Chapter 13.4 - Physical Properties of Alkanes

In the previous section, we completed a discussion on the preparation of alkanes. In this section, we will see the properties of alkanes.

Physical properties of Alkanes

The first physical property that we will discuss, is about the non-polar nature of alkanes. It can be written in 3 steps:
1. We have seen that some molecules are polar in nature. We know the reason for such polarity [see fig.4.223 in section 4.40]
2. In the case of alkanes, there are only two types of bonds: C-C bonds and C-H bonds.
• The difference in electronegativity between C and H atoms is very small. So neither C nor H can pull the electron clouds. Thus both C-C bonds and C-H bonds are non-polar.
3. So we can write:
Alkanes are almost non-polar molecules.

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The second physical property that we will discuss, is about the state of various alkanes. It can be written in 7 steps:
1. Alkanes have weak van der Waal’s forces. We have seen van der Waals’s forces in an earlier section. It is the collection of four forces: (i) London force (ii) Dipole-dipole force (iii) Dipole-induced dipole force (iv) Hydrogen bond [see section 5]
2. When the number of C atoms in the chain is small, there will be a more uniform distribution of charge.
3. If the charge distribution is uniform, there will not be +ve and -ve regions.
• Consequently, there will not be much attraction between the molecules.
4. So, if the number of C atoms is small, the alkanes molecules will be in the gaseous state at room temperature (298 K).
• The first four members (C₁ to C₄) of the alkane series are gases.     
5. When the number of C atoms in the chain is larger, there will be a non- uniform distribution of charge.
6. If the charge distribution is non-uniform, there will be +ve and -ve regions.
• Consequently, there will be attraction between the molecules.
7. So, if the number of C atoms is large, the alkane molecules will be in the liquid or solid states.
• C₅ to C₁₇ are liquids at room temperature (298 K).     
• C₁₈ and higher are solids at room temperature (298 K).

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The third physical property that we will discuss, is about the solubility of alkanes. It can be written in 7 steps:
1. Consider a solution in which the solvent is polar and solute is non-polar.
• Let us name the polar solvent as A and non-polar solute as B
2. Since A is polar, the molecules of A will be attracting each other.
• Consider any two molecules of A, which are close together. They will be acting like a chain.
• B being non-polar, cannot form an electrostatic attraction with the two A molecules.
• So the molecule B will not be able to break the chain and occupy the space between the two A molecules.
• Even if we forcibly put the B between the two A molecules, the B will be expelled.
• So it is clear that, non-polar substances will not dissolve in polar solvents.
3. On the other hand, if B is a polar substance just like A, then B can occupy space in between the two A molecules to form a new chain.
• So if B is polar, it will dissolve in A.
• We can write: Polar substances will dissolve in polar solvents.
4. Consider the situation when both A and B are non-polar.
• In such a situation, neither A or B has permanent polarity. But London forces is applicable to both of them.
• Due to London forces, both A and B have equal chances of forming electrostatic attraction with each other. So B will dissolve in A
5. Thus we see that:
   ♦ Non-polar substances do not dissolve in polar solvents.
   ♦ Polar substances dissolve in polar solvents.
   ♦ Non-polar substances dissolve in non-polar solvents.
• Based on this, we can write:
Like dissolves like.
6. We have seen that alkanes are non-polar. We already know that water is polar.
So it is clear that, alkanes are insoluble in water.
7. Grease is composed of alkanes. Petrol is also composed of alkanes. Both being non-polar, grease will dissolve in petrol. That is why petrol is used in dry cleaning, to remove grease stains from clothes.

