Chapter 9.1 - Properties of Dihydrogen

In the previous section, we saw some basic details about dihydrogen. In this section, we will see properties of dihydrogen.

Physical properties of dihydrogen

• Following are the main physical properties of dihydrogen:
    ♦ Dihydrogen is colorless, odorless and tasteless.
    ♦ It is a combustible gas.
    ♦ It is lighter than air.
    ♦ It is insoluble in water.
• Other physical properties like melting point, boiling point etc., can be obtained from standard tables.

Chemical properties of dihydrogen

This can be explained in 9 steps:
1. Bond dissociation enthalpy plays a major role in determining the chemical properties of any molecule.
• We have seen the details about bond dissociation enthalpy in a previous section 4.8.
2. If the bond dissociation enthalpy of a molecule is high, it will not be easy to break that molecule into individual atoms. If it does not break into individual atoms, reaction with other atoms or molecules is not possible.
3. If the bond dissociation enthalpy of a molecule is low, it will be easy to break it into individual atoms. If it breaks into individual atoms, reaction with other atoms or molecules is possible.
4. In our present case, the bond dissociation enthalpy of H₂ molecule is very high.
• That means, it is very difficult to break the H-H single bond.
• It is the strongest single bond between any two atoms of the same element.
• This can be explained as follows:
    ♦ Consider various single bonds like H-H, Cl-Cl, F-F, etc.,
    ♦ They are all single bonds between the same two atoms.
    ♦ The H-H bond will have the highest bond dissociation enthalpy among such molecules.
5. Consider a sample of H₂ kept at a high temperature of 2000 K.
• Only 0.081% of that sample will dissociate into individual H atoms
6. If we want more H₂ molecules of that sample to dissociate, we will have to increase the temperature to a very high value.
• If the temperature is 5000 K, 95.5% of that sample will dissociate into individual H atoms.
7. If the sample is kept at room temperature, those molecules will be inert. That is., they will not take part in reactions.
8. Due to the high bond dissociation enthalpy, we will need electric arc or ultraviolet radiations to accomplish dissociation of H₂ molecules.
9. Once an individual H atom is formed, it will be having a single electron in the outer most shell.
◼ Such an atom will attain stability by any one of the following three methods:
(i) The H atom donates it’s electron to an atom of another element.
• When the donation is done, it becomes H⁺ ion.
• The H⁺ ion then combines with the atom of the other element to become a new compound.
(ii) The H atom accepts an electron from an atom of another element.
• When the acceptance is done, it becomes H^- ion.
• The H^- ion then combines with the atom of the other element to become a new compound.
(iii) The H atom shares it’s electron with another H atom or an atom of another element. This leads to a covalent bond.

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Let us see some reactions involving hydrogen:

Reaction of dihydrogen with halogens

This can be explained in 4 steps:
1. Let us denote halogen molecules in general as X₂, where X is a halogen atom. (Recall that the important halogens are: F, Cl, Br and I)
• Then the reaction can be written as:
H₂ (g) + X₂ (g) → 2HX (g)
2. We already know that, there will be a single covalent bond between H and X. For example, we have seen the formation of HCl in earlier sections.
3. HX is the general form of hydrogen halides.
• For example:
    ♦ HF is a hydrogen halide
    ♦ HCl is a hydrogen halide.
4. Fluorine is so reactive that, it’s reaction with dihydrogen takes place even in the dark.
• But the reaction of I with dihydrogen requires a catalyst.

Reaction of dihydrogen with dioxygen

This can be written in 3 steps:
1. The reaction between dihydrogen and dioxygen is highly exothermic.
2H₂ (g) + O₂ (g) → 2H₂O (l) ΔH^⊝ = -285.9 kJ mol-1`
2. But we will need to supply energy in the form of heat to get the reaction started.
• That is., we need to supply energy to break the bonds in H₂ and O₂ molecules.
3. But once the reaction starts, energy is released. This energy is sufficient to propagate the reaction.
• But the energy released is so huge that, it creates an explosion. That is the reason why we are not able to make water by burning hydrogen with oxygen.
• We can burn only very small quantities of dihydrogen with oxygen in labs. Those labs should have highly advanced safety equipment.

Reaction of dihydrogen with dinitrogen

This can be written in 3 steps:
1. The reaction between dihydrogen and dinitrogen is exothermic.
3H₂ (g) + N₂ (g) → 2NH₃ (g) ΔH^⊝ = -92.6 kJ mol-1`
2. But we will need to supply energy in the form of heat and pressure to get the reaction started.
• That is., we need to supply energy to break the bonds in H₂ and N₂ molecules.
3. This reaction is used in Haber process to manufacture NH₃ (ammonia).

