electrons are delocalized, we insert a circle inside the hexagon.

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In the next section, we will see it aromaticity.

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Chapter 13.8 - Nomenclature of Alkenes

In the previous section, we completed a discussion on conformations of ethane. In this section, we will start a discussion on alkenes.

Let us recall some properties of alkenes that we have seen in earlier chapters. They can be written in 4 steps:
1. Alkenes are unsaturated hydrocarbons. They contain at least one double bond.
2. If there is one double bond in an alkene, it will contain two H atoms less than the corresponding alkane.
• For example:
    ♦ Molecular formula of butane is C₄H₁₀
    ♦ Molecular formula of butene is C₄H₈
• We know that, the general formula for alkanes is C_nH_2n+2
• We also know that, the corresponding alkene has two H atoms less. So the general formula for alkenes is C_nH_2n
3. Chemists in the 19^th century noticed that, the first member of the alkene series (ethene) formed a oily liquid on reaction with chlorine. So the alkenes were generally known as olefins. The word olefin means ‘oil forming’.
4. Ethene is the IUPAC name of the first member. It’s common name is ethylene.

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Structure of double bond

Alkenes have at least one double bond. We have seen the details about that double bond in the previous chapters. [see fig.4.147 of section 4.26 ] Let us recall those details. They can be written in 8 steps:
1. The double bond consists of:
    ♦ One sigma (σ) bond
    ♦ One pi (𝜋) bond.
2. The sigma bond is formed by the head-on overlapping of the sp² hybridized orbitals.
    ♦ The pi bond is formed by the sideways overlapping of the two 2p orbitals.
3. The sigma bond is a strong bond.
    ♦ Bond enthalpy of a sigma bond is 397 kJ mol^-1
• Pi bond is a weak bond.
    ♦ Bond enthalpy of a pi bond is 284 kJ mol^-1
4. The double bond is shorter in bond length.
    ♦ The C-C single bond in alkanes has a bond length of 154 pm.
    ♦ The C=C double bond in alkenes has a bond length of 134 pm.
    ♦ The C-H single bond has a bond length of 110 pm.
• The bond angle between C-H single bond and C=C double bond is 121.7^o
• The bond angle between the two C-H single bonds in a CH₂ group is 116.6^o
• These details are shown in fig.13.46 below:

Fig.13.46

5. In a pi bond, there is sideways overlapping of the p-orbitals. The two p-orbitals combine together to form a 𝜋-cloud. We saw this in figs.4.148 and 4.149 of section 4.26.
6. The electrons in the 𝜋-cloud are loosely held.
• Those loosely held electrons are easily available for reagents which are in search for electrons (reagents in search for electrons are known as electrophilic reagents).
• So electrophilic reagents attack alkenes easily.
7. We can say that:
• Alkenes have lesser stability when compared to alkanes.
• Alkenes can be easily converted into alkanes by combining with electrophilic reagents.
8. In step (3), we wrote that:
    ♦ Bond enthalpy of the sigma bond in a double bond is 397 kJ mol^-1
    ♦ Bond enthalpy of the pi bond in a double bond is 284 kJ mol^-1
• So the bond enthalpy of a C=C double bond as a whole will be (397+284) = 681 kJ mol^-1
• The bond enthalpy of a C-C single bond is 348 kJ mol^-1

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Nomenclature of Alkenes

• We have seen the rules for writing the IUPAC names in an earlier chapter [see section 12.3]
• Let us recall some basic rules which are applicable to alkenes. It can be written in 3 steps:
1. The longest chain should be selected in such a way that, it contains the double bonds.
2. Numbering must be done in such a way that, the double bonds get the lowest possible numbers.
3. The suffix ‘ene’ is used instead of ‘ane’ of alkanes.
• Let us see some examples:

Example 1:
Write the IUPAC name of the structure: CH₃-CH=CH₂
Solution:
1. Applying rule 1, we see that, there are three C atoms in the main chain.
2. Applying rule 2, we see that, ‘prop’ must be used.
3. Applying rule 3, we get propane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: propane – ane + ene = propene
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that the correct way of numbering is from right to left.
    ♦ Numbering from right to left will give number ‘1’ to the double bond.
    ♦ Numbering from left to right will give number ‘2’ to the double bond.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
Prop-1-ene.
7. Here we note an interesting point:
• In propene, the double bond has only two possible positions. Both those positions will give the same name: Prop-1-ene. So we can write the name simply as: Propene.

Example 2:
Write the IUPAC name of the structure: CH₃-CH₂-CH=CH₂
Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: butane – ane + ene = butene
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that the correct way of numbering is from right to left.
    ♦ Numbering from right to left will give number ‘1’ to the double bond.
    ♦ Numbering from left to right will give number ‘3’ to the double bond.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
But-1-ene.

Example 3:
Write the IUPAC name of the structure: CH₃-CH=CH-CH₃
Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: butane – ane + ene = butene
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that numbering can be done either from left to right or from right to left.
    ♦ Numbering from left to right will give number ‘2’ to the double bond.
    ♦ Numbering from right to left will give number ‘2’ to the double bond.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
But-2-ene.

Example 4:
Write the IUPAC name of the structure: CH₂=CH-CH=CH₂
Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: butane – ane + ene = butene
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that numbering can be done either from left to right or from right to left.
    ♦ Numbering from left to right will give numbers '1,3' to the double bonds.
    ♦ Numbering from right to left will give number ‘1,3’ to the double bonds.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
Buta-1,3-diene.
• Note that, since more than one double bond is present, we write 'buta' instead of 'but'
• Similarly, 'penta' is written instead of 'pent', 'hexa' is written instead of hex', so on . . .