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The fourth physical property that we will discuss, is about the boiling points of alkanes. It can be written in 3 steps:
1. The boiling point (b.p) of different alkanes can be obtained from the table in the data book. The table is also shown in the text book.
2. We see that, higher alkanes have higher boiling points.
This can be explained in two steps:
(i) As the number of C atoms increase, the size of the molecules increase. As a result, the magnitude of the inter molecular van der Waals forces will be higher.
(ii) When the magnitude of the forces increase, it will be difficult to separate the molecules from each other.
(iii) Consequently, the b.p increases for higher alkanes.
3. From the table we see another interesting point:
Pentane, 2-Methylbutane and 2,2-Dimethylpropane have the same molecular mass but different boiling points.
• This can be explained in 4 steps:
(i) Pentane, 2-Methylbutane and 2,2-Dimethylpropane are isomers. So they have the same molecular mass.
    ♦ Pentane has a straight chain structure.
    ♦ 2-Methylbutane has one branch.
    ♦ 2,2-Dimethylpropane has two branches.
(ii) When the number of branches increase, the shape of the molecule becomes more and more spherical.
(iii) Spherical shape has lower surface area.
• Due to the lower surface area, the area of contact between the spheres will be small. Consequently, the inter molecular forces will be small.
(iv) If the inter molecular forces are low, the molecules can be easily separated from each other.
• So we can write:
When the shape becomes more and more spherical, the b.p becomes lower and lower.

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The fifth and final physical property needs a mention only:
Alkanes are colourless and odourless

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Chemical properties of Alkanes

• Alkanes are generally inert towards acids, bases, oxidising agents and reducing agents. The reason can be written in 3 steps:
(i) In alkanes, all four valencies of C atoms are satisfied. The single valency of H atoms are also satisfied. So an alkane molecule as a whole, is stable.
(ii) All bonds in alkanes are sigma bonds. We know that, sigma bonds involve linear overlap of orbitals. This creates bonds which are very strong. So it is difficult to break the bonds in alkanes.
(iii) Alkanes are non-polar. So they do not have any additional +ve or -ve charges. This makes it difficult to attack alkane molecules.

• However, alkanes can undergo reactions like substitution, combustion, controlled oxidation etc., We will now see those reactions in detail
I. Substitution reactions
This can be written in 9 steps:
1. In this reaction, one H atom is first removed from the alkane.
• Then a halogen atom takes the place of that H atom.
• Instead of halogen atom, nitro group or sulphonic acid group can also take the place of the H atom.
    ♦ If it is a halogen, the reaction is called halogenation.
    ♦ If it is a nitro group, the reaction is called nitration.
    ♦ If it is a sulphonic acid group, the reaction is called sulphonation.
2. Some times more than one H atoms can be substituted in this way.
3. Halogenation requires higher temperatures (573-773 K)
• However, instead of higher temperatures, we can use diffused sunlight or ultraviolet light also.
• Diffused sunlight is obtained when direct sunlight is scattered by clouds or some artificial means.
• Ultraviolet light can be produced by passing electricity through a suitable gas like mercury vapour.
4. Lower alkanes do not undergo nitration and sulphonation. They can undergo halogenation only.
5. Let us see some examples of halogenation:
(i) $\rm{CH_4~+~Cl_2~ \color {green}{\xrightarrow[{}]{h \nu}} ~ CH_3 Cl~+~HCl}$
• CH₃Cl is Chloromethane.
[$h \nu$ represent the energy provided by radiation (in our present case, radiation is sunlight or ultraviolet light). Here h is the plank's constant and 𝜈 is the frequency. We saw those details in chapter 2]
(ii) $\rm{CH_3 Cl~+~Cl_2~ \color {green}{\xrightarrow[{}]{h \nu}} ~ CH_2 Cl_2~+~HCl}$
• CH₂Cl₂ is Dichloromethane.
(iii) $\rm{CH_2 Cl_2~+~Cl_2~ \color {green}{\xrightarrow[{}]{h \nu}} ~ CH Cl_3~+~HCl}$
• CHCl₃ is Trichloromethane.
(iv) $\rm{CH Cl_3~+~Cl_2~ \color {green}{\xrightarrow[{}]{h \nu}} ~ C Cl_4~+~HCl}$
• CCl₄ is Tetrachloromethane.
(v) $\rm{CH_3 - CH_3~+~Cl_2~ \color {green}{\xrightarrow[{}]{h \nu}} ~ CH_3 - CH_2 Cl~+~HCl}$
• CH₃-CH₂Cl is Chloroethane.
6. The rate of reaction of alkanes with the various halogens varies.
   ♦ Rate of reaction with F is the highest.
   ♦ Rate of reaction with I is the lowest.
• The order is: F₂ > Cl₂ > Br₂ > I₂
(Recall that, rate of reaction is the quantity of products formed in unit time)
7. Reaction with F is violent. We will need special equipment to control the reaction.
8. Reaction with I is a very slow reaction. It is a reversible reaction. The equation is:
CH₄ + I₂ ⇌ CH₃I + HI
• We have discussed about reversible reactions in an earlier chapter [see fig.7.3 in section 7.3]
• We can convert more quantities of CH₄ into CH₃I if we remove the HI.
• For that, we use oxidising agents like HIO₃ or HNO₃. The HI will react with the oxidising agent and form I₂. The equation is:
HIO₃ + 5HI ⟶ 3I₂ + 2H₂O
9. We have seen 1^o carbon atom, 2^o carbon atom etc., in a previous section of this chapter [see step 11 below fig.13.8 in section 13.1]
• Based on that, we can now write an important information. It can be written in 2 steps:
(i) During halogenation of alkanes, the H atoms attached to the 3^o carbon atoms are removed more readily than those attached to the 2^o carbon atoms.
(ii) Similarly, the H atoms attached to the 2^o carbon atoms are removed more readily than those attached to the 1^o carbon atoms.