Reaction of dihydrogen with metals

This can be written in 6 steps:
1. dihydrogen reacts with many metals to form the corresponding hydrides.
2. Let us denote the alkali metal atom by the letter 'M'
• Then the reaction can be written as:
H₂ (g) + 2M (s) → 2MH (s)
3. Metals are electron donors. The alkali metals are strong electron donors.
• They donate the outermost electron and become M⁺.
4. The H atom is forced to accept this electron. Thus it becomes H^-.
(The H atom is forced to accept electrons from metals which are above it in the reactivity series. Some images of the series can be seen here)
5. The M⁺ and H^- combines together to form the ionic compound M⁺H^-.
6. The formation of NaH (sodium hydride) is an example.
    ♦ Here, Na donates it's outermost electron to become Na⁺
    ♦ H accepts this electron to become H^-.
    ♦ Thus the ionic compound Na⁺H^- is formed.

Reaction of dihydrogen with metal ions

This can be written in 3 steps:
1. dihydrogen reduces some metal ions in aqueous solution into the corresponding metals.
2. Metals are electron donors. But those metals which are below hydrogen in the reactivity series, will be forced to accept electron from hydrogen
3. Let us see an example:
H₂ (g) + Pd²⁺(aq) → Pd (s) + 2H⁺ (aq)
Pd (Palladium) is below hydrogen in the reactivity series.

Reaction of dihydrogen with metal oxides

This can be written in 4 steps:
1. dihydrogen reduces some metal oxides into the corresponding metals.
2. Metals are electron donors. The electrons donated by them are accepted by oxygen to form metal oxides.
3. But in the presence of hydrogen, those metals in the oxides, are forced to accept electrons from hydrogen.
• When those metals accept electrons, they are reduced to pure metals.
3. The oxygen is thus released from the oxide.
• Also, hydrogen becomes H⁺ due to the electron donation
• These two combine together to form water.
3. If we denote the metals by the letter 'M', the general equation will be as follows:
yH₂  (g) + M_xO_y (s) → xM (s) + yH₂O (l)
This equation can be explained in 3 steps:
(i) There are y oxygen atoms on the left side. Each O atom requires 2 electrons to complete octet.
(ii) So there should be a total of 2y electrons available.
(iii) These 2y electrons can be supplied by the y number of H₂ molecules because, each H can supply one electron.
4. Let us see an example:
• Dihydrogen reacts with copper(II) oxide to give pure copper and water. The equation is:
H₂ (g) + CuO → Cu (s) + H₂O (l)
• In Copper(II) oxide, Cu is in the oxidation state of +2.
    ♦ The two electrons required are supplied by the two H atoms.
    ♦ The Cu²⁺ gets reduced to Cu.
(Cu²⁺ is forced to accept electrons from H. Note that, H is above Cu in the reactivity series)
• The H⁺ thus formed will share electrons from the O atom and become H₂O molecule
    ♦ One O atom is able to supply two electrons
    ♦ One electron is shared by each of the two H⁺ ions.

Reaction of dihydrogen with organic compounds

• Reaction of dihydrogen with some organic compounds give many commercially important products. We will see two examples:
1. Dihydrogen reacts with vegetable oils to give edible fats. (margarine and vanaspati ghee)
2. Dihydrogen reacts with olefins to give aldehydes.  Aldehydes undergo reduction to give alcohols.
We will see more details in organic chemistry classes.

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Uses of dihydrogen

1. Dihydrogen is used in the manufacture of ammonia. This ammonia is essential for the manufacture of nitric acid and nitrogenous fertilizers. Nitrogenous fertilizers are the only means by which we can supply the required nitrogen for food crops.
2. Dihydrogen is essential for the manufacture of edible fats from vegetable oils.
3. Dihydrogen is essential for the manufacture of methanol.
• Methanol is required for the manufacture of paints, plastics, construction materials etc.,
4. Dihydrogen is essential for the manufacture of metal hydrides.
5. Dihydrogen is essential for the manufacture of halides. Hydrogen chloride is an example.
6. We saw that dihydrogen can reduce metal oxides. So it is used in many metallurgical processes where reduction of metal oxides is involved.
7. Dihydrogen is used for cutting and welding of metals.
• First, an electric arc is used to dissociate dihydrogen into individual H atoms.
• These H atoms recombine to form dihydrogen. Huge energy is released when they recombine.
• This energy helps to raise the temperature of the metal surface up to about 4000 K. Thus welding of different metal parts can be achieved.
8. Dihydrogen is used as a fuel in rocket engines.
9. Dihydrogen has greater energy density than other fuels. For example, one kg of dihydrogen releases more energy than one kg of petrol.
10. Dihydrogen is used in fuel cells to produce electricity.
• Fuel cells are electrochemical cells in which the chemical energy of dihydrogen and dioxygen are converted into electrical energy. We will see more details about this cell in higher classes.

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In the next section, we will see hydrides.