Example 5:
Write the IUPAC name of the structure shown in fig.13.47(a) below:

Fig.13.47

Solution:
1. Applying rule 1, we see that, there are three C atoms in the main chain containing double bond.
2. Applying rule 2, we see that, ‘prop’ must be used.
3. Applying rule 3, we get propane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: propane – ane + ene = propene
4. Applying rule 4, we see that, there is one branch: methyl.
5. Applying rule 5, we see that numbering should be done from left to right. This is shown in fig.b.
    ♦ Numbering from left to right will give number '1' to the double bond.
    ♦ Numbering from right to left will give number ‘2’ to the double bond.
6. Applying rule 6, we get: 2-methyl.
7. Rule 7 is not applicable here because, there is only one branch.
8. Applying rule 8, we get: 2-Methylprop-1-ene.

Example 6:
Write the IUPAC name of the structure shown in fig.13.48(a) below:

Fig.13.48

Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain containing the double bond.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: butane – ane + ene = butene
4. Applying rule 4, we see that, there is one branch: methyl.
5. Applying rule 5, we see that numbering should be done from left to right. This is shown in fig.b.
    ♦ Numbering from left to right will give number '1' to the double bond.
    ♦ Numbering from right to left will give number ‘3’ to the double bond.
6. Applying rule 6, we get: 3-methyl.
7. Rule 7 is not applicable here because, there is only one branch.
8. Applying rule 8, we get: 3-Methylbut-1-ene.

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Now we will see some solved examples:

Solved example 13.7
Write the IUPAC names of the three compounds shown in fig.13.49 below:

Fig.13.49

Solution:
Part (i):
We are given the condensed formula. It is shown in fig.13.50(a) below:

Fig.13.50

• The expanded form is shown in fig.b. Based on the expanded form, we can write 8 steps: 
1. Applying rule 1, we see that, there are ten C atoms in the main chain containing both the double bonds.
2. Applying rule 2, we see that, ‘dec’ must be used.
3. Applying rule 3, we get decane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: decane – ane + ene = decene
4. Applying rule 4, we see that, there are two branches: two methyl branches.
5. Applying rule 5, we see that numbering should be done from left to right. This is shown in fig.b.
    ♦ Numbering from left to right will give numbers '3,6' to the double bonds.
    ♦ Numbering from right to left will give numbers ‘4,7’ to the double bond.
6. Applying rule 6, we get: 2-methyl and 8-methyl.
7. Applying rule 7, we get: 2,8-dimethyl.
8. Applying rule 8, we get: 2,8-Dimethyldeca-3,6-diene.

Part (ii):
We are given the bond line formula. It is shown in fig.13.51(a) below:

Fig.13.51

• The expanded form is shown in fig.b. Based on the expanded form, we can write 6 steps: 
1. Applying rule 1, we see that, there are eight C atoms in the main chain.
2. Applying rule 2, we see that, ‘oct’ must be used.
3. Applying rule 3, we get octane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: octane – ane + ene = octene
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that numbering can be done either from left to right or from right to left. This is shown in fig.b.
    ♦ Numbering from left to right will give numbers '1,3,5,7' to the double bonds.
    ♦ Numbering from right to left will give the same numbers ‘1,3,5,7’ to the double bonds.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
Octa-1,3,5,7-tetraene.

Part (iii):
We are given the condensed formula. It is shown in fig.13.52(a) below:

Fig.13.52

• The expanded form is shown in fig.b. Based on the expanded form, we can write 8 steps: 
1. Applying rule 1, we see that, there are five C atoms in the main chain containing the double bond.
2. Applying rule 2, we see that, ‘pent’ must be used.
3. Applying rule 3, we get pentane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: pentane – ane + ene = pentene
4. Applying rule 4, we see that, there is one branch: propyl.
5. Applying rule 5, we see that numbering should be done from left to right. This is shown in fig.b.
    ♦ Numbering from left to right will give number '1' to the double bond.
    ♦ Numbering from right to left will give number ‘4’ to the double bond.
6. Applying rule 6, we get: 2-propyl.
7. Rule 7 is not applicable because, there is only one branch.
8. Applying rule 8, we get: 2-Propylpent-1-ene.

Part (iv):
We are given the expanded form. It is shown in fig.13.53(a) below:

Fig.13.53

• Based on the expanded form, we can write 8 steps: 
1. Applying rule 1, we see that, there are ten C atoms in the main chain containing the double bond.
2. Applying rule 2, we see that, ‘dec’ must be used.
3. Applying rule 3, we get decane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: decane – ane + ene = decene
4. Applying rule 4, we see that, there are three branches: two methyl branches and one ethyl branch.
5. Applying rule 5, we see that numbering should be done from right to left. This is shown in fig.b.
    ♦ Numbering from left to right will give number '6' to the double bond.
    ♦ Numbering from right to left will give number ‘4’ to the double bond.
6. Applying rule 6, we get: 2-methyl, 4-ethyl and 6-methyl.
7. Applying rule 7, we get: 4-ethyl-2,6-dimethyl.
(Note that, ethyl gets preference over methyl when we consider the alphabetical order)
8. Applying rule 8, we get: 4-Ethyl-2,6-dimethyldec-4-ene.

Solved example 13.8
Calculate the number of sigma (𝜎) and pi (π) bonds in the four structures in the above solved example 13.7
Solution:
Part (a):
1. From the expanded form in fig.13.50(b), we can count the number of bonds:
   ♦ Counting the number of single bonds, we get: 31 Nos.
        ✰ Number of C-H single bonds = 22 Nos. 
        ✰ Number of C-C single bonds = 9 Nos. 
   ♦ Counting the number of double bonds, we get: 2 Nos. 
        ✰ Number of C=C double bonds = 2 Nos.
2. Number of σ bonds can be calculated in 3 steps:
(i) We know that, all single bonds will be σ bonds.
• So we can write:
    ♦ Number of $\sigma_{C-H}$ = 22
    ♦ Number of $\sigma_{C-C}$ = 9
(ii) We know that, all double bonds will be have one σ bond each.
• The double bond is between two C atoms. In our present case, there are two double bonds.
• So we can write:
    ♦ Number of $\sigma_{C=C}$ = 2
(iii) Now we can write the total numbers:
• Total number of σ bonds:
    ♦ $\sigma_{C-H}$ =22
    ♦ $\sigma_{C-C}$ = 9
    ♦ $\sigma_{C=C}$ = 2
3. Number of π bonds can be calculated in 2 steps:
(i) We know that, all double bonds will be have one π bond each.
• We see that the two double bonds in this structure are between C atoms.
    ♦ So the number of $\pi_{C=C}$ from the two double bonds = 2
(ii) Now we can write the total numbers:
• Total number of π bonds:
    ♦ $\pi_{C=C}$ = 2