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In the next section we will see the mechanism of halogenation.

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Chapter 12.13 - The Inductive Effect

In the previous section, we saw the details about nucleophiles and electrophiles. In this section, we will see electron movement in organic reactions.

• The movement of electrons in organic chemical reactions can be shown by curved-arrow notation. This can be explained using some examples.
Example 1:
This can be written in 3 steps:
1. Fig.12.86 below shows the reaction between OH^- ion and CH₃Br to produce methanol and Br^- ion.
• We see that, all atoms on reactant side and product side have octet.

Fig.12.86

2. Consider the green curved arrow.
    ♦ It starts from the two yellow dots on the top of the O atom of the OH^- ion.
    ♦ It ends at the C of CH₃Br.
• That means, C gets two electrons of O.
• The C atom uses those two electrons to form the bond with O. This can be seen in the methanol molecule on the product side.
• So the green curved arrow helps us to track the movement of electrons.
3. Consider the magenta curved arrow.
    ♦ It starts from the bond between C and Br.
    ♦ It ends at the Br.
• That means, Br gets the two electrons in that bond.
• The Br then leaves with the two newly acquired electrons. It has octet. But the octet is attained using the one extra electron. So it becomes Br^-.
• So the magenta curved arrow also helps us to track the movement of electrons.

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Example 2:
• Each curved arrow indicates the movement of two electrons.
• If only one electron is moved, we use half headed curved arrows. Such arrow heads have the shape of fish hooks. In this example, we will see such a case. It can be written in steps:
1. Fig.12.87 below shows the dissociation of CH₃Cl.
• We see that, all atoms on reactant side have octet.
• But the species on the product side do not have octet. They have unpaired electrons.

Fig.12.87

2. Consider the green half headed curved arrow.
    ♦ It starts from the bond between C and Cl.
    ♦ It ends at the C of CH₃Cl.
• That means, C gets one electron from the bond.
    ♦ The resulting species is: $\mathbf{\rm{H_3\overset{.}{C}}}$
    ♦ The dot indicates unpaired electron.
• So the green half headed curved arrow helps us to track the movement of one electron.
3. Consider the magenta half headed curved arrow.
    ♦ It starts from the bond between C and Cl.
    ♦ It ends at the Cl of CH₃Cl.
• That means, Cl gets one electron from the bond.
    ♦ The resulting species is: $\mathbf{\rm{\overset{.}{Cl}}}$
    ♦ The dot indicates unpaired electron.
• So the magenta half headed curved arrow also helps us to track the movement of one electron.