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Chapter 6.9 - Lattice Enthalpy

In the previous section, we saw Hess's law and bond enthalpy. In this section, we will see lattice enthalpy

Lattice enthalpy can be explained in 7 steps:
1. Consider the ionic compound NaCl
   ♦ We know that, NaCl has a crystal structure
   ♦ The crystal is made up of Na⁺ and Cl^- ions
   ♦ Those ions are arranged in a symmetric 3D form
   ♦ This 3D form is called crystal lattice
   ♦ (some images can be seen here)
2. The crystal lattice of NaCl is very stable because of the strong electrostatic attractions between the Na⁺ and Cl^- ions
• We want to know how much energy will be required for the following process:
   ♦ Separating all the Na⁺ and Cl^- ions in one mol NaCl
◼  The resulting individual ions should be scattered far away from each other so that, there will be no attractive or repulsive forces between the resulting ions. In other words, the resulting ions must be in the gaseous state
3. It is obvious that:
• When the lattice is turned into gaseous ions, we will need to supply energy. In other words, it will be an endothermic process
• For the reverse process, that is., when the gaseous ions come together and form the lattice, we will receive energy. In other words, it will be an exothermic process  
◼  This energy that we supply/receive is called lattice enthalpy
4. Lattice enthalpy can be defined in two ways:
◼  Lattice dissociation enthalpy
The lattice dissociation enthalpy is the enthalpy change needed to convert 1 mole of solid crystal lattice into its scattered gaseous ions
   ♦ We will denote it by the symbol: ΔH^⊖_lattice(d)
◼  Lattice formation enthalpy
The lattice formation enthalpy is the enthalpy change when 1 mole of solid crystal lattice is formed from its separated gaseous ions
   ♦ We will denote it by the symbol: ΔH^⊖_lattice(f)
5. It is obvious that:
   ♦ Both ΔH^⊖_lattice(d) and ΔH^⊖_lattice(f) will have the same magnitude
   ♦ The sign of ΔH^⊖_lattice(d) will be positive
   ♦ The sign of ΔH^⊖_lattice(f) will be negative
6. To find ΔH^⊖_lattice(d) of NaCl means, to find X in the following thermochemical equation:
   ♦ NaCl(s) → Na⁺(g) + Cl^-(g); ΔH^⊖_lattice(d) = X
7. To find ΔH^⊖_lattice(f) of NaCl means, to find X in the following thermochemical equation:
   ♦ Na⁺(g) + Cl^-(g) → NaCl(s); ΔH^⊖_lattice(f) = X

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• So we will see how ΔH^⊖_lattice(d) or ΔH^⊖_lattice(f) is determined
• It is not possible to carry out the experiment in (6) above. Neither is it possible to carry out the experiment in (7)
◼  That means, we cannot measure lattice enthalpy from any experiments
• So we use an indirect method. It can be written in 22 steps:
1. The cyan horizontal lines in fig.6.13 below indicates various states in an experiment
• The arrows indicate various processes which transform the system from one state to another

Fig.6.13

2. Consider the thick cyan horizontal line. It is the datum line. It is the state from which we begin our calculations
• At this state, the system consists of:
   ♦ One mol Na atoms
         ✰ These atoms are in the solid state
   ♦ Half mol Cl₂ molecules
         ✰ These molecules are in the gaseous state
◼  The above solid and gaseous states are obvious because:
   ♦ At standard temperature and pressure
         ✰ a sample of sodium consists of Na atoms in the solid state
         ✰ a sample of chlorine consists of Cl₂ molecules in the gaseous state
3. From the datum line, we begin our first process
• Our first process is to convert the sodium into gaseous state
• So it will be a sublimation process
4. From the data book, we have:
One mol Na(s) requires 108.4 kJ energy, for complete sublimation
• In other words, ΔH^⊖_sub for Na(s) = 108.4 kJ mol^-1
• So the thermochemical equation for this process will be:
Na(s) + 1/2Cl₂(g) → Na(g) + ½ Cl₂(g); ΔH^⊖_sub = 108.4 kJ mol^-1
• This process is numbered as I in the fig.12.13 above
5. So when the process I is complete, we will have:
   ♦ One mol Na atoms in the gaseous state
   ♦ Half mol Cl₂ molecules in the gaseous state

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6. Our second process is to convert each of the gaseous Na atoms into gaseous Na⁺ ions
• So it will be an ionization process
7. From the data book, we have:
One mol Na(g) requires 495.6 kJ energy for complete ionization
• In other words, ΔH^⊖_i for Na(g) = 495.6 kJ mol^-1
• So the thermochemical equation for this process will be:
Na(g) + 1/2Cl₂(g) → Na⁺(g) + ½ Cl₂(g); ΔH^⊖_i = 495.6 kJ mol^-1
• This process is numbered as II in the fig.12.13 above
8. So when the process II is complete, we will have:
   ♦ One mol Na⁺ ions in the gaseous state
   ♦ Half mol Cl₂ molecules in the gaseous state

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9. Our third process is to separate each of the gaseous Cl₂ molecules into two gaseous Cl atoms
• So it will be a bond dissociation process
10. From the data book, we have:
One mol Cl₂(g) requires 242 kJ energy for complete bond dissociation
• In other words, ΔH^⊖_Cl-Cl = 242 kJ mol^-1
   ♦ In our present case, we have half mol Cl₂ molecules
   ♦ It will give one mol Cl atoms
   ♦ It will require (242/2) = 121 kJ
• So the thermochemical equation for this process will be:
Na⁺(g) + ½ Cl₂(g)  → Na⁺(g) + Cl(g); 1/2ΔH^⊖_Cl-Cl = 121 kJ mol^-1
• This process is numbered as III in the fig.12.13 above
11. So when the process III is complete, we will have:
   ♦ One mol Na⁺ ions in the gaseous state
   ♦ One mol Cl atoms in the gaseous state