Part (b):
1. From the expanded form in fig.13.51(b), we can count the number of bonds:
   ♦ Counting the number of single bonds, we get: 13 Nos.
        ✰ Number of C-H single bonds = 10 Nos. 
        ✰ Number of C-C single bonds = 3 Nos. 
   ♦ Counting the number of double bonds, we get: 4 Nos. 
        ✰ Number of C=C double bonds = 4 Nos.
2. Number of σ bonds can be calculated in 3 steps:
(i) We know that, all single bonds will be σ bonds.
• So we can write:
    ♦ Number of $\sigma_{C-H}$ = 10
    ♦ Number of $\sigma_{C-C}$ = 3
(ii) We know that, all double bonds will be have one σ bond each.
• The double bond is between two C atoms. In our present case, there are four double bonds.
• So we can write:
    ♦ Number of $\sigma_{C=C}$ = 4
(iii) Now we can write the total numbers:
• Total number of σ bonds:
    ♦ $\sigma_{C-H}$ =10
    ♦ $\sigma_{C-C}$ = 3
    ♦ $\sigma_{C=C}$ = 4
3. Number of π bonds can be calculated in 3 steps:
(i) We know that, all double bonds will be have one π bond each.
• We see that the four double bonds in this structure are between C atoms.
    ♦ So the number of $\pi_{C=C}$ from the four double bonds = 4
(ii) Now we can write the total numbers:
• Total number of π bonds:
    ♦ $\pi_{C=C}$ = 4

Part (c):
1. From the expanded form in fig.13.52(b), we can count the number of bonds:
   ♦ Counting the number of single bonds, we get: 22 Nos.
        ✰ Number of C-H single bonds = 16 Nos. 
        ✰ Number of C-C single bonds = 6 Nos. 
   ♦ Counting the number of double bonds, we get: 1 Nos. 
        ✰ Number of C=C double bonds = 1 No.
2. Number of σ bonds can be calculated in 3 steps:
(i) We know that, all single bonds will be σ bonds.
• So we can write:
    ♦ Number of $\sigma_{C-H}$ = 16
    ♦ Number of $\sigma_{C-C}$ = 6
(ii) We know that, all double bonds will be have one σ bond each.
• The double bond is between two C atoms. In our present case, there is one double bond.
• So we can write:
    ♦ Number of $\sigma_{C=C}$ = 1
(iii) Now we can write the total numbers:
• Total number of σ bonds:
    ♦ $\sigma_{C-H}$ =16
    ♦ $\sigma_{C-C}$ = 6
    ♦ $\sigma_{C=C}$ = 1
3. Number of π bonds can be calculated in 3 steps:
(i) We know that, all double bonds will be have one π bond each.
• We see that the one double bond in this structure is between C atoms.
    ♦ So the number of $\pi_{C=C}$ from the one double bonds = 1
(ii) Now we can write the total numbers:
• Total number of π bonds:
    ♦ $\pi_{C=C}$ = 1

Part (d):
1. From the expanded form in fig.13.53, we can count the number of bonds:
   ♦ Counting the number of single bonds, we get: 13 Nos.
        ✰ Number of C-H single bonds = 28 Nos. 
        ✰ Number of C-C single bonds = 12 Nos. 
   ♦ Counting the number of double bonds, we get: 1 No. 
        ✰ Number of C=C double bonds = 1 No.
2. Number of σ bonds can be calculated in 3 steps:
(i) We know that, all single bonds will be σ bonds.
• So we can write:
    ♦ Number of $\sigma_{C-H}$ = 28
    ♦ Number of $\sigma_{C-C}$ = 12
(ii) We know that, all double bonds will be have one σ bond each.
• The double bond is between two C atoms. In our present case, there is one double bond.
• So we can write:
    ♦ Number of $\sigma_{C=C}$ = 1
(iii) Now we can write the total numbers:
• Total number of σ bonds:
    ♦ $\sigma_{C-H}$ =28
    ♦ $\sigma_{C-C}$ = 12
    ♦ $\sigma_{C=C}$ = 1
3. Number of π bonds can be calculated in 3 steps:
(i) We know that, all double bonds will be have one π bond each.
• We see that the one double bond in this structure is between C atoms.
    ♦ So the number of $\pi_{C=C}$ from the one double bonds = 1
(ii) Now we can write the total numbers:
• Total number of π bonds:
    ♦ $\pi_{C=C}$ = 1

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In the next section we will see isomerism in alkenes.

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Chapter 12.18 - Hyperconjugation in Isopropyl Cation

In the previous section, we saw the hyperconjugation in ethyl cation. In this section, we will see hyperconjugation in isopropyl cation. It can be written in steps:

1. Fig.12.109 below shows the 3D view of the isopropyl cation.

Fig.12.109

• C₁ and C₃ are sp³ hybridized. C₂ is sp² hybridized.
• In this situation, C₂ carries the +ve charge. This is shown in the resonance structure I in fig.12.110 further below.
2. Two types of rotation are possible:
    ♦ C₁ can rotate about the red axis.
    ♦ C₃ can rotate about the green axis.
• The red and green axes intersect at C₂. But those two axes have different directions.
3. Suppose that, C₃ remains stationary while C₁ rotates about the red axis.
• Then H_A, H_B and H_C will come successively in alignment with the empty p orbital of C₂.
4. When H_A comes into alignment, hyperconjugation takes place between the C₁ㅡH_A σ bond and the empty p orbital of C₂.
• This is indicated by the red curved-arrow in I.