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• We will see more applications of curved arrows in later sections.
• At present we will see electron displacement effects in covalent bonds. It can be written in 3 steps:
1. Consider a sample of a pure substance like CH₃Br. It is written as 'pure sample' because, the sample contains only CH₃Br molecules.
• Consider any one molecule in that sample. There will be displacement of electrons in the C-Br bond.
• This is because the Br, being more electronegative, pulls the electrons in the bond.
• Thus we see that, displacement of electrons can occur even in pure samples where there is no external influence.
2. Now consider the reaction that we saw in fig.12.86 above. We see that the displacement of electrons occur as shown by the green and magenta curved arrows.
• Here, the sample is not pure. It contains OH^- ions in addition to the CH₃Br molecules.
• We can write: The displacement of electrons occurred due to the action of OH^- ions, which is the attacking reagent.
3. So we see two cases:
(i) Displacement of electrons can occur in pure samples even when there is no external influence.
(ii) Displacement of electrons can also occur due to the action of an attacking reagent.
• Our next aim is to study the various types of electron displacements in detail.

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Inductive effect

This can be explained in 9 steps:
1. Consider a covalent bond between two different atoms.
• If one of those two atoms is more electronegative than the other, there will be a shift of electron density towards the more electronegative atom.
• Such a bond is called a polar covalent bond. We have seen the details in an earlier section 4.10.
2. Fig.12.88 below shows the structure of chloro ethane.
• The C-Cl bond in this molecule is a polar covalent bond.
    ♦ The Cl atom gains some -ve charge (δ^-).
    ♦ The C₁ atom gains some +ve charge (δ⁺).

Fig.12.88

3. The shift of electron density is indicated by arrows. The arrow points in the direction in which the shift occurs.
    ♦ This is shown in fig.12.88(b) above.
4. We have seen how C₁ acquires the +ve charge.
• Now, due to this +ve charge, C₁ is able to pull the electrons in the C₁C₂ bond.
• So C₂ loses some electron density and thus acquires a small +ve charge.
• The +ve charge acquired by C₂ is smaller when compared to the δ⁺ of C₁.
    ♦ So the +ve charge of C₂ is denoted as δδ⁺.
◼ So we can write:
The polar covalent bond C₁Cl, induces polarity in the adjacent bonds.
5. We know that all the bonds in CH₃CH₂Cl are σ bonds. So we can write a definition for inductive effect. It can be written in 3 steps:
(i) Consider a covalent bond (σ bond) which does not have any electronegative atoms on either of it’s ends.
(ii) Even though it does not have any electronegative atom, it get polarized. This is due to the polarization in an adjacent σ bond.
(iii) This effect is known as inductive effect.
6. Inductive effect is passed on to subsequent bonds also.
• But the effect decreases rapidly with the increase in the number of intervening bonds.
• For example, in fig.12.88, suppose that, it is chloro pentane instead of chloro ethane. Then the chain will be:C₅ㅡC₄ㅡC₃ㅡC₂ㅡC₁ㅡCl
7. The C₁ㅡCl bond is the original polar covalent bond.
• The other bonds C₁ㅡC₂, C₂ㅡC₃, C₃ㅡC₄, C₄ㅡC₅ will also undergo polarization. But that polarization will be due to inductive effect.
    ♦ C₁ㅡC₂ will experience the highest inductive effect.
        ✰ Because, it is nearest to the polar bond.
        ✰ It has zero number of intervening bonds.
    ♦ C₂ㅡC₃ will experience a lesser inductive effect.
        ✰ It has one intervening bond.
    ♦ C₃ㅡC₄ will experience a still lesser inductive effect.
        ✰ It has two intervening bonds.
    ♦ C₄ㅡC₅ will experience a still lesser inductive effect.
        ✰ It has three intervening bonds.
• If the number of intervening bonds is greater than three, the inductive effect is considered to be negligible.
8. In the above steps, we saw that inductive effect is caused due to the electron pulling by Cl atom. Atoms similar to Cl, which have high electronegativity, will cause the such inductive effect.
• The reverse can also happen. This can be explained in 4 steps:
(i) Suppose that, in fig.12.88 above, instead of CH₃CH₂Cl, we have CH₃CH₂Z. Where Z is a highly electropositive atom.
(ii) Then the Z will push the electron density towards C₁. The C₁ will acquire a small -ve charge (δ^-)
(iii) Due to this newly acquired -ve charge, the C₁ will push the electrons in the C₁ㅡC₂ bond. Thus the C₂ will also acquire a small -ve charge (δδ^-)
This is shown in fig.12.88(c) above.
(iv) The C₁ㅡZ is the actual polar covalent bond. The polarization in C₁ㅡC₂ is due to inductive effect.
9. We see that:
• Electronegative atoms like Cl can cause inductive effect due to their ability to withdraw electrons.
• Electropositive atoms can also cause inductive effect due to their ability to donate electrons.
◼ Based on this, the substituents are classified into two groups:
(i) Electron-withdrawing group
(ii) Electron-donating group
• Halogens, groups like nitro (ㅡNO₂), cyano (ㅡCN), carboxy (ㅡCOOH), ester (ㅡCOOR) etc., fall in the electron-withdrawing group.
• Groups like methyl (ㅡCH₃), ethyl (ㅡCH₂CH₃) fall in the electron-donating group.