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12. Our fourth process is to convert each of the gaseous Cl atoms into gaseous Cl^- ions
• So it will be an electron accepting process
• We have seen electron gain enthalpy in an earlier chapter (details here)
13. From the data book, we have:
One mol Cl(g) releases 348.6 kJ energy when all the atoms accept one electron each
• In other words, ΔH^⊖_eg for Cl(g) = 348.6 kJ mol^-1
• So the thermochemical equation for this process will be:
Na⁺(g) + Cl(g) → Na⁺(g) + Cl^-(g); ΔH^⊖_eg = -348.6 kJ mol^-1
• This process is numbered as IV in the fig.12.13 above
14. So when the process IV is complete, we will have:
   ♦ One mol Na⁺ ions in the gaseous state
   ♦ One mol Cl^- ions in the gaseous state

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15. The completion of process IV is a milestone
• Consider the products obtained at the end of this process:
   ♦ One mol Na⁺ ions in the gaseous state
   ♦ One mol Cl^- ions in the gaseous state
◼  It is from these products that, we calculate the of ΔH^⊖_lattice(f) NaCl. This fact is clear from the 'definition of ΔH^⊖_lattice(f)' that we wrote in (4) at the beginning of this section
16. So, our fifth process is to make the system release an energy equal to ΔH^⊖_lattice(f)
• When this energy is released, we will get NaCl(s)
• The ΔH^⊖_lattice(f) value of NaCl is available in the data book
   ♦ But remember that, our aim itself is to find this ΔH^⊖_lattice(f)
    ♦ For the time being, we will ignore the value given in the data book, and find it ourselves
• The thermochemical equation of this process will be:
Na⁺(g) + Cl^-(g) → NaCl(s); ΔH^⊖_lattice(f) = X kJ mol^-1
• This process is numbered as V in the fig.12.13 above
• Our aim is to find X

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17. To find X, we split it into X₁ and X₂
• This is shown in fig.6.13
◼  It is clear that:
After completing process IV, if the system releases X₁, the datum line can be reached
• The process in which X₁ is released, is marked as V₁
◼  When this X₁ is released, we can say:
The net energy absorbed/released by the system becomes zero
• Thus we get:
108.4 + 495.6 + 121 - 348.6 - X₁ = 0
⇒ X₁ = 376.4
18. So when the process V₁ is complete, we will have:
   ♦ One mol Na(s)
   ♦ Half mol Cl₂(g)

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19. The completion of process V₁ is another milestone
• We have reached back at the datum
• From here, we have to release some more energy
   ♦ This energy is marked as X₂
• The process in which X₂ is released, is marked as V₂
20. We have to find the magnitude of X₂
   ♦ For this, the red arrow gives us a valuable clue. It can be explained in 4 steps:
(i) Consider the products obtained at the end of process V₁:
   ♦ One mol Na(s)
   ♦ Half mol Cl₂(g)
◼  It is from these products that, we calculate the of ΔH^⊖_(f) NaCl(s)
◼  This fact is clear from the definition of ΔH^⊖_(f) that we wrote in an earlier section
   ♦ It is the enthalpy of formation from constituent elements in the standard states
   ♦ It is available in the data book: -411.2 kJ mol^-1
◼  It is clear that:
From the datum line, if the system releases 411.2 kJ, the bottom most cyan line can be reached
(ii) The process in which this ΔH^⊖_(f) is released, is marked with the downward red arrow
   ♦ The down ward red arrow reaches upto the bottom most cyan line
   ♦ After completing process V₁ also, we have to reach the bottom most cyan line
(iii) So we can write:
   ♦ The process represented by the downward red arrow
   ♦ is equivalent to
   ♦ The process V₂
(iv) Thus we get: X₂ = 411.2 kJ
21. So we have calculated both X₁ and X₂
   ♦ Then X = (X₁ + X₂) = (376.4 + 411.2) = 787.6 kJ
   ♦ Thus we can write: ΔH^⊖_lattice(f) of NaCl(s)= -787.6 kJ mol^-1
22. The cyclic process shown in fig.6.13 is known as the Born-Haber cycle
◼  It is based on the Hess's law of constant heat summation
◼  Whatever be the path, the energy of the system at a particular state, will be the same
◼  This method can be used to determine those ΔH values which are impossible to find experimentally

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• Note:
The above result of -787.6 kJ mol^-1 can be obtained using another approach also. It can be explained in 4 steps:
1. From the datum line, we travel along the path: I - II - III - IV - V
• We reach the bottom most cyan line
• The net energy along this path = (108.4 + 495.6 + 121 - 348.6 - X) = (376.4 - X)
2. From the datum line, we travel along the red arrow
• We reach the bottom most cyan line
• The net energy along this path = -411.2
3. In both the paths:
   ♦ The starting states are the same: The datum line
   ♦ The ending states are the same: The bottom most cyan line
4. So, according to Hess's law, the energies must be the same
• Thus we can write: (376.4 - X) = -411.2
⇒ X = 787.6 kJ mol^-1

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• In the next section, we will see enthalpy of solution


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Chapter 6.8 - Application of Bond Enthalpy

In the previous section, we saw Hess's law and bond enthalpy. In this section, we will see some applications of bond enthalpy.