Fig.12.110

• As a result, the C₁ㅡC₂ single bond becomes C₁=C₂ double bond. This double bond is shown in II. The H_A loses the bond with C₁. Due to the loss of an electron, H_A gets a +ve charge. This is also shown in II.
5. So now we know how II is obtained. We can analyze the curved-arrows in II.
• The magenta curved-arrow in II comes into play when H_B comes into alignment with the empty p-orbital of C₂. Hyperconjugation takes place between the C₁ㅡH_B σ bond and the empty p orbital. Thus we get the structure III.
6. In this way, by following the route shown by the double headed green arrows, we can understand all the seven resonance structures of the isopropyl cation. Note that, IV, V, VI and VII are obtained when C₁ remains stationary and C₃ rotates.
7. Now we can write about the stability of the isopropyl cation. It can be written in three steps:
(i) Originally, the isopropyl cation does not have any double bond. The + charge is carried by the C₂ atom. This is one of the seven resonance structures.
(ii) Due to hyperconjugation, the single bonds between the three C atoms become double bonds. The + charge is carried successively by six H atoms.
(iii) So in the hybrid structure, the + charge is distributed among seven atoms:
One C atom and six H atoms.
• Such a distribution of charge, gives greater stability to the isopropyl cation.
8. In an earlier section, we wrote about the stability of isopropyl cation. (step 9 below fig.12.70 of section 12.10). The above seven steps helps us to explain it's stability.
9. We can write a comparison between the stability of various cations. It can be written in 5 steps:
(i) In ethyl cation, the +ve charge is distributed among four atoms.
(ii) In isopropyl cation, the +ve charge is distributed among seven atoms.
(iii) So isopropyl cation will have greater stability when compared to ethyl cation.
(iv) In tertiary butyl cation (CH₃)₃C, the +ve charge is distributed among even more number of atoms.
(v) Thus we get the order:
Tertiary butyl cation is the most stable, followed by isopropyl cation, followed by ethyl cation, followed by methyl cation.

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◼ We see that, the methyl cation has the least stability. The reason can be written in 2 steps:
1. Fig.12.71 of section 12.10 shows the 3D view of methyl cation. We see that, all the CㅡH σ bonds lie in a plane. But the empty p-orbital is perpendicular to that plane.
2.So none of the CㅡH σ bonds can ever come into alignment with the empty p-orbital.
• We can write:
Hyperconjugation can never occur in methyl cation. So it has the least stability when compared to the other cations.

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Next we will see the hyperconjugation in propene. It can be written in 6 steps:
1. Fig.12.111 below shows the 3D view of propene.

Fig.12.111

• C₁ is sp³ hybridized. C₂ and C₃ are sp² hybridized.
• The p-orbitals of C₂ and C₃ are not empty. They contain one electron each. So a 𝜋 bond is formed between C₂ and C₃ by the lateral overlap of those p-orbitals. This lateral overlap is indicated by the two double headed yellow arrows.
• Thus we see the double bond C₂=C₃ in the resonance structure I in fig.12.112 further below.
(The p-orbital of C₂ of the isopropyl cation that we saw previously, was empty because, the cation has lost one electron. But in our present case of propene, no electron is lost)
2. C₁ can rotate about the red axis.
• Then H_A, H_B and H_C will come successively in alignment with the p-orbital of C₂.
3. Hyperconjugation takes place between the C₁ㅡH_A σ bond and the p-orbital of C₂.
• That is., the two electrons in the C₁ㅡH_A σ bond gets delocalized into the p-orbital.
• This is indicated by the red curved-arrow in I.

Fig.12.112

• As a result, the C₁ㅡC₂ single bond becomes C₁=C₂ double bond. This double bond is shown in II. The H_A loses the bond with C₁. Due to the loss of an electron, H_A gets a +ve charge. This is also shown in II.
• But due to the red arrow in I, C₂ has gained two extra electrons. It does not need two extra electrons. It has already octet. So the magenta arrow comes into play. Two electrons are transferred to C₃. The C₃ gains two electrons and a -ve charge.
4. So now we know how II is obtained. We can analyze the curved-arrows in II.
• The magenta curved-arrow in II comes into play when H_B comes into alignment with the p-orbital of C₂. Hyperconjugation takes place between the C₁ㅡH_B σ bond and the p orbital.
• The red curved-arrow indicates that, H_A regains it's two electrons. Thus we get the structure III.
5. In this way, by following the route shown by the double headed green arrows, we can understand all the four resonance structures of propene.
6. Note that, the +ve charge is distributed among three H atoms.
• The negative charge is not distributed. It permanently resides at C₃.
• The actual propene molecule is the hybrid of the four resonance structures. So the hybrid structure will be a polarized structure. One end of that structure has a +ve charge and the other end has a -ve charge.
• In other words, all molecules in a sample of propene will be in a polarized state.

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• We saw three examples for hyperconjugation. Ethyl cation, isopropyl cation and propene. Based on those examples, we can now write a definition for hyperconjugation. It can be written in 4 steps:
1. Hyperconjugation involves delocalization of the electrons in the CㅡH σ bond of an alkyl group.
[Recall that, in all the examples that we saw, the delocalization started from the CㅡH σ bond of the alkyl group  (methyl group: ㅡCH₃)]
2. In some cases, the delocalized electrons move into an unshared p-orbital.
[Recall that, this happened in ethyl cation and isopropyl cation]
3. In some other cases, the delocalized electrons move into a nearby unsaturated system (double or triple bond).
[Recall that, this happened in propene]
4. Hyperconjugation is a permanent effect.

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• We have completed a discussion on mechanisms of organic reactions. These mechanisms will help us to understand the following types of reactions:
    ♦ Substitution reactions
    ♦ Addition reactions
    ♦ Elimination reactions
    ♦ Rearrangement reactions
• We will see these reactions in later sections.

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• The link below gives the folder containing additional solved examples on this chapter.
• Part 2 is related to reaction mechanism.

Additional solved examples

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• In the next section we will see the methods of purification of organic compounds.