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• Recall what we saw in section 12.11:
The alkyl groups attached to the 'C atom with unpaired electron' helps to stabilize the radical. (step 9, below fig.12.76 in section 12.11)
• Now we are in a position to give an explanation for this 'stabilizing action'. It can be written in 4 steps:

1. Just now, we saw that:
Groups like methyl (ㅡCH₃), ethyl (ㅡCH₂CH₃) fall in the electron-donating group. 2. The ethyl free radical has one methyl group attached to the 'C atom with unpaired electron'.
• This methyl group donates electron density to the 'C atom with unpaired electron'.
• So that C atom will not feel that much necessity to acquire a new electron.
• As a result, the radical as a whole, will get some stability.
3. The isopropyl free radical has two methyl groups attached to the 'C atom with unpaired electron'.
• Those two methyl groups donate electron densities to the 'C atom with unpaired electron'.
• So that C atom will not feel that much necessity to acquire a new electron.
• As a result, the radical as a whole, will get some stability.
• Since two methyl groups donate electron densities, the stability will be greater than that in ethyl free radical.
4. The tertiary butyl free radical has three methyl groups attached to the 'C atom with unpaired electron'.
• Those three methyl groups donate electron densities to the 'C atom with unpaired electron'.
• So that C atom will not feel that much necessity to acquire a new electron.
• As a result, the radical as a whole, will get some stability.
• Since three methyl groups donate electron densities, the stability will be greater than that in isopropyl free radical.

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Now we will see some solved examples

Solved example 12.16
Which bond is more polar in the following pairs of molecules:
(a) H₃CㅡH, H₃CㅡBr
(b) H₃CㅡNH₂, H₃CㅡOH
(b) H₃CㅡOH,  H₃CㅡSH
Solution:
Part (a):
• We have to compare two bonds: CㅡH and CㅡBr
• CㅡBr will be more polar because, Br is more electronegative than H
Part (b):
• We have to compare two bonds: CㅡN and CㅡO
• CㅡO will be more polar because, O is more electronegative than N
Part (c):
• We have to compare two bonds: CㅡO and CㅡS
• CㅡO will be more polar because, O is more electronegative than S

Solved example 12.17
In which CㅡC bond of CH₃CH₂CH₂Br, the inductive effect is expected to be the least?
Solution:
• We can expand the given formula as: CH₃ㅡCH₂ㅡCH₂ㅡBr
• C atoms are to be numbered (starting from the substituent Br) from right to left.
• C₁ㅡBr is the original polar covalent bond.
• C₁ㅡC₂ will experience inductive effect.
   ♦ This bond has zero number of intervening bonds.
   ♦ So this bond will experience the maximum inductive effect .
• C₂ㅡC₃ will experience inductive effect.
   ♦ This bond has one intervening bond.
   ♦ So this bond will experience the least inductive effect.

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In the next section, we will see resonance in organic molecules.