• The bond enthalpy values given in the data book can be used to find ΔH^⊖_r
• This can be explained in 11 steps using an example
1. Consider the following reaction:
CO(g) + H₂O(g) → CO₂(g) + H₂(g)
• Steam is made to react with carbon monoxide.
2. Let us see the making and breaking of bonds in this reaction:
• When the reaction proceeds:
    ♦ Bonds in the reactant side are broken.
    ♦ As a result, the reactant molecules separate into individual atoms.
    ♦ Those individual atoms rearrange and make new bonds.
    ♦ The new bonds result in product molecules.
3. We want answers to these questions:
• Which all bonds are broken in the reactant side?
    ♦ How many bonds of each type are broken?
• Which all bonds are made in the product side?
    ♦ How many bonds of each type are made?
4. It will be easier to find the answers to the above questions, if we  rewrite the equation in (1) using structural formulae. This is shown below:
C≡O + H-O-H → O=C=O + H-H
5. It is clear that:
• On the reactant side:
    ♦ One C≡O bond is broken.
    ♦ Two O-H bonds are broken.
• On the product side:
    ♦ Two C=O bonds are made.
    ♦ One H-H bond is made.
6. We know the following two facts:
(i) When bonds are broken, energy is absorbed.
    ♦ So, we need to supply energy to break the bonds in the reactant side.
(ii) When bonds are made, energy is released.
    ♦ So we will receive energy when bonds are made in the product side.
7. Let us calculate each of the above two items in our present case.
• From the data book, we have:
    ♦ C≡O has a bond enthalpy of 1072 kJ mol^-1
    ♦ O-H has a bond enthalpy of 463 kJ mol^-1
    ♦ C=O has a bond enthalpy of 799 kJ mol^-1
    ♦ H-H has a bond enthalpy of 436 kJ mol^-1
• So we get:
(i) Total energy required for breaking all the bonds in the reactant side
= (1 × 1072) + (2 × 463) = 1998
(ii) Total energy released when making all the bonds in the product side
= (2 × 799) + (1 × 436) = 2034
8. We see that:
   ♦ Energy required to break the bonds in the reactant side
   ♦ is lesser than
   ♦ Energy released when bonds are made in the product side.
• So we receive a net energy of (2034-1998) = 36 kJ mol^-1
• Since energy is released, it is an exothermic reaction. The enthalpy value should be given a negative sign.
9. Now we can write the thermochemical equation:
CO(g) + H₂O(g) → CO₂(g) + H₂(g); ΔH^⊖_r = -36.0 kJ mol^-1
10. Based on the above discussion, we can write a general equation to find ΔH^⊖_r
Eq.6.10: $\mathbf\small{\rm{\Delta {H^{\ominus }}_r=\sum{Bond\;Enthalpies_{reactants}}-\sum{Bond\;Enthalpies_{products}}  }}$
11. Consider the two terms on the right side of Eq.6.10
• If the first term is larger, it means that, the bonds in the reactant side have greater energy.
   ♦ We will have to supply a net energy to break the bonds.
   ♦ ΔH^⊖_r will have a positive sign.
   ♦ The reaction will be endothermic.
• If the second term is larger, it means that, the bonds in the product side have greater energy.
   ♦ We will receive a net energy.
   ♦ ΔH^⊖_r will have a negative sign.
   ♦ The reaction will be exothermic.
• In our present case, the second term is larger.
• We see that, the Eq.6.10 automatically gives us the appropriate sign.

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Next we will see a simple case. It is called 'simple case' because, only a very few bonds are broken and made. It can be written in steps:
1. Consider the reaction shown in fig.6.11 below:

• It is the reaction between propene and hydrogen to give propane.
2. Let us see the making and breaking of bonds in this reaction:
• When the reaction proceeds:
    ♦ Two bonds in the reactant side are broken.
          ✰ They are marked with red color in fig.6.12 below.
    ♦ Three bonds are newly formed in the product side.
          ✰ They are marked with green color in the fig.6.12

 