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Chapter 12 - Basic Principles in Organic Chemistry

In this chapter, we will see some basic principles and techniques in organic chemistry.

Catenation

• Carbon exhibits the property of catenation.
• It is the property by which an element can bind with itself through covalent bonds to form long chains, branched chains or rings.
   ♦ Due to this property, carbon can form a large number of compounds.
• Since the number of such compounds is large, they are studied under a separate branch of chemistry called, organic chemistry.

Relation between structure of an organic molecule and it’s chemical properties

This can be explained in 5 steps:
1. We have seen that, the C atom in CH₄ is sp³ hybridized. (Fig.4.136 of section 4.25)
• There will be four orbitals. Those four orbitals will be directed towards the four corners of a tetrahedron.
2. In C₂H₄, the C atoms are sp² hybridized. (Fig.4.149 of section 4.26)
• The sp² hybrid orbitals have greater s character than sp³ hybrid orbitals.
3. Similarly, the sp hybrid orbitals of C₂H₂ have still more s character. (Fig.4.157 of section 4.27)
4. This greater or lesser s character will influence the chemical and physical properties of molecules. We will see more details in later sections.
5. Another aspect is the presence of π bonds.
• We have seen that in C₂H₄, all atoms lie on the same plane. But the p orbitals are oriented out of plane in the perpendicular direction to the plane. (Fig.4.150 of section 4.26)
• These perpendicular electron clouds are easily available for the attacking reagents. This will influence the general chemical properties of C₂H₄.

As we proceed with our discussion, we will see more examples.

Let us see a solved example

Solved example 12.1
How many σ and π bonds are present in (a) HC☰CCH＝CH₃ (b) CH₂＝C＝CHCH₃
Solution:
Part (a):
1. We are given the condensed formula. Based on that, we will draw the structural formula. In the structural formula, all the single, double and triple bonds will become visible. It is shown in fig.12.1(a) below. Note that, we will learn the details about how to draw those structural formulae, in later sections of this chapter.

Fig.12.1

2. Number of bonds:
   ♦ Counting the number of single bonds, we get: 8 Nos.
        ✰ Number of C-H single bonds = 6 Nos. 
        ✰ Number of C-C single bonds = 2 Nos. 
   ♦ Counting the number of double bonds, we get: 1 No. 
        ✰ Number of C=C double bonds = 1 No. 
   ♦ Counting the number of triple bonds, we get: 1 No. 
        ✰ Number of C☰C triple bonds = 1 No. 
3. Number of σ bonds can be calculated in 4 steps:
(i) We know that, all single bonds will be σ bonds.
• So we can write:
    ♦ Number of $\sigma_{C-H}$ = 6
    ♦ Number of $\sigma_{C-C}$ = 2
(ii) We know that, all double bonds will be have one σ bond each.
• The double bond is between two C atoms. In our present case, there is one double bond.
• So we can write:
    ♦ Number of $\sigma_{C=C}$ = 1
(iii) We know that, all triple bonds will be have one σ bond each.
• The triple bond is between two C atoms. In our present case, there is one triple bond.
• So we can write:
    ♦ Number of $\sigma_{C☰C}$ = 1
(iv) Now we can write the total numbers:
• Total number of σ bonds:
    ♦ $\sigma_{C-H}$ = 6
    ♦ $\sigma_{C-C}$ = 2
    ♦ $\sigma_{C=C}$ = 1
    ♦ $\sigma_{C☰C}$ = 1
4. Number of π bonds can be calculated in 3 steps:
(i) We know that, all double bonds will be have one π bond each.
• We see that the only one double bond in this molecule is between two C atoms.
    ♦ So the number of $\pi_{C=C}$ from the one double bond = 1
(ii) We know that, all triple bonds will be have two π bonds each.
• We see that the only one triple bond in this molecule is between two C atoms.
    ♦ So the number of $\pi_{C☰C}$ from the one double bond = 2
(iii) Now we can write the total numbers:
• Total number of π bonds:
    ♦ $\pi_{C=C}$ = 1
    ♦ $\pi_{C☰C}$ = 2

Part (b):
1. We are given the condensed formula. Based on that, we will draw the structural formula. In the structural formula, all the single, double and triple bonds will become visible. It is shown in fig.12.1(b) above.
2. Number of bonds:
   ♦ Counting the number of single bonds, we get: 7 Nos.
        ✰ Number of C-H single bonds = 6 Nos. 
        ✰ Number of C-C single bonds = 1 Nos. 
   ♦ Counting the number of double bonds, we get: 2 No. 
        ✰ Number of C=C double bonds = 2 No. 
3. Number of σ bonds can be calculated in 4 steps:
(i) We know that, all single bonds will be σ bonds.
• So we can write:
    ♦ Number of $\sigma_{C-H}$ = 6
    ♦ Number of $\sigma_{C-C}$ = 1
(ii) We know that, all double bonds will be have one σ bond each.
• The double bond is between two C atoms. In our present case, there are two double bonds.
• So we can write:
    ♦ Number of $\sigma_{C=C}$ = 2
(iii) Now we can write the total numbers:
• Total number of σ bonds:
    ♦ $\sigma_{C-H}$ = 6
    ♦ $\sigma_{C-C}$ = 1
    ♦ $\sigma_{C=C}$ = 2
4. Number of π bonds can be calculated in 3 steps:
(i) We know that, all double bonds will be have one π bond each.
• We see that the two double bonds in this molecule is between C atoms.
    ♦ So the number of $\pi_{C=C}$ from the two double bonds = 2

Solved example 12.2
What is the type of hybridization of each carbon atom in the following compounds?
(a) CH₃Cl,   (b) (CH₃)₂CO,  (c) CH₃CN,  (d) HCONH₂,  (e) CH₃CH=CHCN
Solution:
• We are given the condensed formulae. Based on that, we will draw the structural formulae. In the structural formulae, all the single, double and triple bonds will become visible. They are shown in figs.12.2 and 12.3 below. Note that, we will learn the details about how to draw those structural formulae, in later sections of this chapter.