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Chapter 4.10 - Polarity of Bonds

In the previous section, we saw the basics about resonance structures. In this section, we will see polarity of bonds

1. We have seen two types of bonds:
    ♦ Ionic bonds
          ✰ This involves transfer of electrons
    ♦ Covalent bonds
          ✰ This involves sharing of electrons
2. Consider ionic bonds
• In some ionic bonds, the cation is able to pull the ‘extra electrons of the anion’
    ♦ Then, the ‘extra electrons of the anion’ will no longer belong to the anion alone
    ♦ They will belong to both the anion and the cation
• Thus a little 'covalent character' is induced in that ionic bond
■ An ionic bond with no covalent characteristics is called an ‘ideal ionic bond’
3. Consider covalent bonds
• In some covalent bonds, the ‘more electronegative atom’ is able to pull the ‘shared pairs of electrons’
    ♦ Then, that ‘more electronegative atom’ acquires a 'partial negative charge'
    ♦ Also, the 'less electronegative atom' acquires a 'partial positive charge'
• Thus a little 'ionic character' is induced in that covalent bond
■ A covalent bond with no ionic characteristics is called an ‘ideal covalent bond’
4. In reality,
• No ionic bond is an ideal ionic bond
    ♦ All ionic bonds will have some covalent character
• No covalent bond is an ideal covalent bond
    ♦ All covalent bonds will have some ionic character

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Polarity in covalent bonds

1. Consider the homonuclear diatomic molecules like H₂, O₂, Cl₂, N₂, F₂ etc.,
• In those molecules, the shared pairs of electrons are ‘attracted equally’ by the two atoms
    ♦ So the shared pairs will be exactly midway between the two nuclei
■ In such molecules, the covalent bonds are called non-polar covalent bonds
2. Consider heteronuclear diatomic molecules like HCl, HF etc.,
• Let us compare H with Cl and F
    ♦ Cl and F are more electronegative
• So in those molecules, the shared pair gets displaced more towards the more electronegative atom
■ In such molecules, the covalent bonds are called polar covalent bonds

We will see a solved example
Solved example 4.6
Arrange the bonds in order of increasing ionic character in the molecules: LiF, K₂O, N₂, SO₂ and ClF₃

Solution:
1. Consider LiF
(i) Electronegativity values:
Li - 1.0, F - 4.0
(ii) Difference in electronegativity values = (4.0 - 1.0) = 3.0
2. Consider K₂O
(i) Electronegativity values:
K - 0.8, O - 3.5
(ii) Difference in electronegativity values = (3.5 - 0.8) = 2.7
3. Consider N₂
(i) Electronegativity values:
N - 3.0
(ii) Difference in electronegativity values = (3.0 - 3.0) = 0.0
4. Consider SO₂
(i) Electronegativity values:
S - 2.5, O - 3.5
(ii) Difference in electronegativity values = (3.5 - 2.5) = 1.0
5. Consider ClF₃
(i) Electronegativity values:
Cl - 3.0, F - 4.0
(ii) Difference in electronegativity values = (4.0 - 3.0) = 1.0
6. Arranging the differences in increasing order, we get:
0.0 < [1.0 = 1.0] < 2.7 < 3.0
7. Greater the difference, greater is the ionic character. So we get:
N₂ < [SO₂ = ClF₃] < K₂O < LiF
8. Consider [SO₂ = ClF₃]
• The F is more electronegative than O
    ♦ Also, in SO₂, two O atoms are pulling onto S
    ♦ But in ClF₃, three F atoms are pulling onto Cl
• So ClF₃ is more ionic
• Thus the correct order is:
N₂ < SO₂ < ClF₃ < K₂O < LiF