Fig.6.12

3. Since only a very few bonds are broken and made, we can easily answer our earlier questions:
• Which all bonds are broken in the reactant side?
    ♦ How many bonds of each type are broken?
Answer:
One C=C bond and one H-H bond are broken.
• Which all bonds are made in the product side?
    ♦ How many bonds of each type are made?
Answer:
One C-C bond and two C-H bonds are made.
4. We know the following two facts:
(i) When bonds are broken, energy is absorbed.
    ♦ So, we need to supply energy to break the bonds in the reactant side.
(ii) When bonds are made, energy is released.
    ♦ So we will receive energy when bonds are made in the product side.
7. Let us calculate each of the above two items in our present case.
• From the data book, we have:
    ♦ C=C has a bond enthalpy of 614 kJ mol^-1
    ♦ H-H has a bond enthalpy of 436 kJ mol^-1
    ♦ C-C has a bond enthalpy of 348 kJ mol^-1
    ♦ C-H has a bond enthalpy of 413 kJ mol^-1
• So we get:
(i) Total energy required for breaking bonds in the reactant side
= (1 × 614) + (1 × 436) = 1050
(ii) Total energy released when making all the bonds in the product side
= (1 × 348) + (2 × 413) = 1174
8. We see that:
   ♦ Energy required to break the bonds in the reactant side
   ♦ is lesser than
   ♦ Energy released when bonds are made in the product side.
• So we receive a net energy of (1174-1050) = 124 kJ mol^-1
• Since energy is released, it is an exothermic reaction. The enthalpy value should be given a negative sign.
9. Now we can write the thermochemical equation:
C3H6(g) + H₂(g) → C3H₈(g); ΔH^⊖_r = -124.0 kJ mol^-1
10. Based on the above discussion, we can write a general equation to find ΔH^⊖_r for simple cases
Eq.6.11:
$\mathbf\small{\rm{\Delta {H^{\ominus }}_r=\sum{Broken \; Bond\;Enthalpies}-\sum{Made \; Bond\;Enthalpies}}}$
11. Consider the two terms on the right side of Eq.6.11.
• If the first term is larger, it means that, the 'total energy required to break bonds' is greater.
   ♦ We will have to supply a net energy to break the bonds.
   ♦ ΔH^⊖_r will have a positive sign.
   ♦ The reaction will be endothermic.
• If the second term is larger, it means that, the 'total energy released when bonds are made' is greater.
   ♦ We will receive a net energy.
   ♦ ΔH^⊖_r will have a negative sign.
   ♦ The reaction will be exothermic.
• In our present case, the second term is larger.
• We see that, the Eq.6.11 automatically gives us the appropriate sign.

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• Now we have a basic idea about the application of bond enthalpy
• In the next section, we will see lattice enthalpy

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Chapter 6.7 - Hess's Law of Constant Heat Summation

In the previous section, we saw enthalpies of fusion, vaporization, sublimation and combustion. Next we will see enthalpies of atomization, bond, lattice and solution. But before that, we have to see Hess’s law of constant heat summation. We will see it in this section.

Basics of Hess's law can be written in 10 steps:
1. Consider a chemical reaction between two reactants R₁ and R₂
• Let the products be P₁ and P₂
• We can write:
R₁ + R₂ → P₁ + P₂; ΔH^⊖_r = ±X
    ♦ ‘X’ is the magnitude of the enthalpy change.
    ♦ sign may be ‘+’ or ‘-’
        ✰ sign depends on whether the reaction is exothermic or endothermic.
2. Let state A be the initial state of the reaction.
    ♦ At A, the only molecules present, will be those of R₁ and R₂
3. Let state B be the final state of the reaction.
    ♦ At B, the only molecules present, will be those of the products P₁ and P₂
4. Then we can represent the above reaction as shown in fig.6.10(a) below:

Fig.6.10

Fig.a indicates a transformation from A to B, with an accompanying enthalpy change of ±X

5. Let us consider an alternate path:
• Imagine that, from the initial state A, the system first transforms to another state C.
• And from that state C, it transforms to the final state B.
• We can represent this as shown in fig.6.10(b) above.
6. We see that:
• The first transformation from A to C is accompanied by an enthalpy change of ±Y
• The second transformation from C to D is accompanied by an enthalpy change of ±Z
◼ Then, according to Hess’s law, ±X will be the algebraic sum of ±Y and ±Z
    ♦ That is., (±X) = [(±Y) + (±Z)]
7. In fig.b:
    ♦ the single transformation A → B
    ♦ is replaced by
    ♦ two transformations: A → C and C → B
• The original path is indicated by the white arrow.
• The alternate path is indicated by the two magenta arrows.
8. In some cases
    ♦ the single transformation A → B
    ♦ is replaced by
    ♦ three transformations: A → C, C → D and D → B
• This is shown in fig.6.11(c)
• Applying Hess’s law, we get: (±X) = [(±Y) + (±Z) + (±K)]
• The original path is indicated by the white arrow.
• The alternate path is indicated by the three magenta arrows.
9. In this way, the single transformation A → B can be replaced by any number of intermediate transformations. Whatever be the 'number of intermediate transformations', Hess's law will be valid
• All we need to do is: Find the algebraic sum of the intermediate enthalpies.
10. But before finding the algebraic sum, we have to establish the alternate path.
• With some practice, we will be able to do it easily.
• An example is shown below:

Links to some more examples are given below:

Example 2 

Example 3

Example 4 

Example 5

Example 6  

Example 7   

Example 8  

Example 9

◼  Based on the above discussion, we can write the Hess's Law of Constant Heat Summation:
The law states that, regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law shows that enthalpy is a state function.