Fig.12.2

Part (a):
• Consider the C atom in fig.12.2(a) above.
• There are four single bonds around it. This is possible only if there are four orbitals around that C atom.
• Only sp³ hybridization can give four orbitals. The Cl and H atoms will enter into σ bonding with the four sp³ hybrid orbitals.
• So we can write:
The type of hybridization of the C atom is sp³

Part (b):
1. Consider the left side C atom in fig.12.2(b) above.
• There are four single bonds around it. This is possible only if there are four orbitals around that C atom.
• Only sp³ hybridization can give four orbitals. The middle C and H atoms will enter into σ bonding with the four sp³ hybrid orbitals.
• So we can write:
The type of hybridization of the left side C atom is sp³
2. Consider the middle side C atom in fig.12.2(b) above.
• There are two single bonds and one double bond around it. This is possible only if there are three orbitals around that C atom.
• Only sp² hybridization can give three orbitals. The left and right C atoms will enter into σ bonding with two of the three sp² hybrid orbitals.
• The O atom will enter into σ bonding with the third sp² hybrid orbital. Also, the p orbital of the C atom will enter into π bonding with O. This creates the double bond with O.
• So we can write:
The type of hybridization of the middle C atom is sp²
3. Consider the right side C atom.
• By symmetry, we can write the same steps as in (1).
• We can write:
The type of hybridization of the right side C atom is sp³

Part (c):
1. Consider the left side C atom in fig.12.2(c) above.
• There are four single bonds around it. This is possible only if there are four orbitals around that C atom.
• Only sp³ hybridization can give four orbitals. The right side C and H atoms will enter into σ bonding with the four sp³ hybrid orbitals.
• So we can write:
The type of hybridization of the left side C atom is sp³.
2. Consider the right side C atom in fig.12.2(c) above.
• There is one single bond and one triple bond around it. This is possible only if there are two orbitals around that C atom.
• Only sp hybridization can give two orbitals. The left C atom will enter into σ bonding with one of the two sp hybrid orbitals.
• The N atom will enter into σ bonding with the other sp hybrid orbital. Also, the p orbitals of the C atom will enter into two π bonding s with N. This creates the triple bond with N.
• So we can write:
The type of hybridization of right side C atom is sp.

Fig.12.3

Part (d):
• Consider the C atom in fig.12.3(d) above.
• There are two single bonds and one double bond around it. This is possible only if there are three orbitals around that C atom.
• Only sp² hybridization can give three orbitals. The left H atom and right NH₂ group will enter into σ bonding with two of the three sp² hybrid orbitals.
• The O atom will enter into σ bonding with the third sp² hybrid orbital. Also, the p orbital of the C atom will enter into π bonding with O. This creates the double bond with O.
• So we can write:
The type of hybridization of the C atom is sp²

Part (e):
1. Consider the first (from left) C atom in fig.12.3(e) above.
• There are four single bonds around it. This is possible only if there are four orbitals around that C atom.
• Only sp³ hybridization can give four orbitals. The second C atom and H atoms will enter into σ bonding with the four sp³ hybrid orbitals.
• So we can write:
The type of hybridization of the left side C atom is sp³
2. Consider the second C atom.
• There are two single bonds and one double bond around it. This is possible only if there are three orbitals around that C atom.
• Only sp² hybridization can give three orbitals. The first C atom and the H atom will enter into σ bonding with two of the three sp² hybrid orbitals.
• The third C atom will enter into σ bonding with the third sp² hybrid orbital. Also, the p orbital of the C atom will enter into π bonding with the third C. This creates the double bond with the third C.
• So we can write:
The type of hybridization of the second C atom is sp²
3. Consider the third C atom.
• There are two single bonds and one double bond around it. This is possible only if there are three orbitals around that C atom.
• Only sp² hybridization can give three orbitals. The fourth C atom and the H atom will enter into σ bonding with two of the three sp² hybrid orbitals.
• The second C atom will enter into σ bonding with the third sp² hybrid orbital. Also, the p orbital of the C atom will enter into π bonding with the second C. This creates the double bond with the second C.
• So we can write:
The type of hybridization of the third C atom is sp²
4. Consider the fourth C atom.
• There is one single bond and one triple bond around it. This is possible only if there are two orbitals around that C atom.
• Only sp hybridization can give two orbitals. The third C atom will enter into σ bonding with one of the two sp hybrid orbitals.
• The N atom will enter into σ bonding with the other sp hybrid orbital. Also, the p orbitals of the C atom will enter into two π bonding s with N. This creates the triple bond with N.
• So we can write:
The type of hybridization of the fourth C atom is sp.

Solved example 12.3
Write the state of hybridization of carbon in the following compounds and shape of each of the molecules.
(a) H₂C=O,   (b) CH₃F,  (c) HC☰N
Solution:
• We are given the condensed formulae. Based on that, we will draw the structural formulae. In the structural formulae, all the single, double and triple bonds will become visible. They are shown in figs.12.2 and 12.3 below. Note that, we will learn the details about how to draw those structural formulae, in later sections of this chapter.