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Dipole moment

1. Consider a polar covalent bond

• The more electronegative atom acquires a partial negative charge indicated by Q^- 
• The less electronegative atom acquires a partial positive charge indicated by Q⁺
• Magnitudes of both charges will be equal to 'Q'
    ♦ But their signs will be opposite
2. So in a molecule with polar covalent bond, we have:
    ♦ A positive charge Q⁺ at one end of the molecule
          ✰ This is the end where the less electronegative atom is situated
    ♦ A negative charge Q^- at the other end of the molecule
          ✰ This is the end where the more electronegative atom is situated
3. Note down the ‘distance between the centers of the two charges’
• Let this distance be 'r'
■ Then the product (Q × r) is called the dipole moment possessed by the molecule
4. Dipole moment is denoted by the Greek letter '𝝁'
• So mathematically, we can write: 𝝁 = Q r
    ♦ The charge Q is measured in coulombs (C)
    ♦ The distance r is measured in meters (m)
    ♦ So the unit of 𝝁 will be coulomb meter (C m)
5. But When expressed in C m, the dipole moment will be very small
(With a lot of zeros after the decimal point)
• So we usually use the unit Debye 
    ♦ It’s symbol is D
• 1 D is equal to 3.33564 × 10^-30 C m
6. Dipole moment is a vector quantity
• So it has both magnitude and direction
7. Whenever two opposite charges are placed at a distance apart, we can calculate the dipole moment
• In physics classes, we come across this situation on many occasions
• There, we denote the dipole moment using a small arrow
    ♦ The tail of the arrow is at the center of the negative charge
    ♦ The head points towards the center of the positive charge
8. But in chemistry, we follow a different notation
• Instead of the arrow, we use a crossed arrow
    ♦ This crossed arrow is put on the Lewis structure of the molecule
          ✰ The cross is at the positive end
          ✰ The arrow head points towards the negative end
9. Using this notation, the dipole moment of HF is shown in fig.4.65 below:

Fig.4.65

• The crossed arrow is shown in yellow color
• The crossed arrow symbolizes 'a direction'
    ♦ It is the direction in which the electrons shift
• In our present case, the electron shifts from H to the 'more electronegative F'




10. HF is a diatomic molecule. Next we will see some polyatomic molecules
• Consider the water molecule
• Though we sometimes represent the water molecule as H-O-H, the two O-H bonds do not lie on the same line
    ♦ If the two bonds lie on the same line, the angle between them would be 180^o
    ♦ This is shown in fig.4.66(b) below:

Fig.6.66

• But the actual angle is 104.5^o 
    ♦ This is shown in fig.4.66(c) above
11. We will have two dipole moments
    ♦ The dipole moment between the 'left H' and O
    ♦ The dipole moment between the 'right H' and O
• They are shown in yellow color in fig.4.66(d) above
• The dipole moments are due to the 'pulling of electrons' by the O atom
12. But we want the dipole moment of the ‘molecule as a whole’
■ For that, we calculate the vector sum
• We have seen that, dipole moments are vector quantities
    ♦ So they can be added using 'principles of vector addition'
          ✰ Details can be seen here
13. Let us add the two dipole moment vectors:
(i) Fig.4.67(a) below shows the left side dipole moment

Fig.6.67

• It is resolved into rectangular components
    ♦ The horizontal component is shown in red color
    ♦ The vertical component is shown in green color
(ii) For resolving the vector into it's rectangular components, we need the value of 𝞱 shown in fig.4.67(a) 
• It can be easily calculated as shown in fig.4.67(b)
• All we need to do is: Draw a right triangle indicated by the magenta dashed lines
(iii) Once we know 𝞱, the rectangular components can be calculated as follows:
    ♦ The horizontal component (red) in fig.6.67(a) is calculated using the cosine of 𝞱
    ♦ The vertical component (green) in fig.6.67(a) is calculated using the sine of 𝞱
14. Next we repeat the same procedure for the right dipole moment
• That is.,
    ♦ We find the horizontal component (red) in fig.6.67(c)
    ♦ We find the vertical component (green) in fig.6.67(c)
15. Now we can do the vector addition
• We add the similar components:
    ♦ We add the reds in fig.a and fig.b
    ♦ We add the greens in fig.a and fig.b
16. The two dipole moments in the earlier fig.6.66(d) are equal in magnitude and have the same 𝞱
• So the two reds will have the same magnitudes
• The two greens will also have the same magnitude
17. While adding, we see that:
• The two reds have opposite directions
    ♦ So they will cancel each other
■ So the final resultant will not have a horizontal component
• The two greens have the same direction
    ♦ So they will add up
18. Since the reds cancel each other, we need to consider the greens only
• The resultant obtained from the two greens is shown in fig.6.67(d)
    ♦ It is shown as a thick green crossed arrow
    ♦ It is the resultant dipole moment  
19. Scientists have determined this resultant
• It's magnitude is: 1.85 D