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• Next we will see enthalpy change for atomization. It can be explained in 5 steps:
1. A sample of hydrogen gas will consist of H₂ molecules. By supplying enough energy, we can separate each of those H₂ molecules into to individual H atoms.
2. Using calorimetry, scientists have calculated that:
    ♦ 1 mole of H₂ gas requires
    ♦ 435.0 kJ energy
3. Based on this, we can write the thermochemical equation:
H₂(g) → 2H (g); ΔH^⊖_a = 435.0 kJ mol^-1
• Note that, instead of 'r', the subscript is 'a'. This is to indicate 'atomization'.
• Another example is the atomization of chlorine. The thermochemical equation is:
Cl₂(g) → 2Cl (g); ΔH^⊖_a = 243.0 kJ mol^-1
4. The above two examples are diatomic molecules. We can consider polyatomic molecules also:
CH₄(g) → C(g) + 4H(g); ΔH^⊖_a = 1665.0 kJ mol^-1
• In the product side, there are only individual atoms:
    ♦ One C atom and four H atoms.
5. In the case of sublimation, we know that, the products are in gaseous form.
• If that gaseous form consists only of individual atoms, we can write:
Enthalpy of atomization will be same as the enthalpy of sublimation.
• An example is the sublimation of Na
    ♦ Na(s) → Na (g); ΔH^⊖_sub = 108.4 kJ mol^-1
    ♦ Na(s) → Na (g); ΔH^⊖_a = 108.4 kJ mol^-1 

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Next we will see enthalpy change in the following two cases:
    ♦ Existing bonds between atoms are broken.
    ♦ New bonds between atoms are formed.
• It can be explained in 8 steps:
1. A sample of hydrogen gas will consist of H₂ molecules. By supplying enough energy, we can separate each of those H₂ molecules into to individual H atoms. For such a separation, we will have to break the H-H bond in each of those H₂ molecules.
2. Using calorimetry, scientists have calculated that:
    ♦ 1 mole of H₂ gas requires
    ♦ 435.0 kJ energy
• If one mol H₂ molecules are converted completely into H atoms, we can be sure that, one mol 'H-H bonds' are broken.
3. Based on this, we can write the thermochemical equation:
H₂(g) → 2H(g); ΔH^⊖_H-H = 435.0 kJ mol^-1
• Note that, instead of 'r', the subscript is 'H-H'. This is to indicate 'the making or breaking of the single bond between two H atoms'.
('Making of bonds' is opposite of 'breaking of bonds'
    ♦ When bonds are broken, a certain energy will be absorbed.
    ♦ If the same bonds are made, the same energy will be released)
• Another example is the enthalpy of Cl-Cl bond. The thermochemical equation is:
Cl₂(g) → 2Cl (g); ΔH^⊖_Cl-Cl = 243.0 kJ mol^-1
4. It is clear that, for diatomic molecules, bond enthalpy will be equal to ΔH^⊖_a
• This is because,
    ♦ breaking all the bonds of diatomic molecules
    ♦ is same as
    ♦ changing the molecules into individual atoms
5. But for polyatomic molecules, bond enthalpy will not be same as ΔH^⊖_a
• The reason can be explained in 8 steps, using CH₄ as an example:
(i) We have the thermochemical equation for the atomization of CH₄:
CH₄(g) → C(g) + 4H(g); ΔH^⊖_a = 1665.0 kJ mol^-1
• That means, to convert one mol CH₄ into individual C and H atoms, we need to supply 1665 kJ energy.
(ii) We know that, in one CH₄ molecule, there are four C-H bonds.
• So in one mol CH₄, there will be four mol C-H bonds
• That means, to break four mol C-H bonds, we need to supply 1665.0 kJ
(iii) So it seems that, to break one mol C-H bonds, we need to supply (¹⁶⁶⁵⁄₄) = 416 kJ
• It seems that, we can write: ΔH^⊖_C-H = 416.0 kJ mol^-1
• But this result is not valid. Let us see the reason:
(iv) Consider the following thermochemical equation:
CH₄(g) → CH₃(g) + H(g); ΔH^⊖_r = 427.0 kJ mol^-1
• It indicates that
    ♦ One mol CH₄ molecules is taken
    ♦ In each of those molecules, one of the four C-H bonds is broken.
    ♦ Thus one mol H atoms are set free.
    ♦ In effect, one mol C-H bonds are broken.
• But we see that, the energy absorbed is 427.0 kJ. This is different from the result in (iii)
(v) There are even more differences:
• Consider the following thermochemical equation:
CH₃(g) → CH₂(g) + H(g); ΔH^⊖_r = 439.0 kJ mol^-1
• It indicates that
    ♦ One mol CH₃ molecules is taken.
    ♦ In each of those molecules, one of the three C-H bonds is broken.
    ♦ Thus one mol H atoms are set free.
    ♦ In effect, one mol C-H bonds are broken.
• But we see that, the energy absorbed is 439.0 kJ. This is different from the result in (iii) and (iv)
(vi) Scientists have discovered the reason for the difference between the results in (iv) and (v):
    ♦ In (iv), the first H atom is set free.
    ♦ Once that H atom leaves, the remaining three are held more tightly by the C atom.
    ♦ So to release a second H atom, more energy will be required.
    ♦ Thus the energy in (iv) is greater than that in (iii)
(vii) We can continue like this until the remaining two H atoms are also set free from the C atom:
• CH₂(g) → CH(g) + H(g); ΔH^⊖_r = 452.0 kJ mol^-1
• CH₃(g) → CH₂(g) + H(g); ΔH^⊖_r = 347.0 kJ mol^-1
(viii) So there are five possible values for ΔH^⊖_C-H. They are:
    ♦ The value obtained by dividing ΔH^⊖_a by 4: 416.0
    ♦ The value when the first H atom is released: 427.0
    ♦ The value when the second H atom is released: 439.0
    ♦ The value when the third H atom is released: 452.0
    ♦ The value when the fourth H atom is released: 347.0
6. To make matters worse, the above five values do not complete the 'list of possible values'. There are even more. The reason can be explained in 3 steps:
(i) Consider the compound CH₃CH₂Cl
• We know that, there are some C-H bonds in this compound.
   ♦ The energy required to break them are different from the five values that we saw.
(ii) Consider the compound CH₃NO₂
• We know that, there are some C-H bonds in this compound.
   ♦ The energy required to break them are different from the five values that we saw.
   ♦ The energy required to break them are different from the values in CH₃CH₂Cl also.
(iii) So it is clear that, the value for C-H bond differs from compound to compound also.
7. But there is nothing to worry about. After considering the different possible values, scientists have agreed upon an average value. It is: 413 kJ mol^-1
• That means, we can write: ΔH^⊖_C-H = 413 kJ mol^-1
• This is the value that we will find in the data book.
• Whenever we encounter a C-H bond, we can use '413 kJ mol^-1' in the calculations.
8. Just like in C-H, we will get different values in other bonds like C-Cl, N-O etc., also.
• Scientists have given the appropriate values that can we can use in such cases also. They are available in the data book. Some examples are given below:
   ♦ For C-Cl, the value is 328 kJ mol^-1
   ♦ For N-O, the value is 201 kJ mol^-1