Fig.12.4

Part (a):
1. Consider the C atom in fig.12.4(a) above.
• There are two single bonds and one double bond around it. This is possible only if there are three orbitals around that C atom.
• Only sp² hybridization can give three orbitals. The left and right H atoms will enter into σ bonding with two of the three sp² hybrid orbitals.
• The O atom will enter into σ bonding with the third sp² hybrid orbital. Also, the p orbital of the C atom will enter into π bonding with O. This creates the double bond with O.
• So we can write:
The type of hybridization of the C atom is sp²
2. In sp² hybridization, there are three orbitals. Those orbitals are directed towards the three corners of an equilateral triangle.
• So we can write:
H₂C=O will have a trigonal planar structure

Part (b):
1. Consider the C atom in fig.12.4(b) above.
• There are four single bonds around it. This is possible only if there are four orbitals around that C atom.
• Only sp³ hybridization can give four orbitals. The F and H atoms will enter into σ bonding with the four sp³ hybrid orbitals.
• So we can write:
The type of hybridization of the C atom is sp³
2. In sp³ hybridization, there are four orbitals. Those orbitals are directed towards the four corners of a tetrahedron.
• So we can write:
CH₃F will have a tetrahedral structure

Part (c):
1. Consider the C atom in fig.12.4(c) above.
• There is one single bond and one triple bond around it. This is possible only if there are two orbitals around that C atom.
• Only sp hybridization can give two orbitals. The left C atom will enter into σ bonding with one of the two sp hybrid orbitals.
• The N atom will enter into σ bonding with the other sp hybrid orbital. Also, the p orbitals of the C atom will enter into two π bonding s with N. This creates the triple bond with N.
• So we can write:
The type of hybridization of right side C atom is sp.
2. In sp hybridization, there are two orbitals. Those orbitals are directed along a straight line with an angle of 180^o between them.
• So we can write:
HC☰N will have a linear structure.

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In the next section, we will see Structural representations of Organic compounds.


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Chapter 4.29 - Structure of Trichloro boron and Beryllium chloride

• In the previous section 4.28, we saw the structure of NH₃ and H₂O. In this section, we will see the structure of BCl₃ (Trichloro boron) and BeCl₂ (Beryllium chloride)

Structure of BCl₃
• Fig.4.164(a) below shows the Lewis dot structure of BCl₃
• The details about the model of BCl₃ can be written in 10 steps:
1. The B atom in BCl₃ is sp² hybridized
• Let us see how this sp² hybridization is achieved. It can be written in 2 steps:
(i) Fig.4.164(b) below shows the orbitals in the valence shell of B

Fig.4.164

(ii) When enough energy is given, one electron in the 2s orbital jumps to the 2p_y orbital
• Thus we get three half filled orbitals. This is shown in fig.c
(iii) The 2s orbital mixes together with 2p_x and 2p_y. This is shown in fig.d
(iv) Since there is one s-orbital and two p-orbitals, it is a sp² hybridization
2. We know that:
    ♦ In sp² hybridization, there will be three resulting hybrid orbitals
          ✰ Together, they form a triangular planar shape
          ✰ The larger lobes will be directed towards the corners of a triangle
• This is shown in fig.4.165 below:

Fig.4.165

• Let us write four 'important points to remember' about fig.4.165:
(i) The three sp² hybrid orbitals are at an angular distance of 120^o apart 
(ii) The nucleus of the B atom is shown as a small red sphere
    ♦ This sphere is situated at the origin of the coordinate axes
(iii) One of the orbitals lies exactly along the x-axis (red axis)
(iv) We know that, all the three sp² hybrid orbitals will lie on a plane
    ♦ So in fig.4.165, all the three orbitals are lying on the xy-plane
3. So we can write:
    ♦ The 'sp² hybridized B atom' consists of two items:
          ✰ The three hybrid orbitals
          ✰ The nucleus (small red sphere)
 • In the fig.4.165, the 'smaller lobes of hybrid orbitals' are not shown
    ♦ This is to reduce congestion and thus obtain greater clarity
4. Distribution of electrons in the hybrid orbitals:
• This can be written in 3 steps:
(i) We know that, the sp² hybrid orbitals are formed from 'one s orbital' and 'two p orbitals'
(ii) In our present case of B atom, they are: 'one 2s orbital' and 'two 2p orbitals'
(iii) Before the hybridization, the orbitals mentioned in (ii) carry a total of three electrons
    ♦ After hybridization, the orbitals mentioned in (ii) will no longer exist
    ♦ Then what will happen to the three electrons?
Answer:
• The three electrons will be distributed among the three sp² hybrid orbitals
    ♦ So each hybrid orbital will carry one electron
    ♦ This is indicated by the arrows in fig.4.165 above

5. So in fig.4.165, we have:

    ♦ Three half filled hybrid orbitals
• Now bonding with Cl atoms can begin
• First we will write some basic details about the Cl atom. it can be written in 2 steps:
(i) The electronic configuration of Cl is: [Ne]3s²3p⁵
• Expanding the valence orbitals, the configuration becomes: [Ne]3s² 3p_x² 3p_y² 3p_z¹
(ii) It is clear that, the 3p_z of Cl is half filled
• This 3p_z needs one more electron
• This is shown in fig.4.166 below:

Fig.4.166

• The p_x orbital is shown in red color
    ♦ It has two arrows
• The p_y orbital is shown in green color
    ♦ It has two arrows
• The p_z orbital is shown in blue color
    ♦ It has only one arrow
• The 3s orbital is  represented by the cyan sphere 
    ♦ It has two arrows
6. One Cl atom will come and overlap with each of the three half filled orbitals of the B atom
• So a total of three Cl atoms will be coming
    ♦ In each of those Cl atoms, it is the p_z which enters into bonding
    ♦ This is shown in fig.4.167 below:

Fig.4.167

• In the fig.4.167 above, the 'p_z orbitals of Cl' have become completely filled
7. Is the structure 3D or 2D?
The answer can be written in 5 steps:
(i) The structure in the above fig.4.167, appears to be a 3D structure
• Because, the red orbitals of the Cl atoms are protruding above and below the xy-plane
(ii) But the red orbitals are mere electron clouds
    ♦ To define the shape of a molecule:
          ✰ We do not consider the positions of electrons or electron clouds
          ✰ We consider only the positions of 'nuclei of atoms' in that molecule
(iii) We see that, all the three Cl nuclei lie in the same plane as the nucleus of B
(Remember that, the nucleus of a Cl atom is at the center of the cyan sphere)
(iv) So the structure is planar. In other words, it is a 2D structure
• We do not need to show the orbitals (electron clouds) of the Cl atoms. So we can delete them
• However, we will retain the p_z orbital. This will convey the fact that, it is 'a p orbital of Cl' that bonds with the B atom
(v) The final structure is shown in fig.4.168(a) below.
• The small yellow spheres represent the nuclei of Cl atoms