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Next, we will see the dipole moment in another polyatomic molecule: BeF₂

1. Here, the two Be-F bonds lie on the same line
• The angle between the two bonds is 180^o
• This is shown in fig.6.68(b) below:

Fig.4.68

2. We will have two dipole moments
    ♦ The dipole moment between the 'left F' and Be
    ♦ The dipole moment between the 'right F' and Be
• They are shown in yellow color in fig.4.68(c) above
• The dipole moments are due to the 'pulling of electrons' by the F atoms
3. But we want the dipole moment of the ‘molecule as a whole’
■ For that, we calculate the vector sum
• We have seen that, dipole moments are vector quantities
    ♦ So they can be added using 'principles of vector addition'
          ✰ Details can be seen here
4. Let us add the two dipole moment vectors:
(i) We see that, both the vectors are horizontal
    ♦ So there is no need to resolve them into rectangular components
(ii) The two vectors are equal in magnitude. But they have opposite directions
    ♦ So they will cancel each other
(iii) So the resultant will be a null vector 
• It is shown in fig.d
• A null vector has zero magnitude and hence no direction
5. We can write:
The dipole moment of BeF₂ is zero

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Next, we will see the dipole moment in another polyatomic molecule: BF₃

1. In this case, there are three bonds
    ♦ All of them are B-F bonds
    ♦ Those bonds do not lie on the same line
2. The orientations  of the bonds can be described in two steps:
(i) From among the three bonds, take any two
(ii) The angle between those two bonds will be 120^o
3. This is a simple orientation
• The '120^o' indicates that, the bonds are distributed uniformly around the central B atom
• This is shown in fig.4.69(b)
[Remember that (3 × 120) = 360. The angle of a full circel is 360^o]

Fig.4.69

 4. We will have three dipole moments
    ♦ The dipole moment between the 'left F' and B
    ♦ The dipole moment between the 'top F' and B
    ♦ The dipole moment between the 'bottom F' and B
• They are shown in yellow color in fig.4.69(c) above
• The dipole moments are due to the 'pulling of electrons' by the F atoms
5. But we want the dipole moment of the ‘molecule as a whole’
■ For that, we calculate the vector sum
• We have seen that, dipole moments are vector quantities
    ♦ So they can be added using 'principles of vector addition'
          ✰ Details can be seen here
6. Let us add the three dipole moment vectors:
• This is a simple case of vector addition. We need not show the detailed steps. It is a 'mental math' problem
• The results of addition can be summarised in 7 steps:
(i) The leftside vector is horizontal. So there is no need to resolve it into rectangular components
(ii) The top vector is inclined. So it can be resolved
    ♦ Let us call the horizontal component as 'red'
    ♦ Let us call the vertical component as 'green'
(iii) The bottom vector is inclined. So it can be resolved
    ♦ Let us call the horizontal component as 'red'
    ♦ Let us call the vertical component as 'green'
(iv) The greens in (ii) and (ii) are equal and opposite
    ♦ So they cancel each other
■ The final resultant will not have any vertical component
(v) The reds in (ii) and (iii) are equal. Also, they act in the same direction
• So they add up
(vi) Consider the two vectors:
    ♦ The resultant vector obtained in (v)
    ♦ The vector mentioned in (i)
• These two vectors are equal and opposite
• So the resultant of these two vectors is a null vector
(viii) Thus the final resultant of all the three vectors is a null vector
• It is shown in fig.4.69(d)
• A null vector has zero magnitude and hence no direction
7. We can write:
• The dipole moment of BF₃ is zero
8. This type of vector addition problems are encountered frequently in physics classes
• If the reader has any doubt, he/she may draw the necessary diagrams like those we saw in fig.6.68 earlier
• It is important to become convinced that, all steps written in (6) are valid
• OR, the reader may use any alternate steps to prove that the resultant is null vector

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In the next section, we will see the dipole moment of ammonia molecule

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