Fig.4.168

8. Thus we get the model of the BCl₃ molecule
• Let us write the salient features of this model. It can be written in 3 steps:
(i) The three Cl atoms are situated at the three corners of a triangle
(ii) The B atom is situated at the 'center of gravity' of the triangle
(iii) The angle between any two bonds is 120^o
9. The 2D representation of the BCl₃ molecule is shown on fig.4.168(b) above
10. Note that, all the bonds in BCl₃ are sigma bonds

Structure of BeCl₂
• Fig.4.169(a) below shows the Lewis dot structure of BeCl₂
• The details about the model of BeCl₂ can be written in 10 steps:
1. The Be atom in BeCl₂ is sp hybridized
• Let us see how this sp hybridization is achieved. It can be written in 2 steps:
(i) Fig.4.169(b) below shows the orbitals in the valence shell of Be

Fig.4.169

(ii) When enough energy is given, one electron in the 2s orbital jumps to the 2p_x orbital
• Thus we get two half filled orbitals. This is shown in fig.c
(iii) The 2s orbital mixes together with 2p_x. This is shown in fig.d
(iv) Since there is one s-orbital and one p-orbital, it is a sp hybridization
2. We know that:
    ♦ In sp hybridization, there will be two resulting hybrid orbitals
          ✰ Together, they form a linear shape
• This is shown in fig.4.170 below:

Fig.4.170

• Let us write four 'important points to remember' about fig.4.170:
(i) The two sp hybrid orbitals are at an angular distance of 180^o apart 
(ii) The nucleus of the Be atom is shown as a small red sphere
    ♦ This sphere is situated at the origin of the coordinate axes
(iii) Both the orbitals lie exactly along the x-axis (red axis)
(iv) We know that, each hybrid orbital have a larger lobe and a smaller lobe
• In the previous cases (except C₂H₂), we deliberately chose not to show the smaller lobes. This was for better clarity
• In our present case, we do not have to hide the smaller lobes because, they are automatically hidden
    ♦ The smaller lobe of the left side orbital is inside the larger lobe of the right side orbital
    ♦ The smaller lobe of the right side orbital is inside the larger lobe of the left side orbital
(Recall that, in C₂H₂ also, we encountered the same situation)
3. So we can write:
    ♦ The 'sp hybridized B atom' consists of two items:
          ✰ The two hybrid orbitals
          ✰ The nucleus (small red sphere)
4. Distribution of electrons in the hybrid orbitals:
• This can be written in 3 steps:
(i) We know that, the sp hybrid orbitals are formed from 'one s orbital' and 'one p orbital'
(ii) In our present case of Be atom, they are: 'one 2s orbital' and 'one 2p orbital'
(iii) Before the hybridization, the orbitals mentioned in (ii) carry a total of two electrons
    ♦ After hybridization, the orbitals mentioned in (ii) will no longer exist
    ♦ Then what will happen to the two electrons?
Answer:
• The two electrons will be distributed among the two sp hybrid orbitals
    ♦ So each hybrid orbital will carry one electron
    ♦ This is indicated by the arrows in fig.4.170 above

5. So in fig.4.170, we have:

    ♦ Two half filled hybrid orbitals
• Now bonding with Cl atoms can begin
• First we will write some basic details about the Cl atom. it can be written in 2 steps:
(We have already seen the two steps when we discussed BCl₃ above. But we will write them again)
(i) The electronic configuration of Cl is: [Ne]3s²3p⁵
• Expanding the valence orbitals, the configuration becomes: [Ne]3s² 3p_x² 3p_y² 3p_z¹
(ii) It is clear that, the 3p_z of Cl is half filled
• This 3p_z needs one more electron. This was shown in fig.4.166 above

• The px orbital is shown in red color
    ♦ It has two arrows
• The py orbital is shown in green color
    ♦ It has two arrows
• The pz orbital is shown in blue color
    ♦ It has only one arrow
• The 3s orbital is  represented by the cyan sphere 
    ♦ It has two arrows
6. One Cl atom will come and overlap with each of the two half filled orbitals of the Be atom
• So a total of two Cl atoms will be coming
    ♦ In each of those Cl atoms, it is the p_z which enters into bonding
    ♦ This is shown in fig.4.171 below:

Fig.4.171

• In the fig.4.171 above, the 'p_z orbitals of Cl' have become completely filled
7. Is the structure 3D or 2D?
The answer can be written in 5 steps:
(i) The structure in the above fig.4.171, appears to be a 3D structure
• Because, the red orbitals of the Cl atoms are protruding above and below the xy-plane
(ii) But the red orbitals are mere electron clouds
    ♦ To define the shape of a molecule:
          ✰ We do not consider the positions of electrons or electron clouds
          ✰ We consider only the positions of 'nuclei of atoms' in that molecule
(iii) We see that, all the three Cl nuclei lie in the same plane as the nucleus of B
(Remember that, the nucleus of a Cl atom is at the center of the cyan sphere)
(iv) So the structure is planar. In other words, it is a 2D structure
• We do not need to show the orbitals (electron clouds) of the Cl atoms. So we can delete them
• However, we will retain the p_z orbital. This will convey the fact that, it is 'a p orbital of Cl' that bonds with the Be atom
(v) The final structure is shown in fig.4.172(a) below.
• The small yellow spheres represent the nuclei of Cl atoms

Fig.4.172

8. Thus we get the model of the BeCl₂ molecule
• Let us write the salient features of this model. It can be written in 4 steps:
(i) The two Cl atoms are situated at the opposite ends of a line
(ii) The Be atom is situated at the 'center of gravity' (midpoint) of the line
(iii) The angle between the two bonds is 180^o
(iv) So it is a linear structure
9. The 2D representation of the BeCl₂ molecule is shown on fig.4.172(b) above
10. Note that, both the bonds in BeCl₂ are sigma bonds

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• In the next section, we will see the hybridization involving d orbitals

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