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• Now we have a basic idea about bond enthalpy.
• Different text books may use different terms for bond enthalpy.
• Bond dissociation enthalpy, bond strength and bond energy are all same as bond enthalpy.
• In the next section, we will see how this bond enthalpy can be put to practical use.


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Chapter 4.22 - Valence Bond Theory

In the previous section 4.21, we completed the discussion on the VSEPR theory. In this section, we will see the Valence bond theory

■ So far, we have seen two methods for describing the structure of molecules: Lewis dot structures and VSEPR theory
• Lewis dot structures give us a basic idea about the bonds in a molecule. But:
    ♦ They do not tell us any thing about bond energies
    ♦ They do not tell us any thing about bond lengths
    ♦ They do not tell us any thing about the actual shapes of molecules
• The VSEPR theory gives us a basic idea about the actual shapes of the molecules. But:
    ♦ They do not tell us any thing about bond energies
    ♦ They do not tell us any thing about bond lengths
• In 1921 The valence bond (VB) theory and molecular orbital (MO) theory was put forward to overcome the above limitations

• First let us see how a bond between two hydrogen atoms is formed. It can be written in 14 steps:
1. Consider a hydrogen atom A
    ♦ Let it’s nucleus be N_A
    ♦ Let it’s electron be e_A
2. Consider another hydrogen atom B
    ♦ Let it’s nucleus be N_B
    ♦ Let it’s electron be e_B
3. Initially, the two hydrogen atoms are at a large distance apart
• So initially, there will be only two forces. We will call them as old forces. They are:
(i) The force between N_A and e_A (attraction)
(ii) The force between N_B and e_B (attraction)
4. Let the two atoms approach each other
• Now additional forces will begin to appear
• There will be 4 additional forces. We will call them as new forces. They are:
(i) The force between N_A and e_B (attraction)
(ii) The force between N_B and e_A (attraction)
(iii) The force between N_A and N_B (repulsion)
(iv) The force between eA and e_B (repulsion)
5. All the forces can be shown using a color scheme:
• The attractive forces can be shown in green color
    ♦ Old attractive forces in dashed green
    ♦ New attractive forces in solid green
• The repulsive forces can be shown in yellow color
    ♦ Old repulsive forces in dashed yellow
    ♦ New repulsive forces in solid yellow
• But we do not have old repulsive forces. So there are only three types of arrows:
    ♦ Old attractive forces in dashed green
    ♦ New attractive forces in solid green
    ♦ New repulsive forces in solid yellow
6. Using this color scheme, the forces are shown in fig.4.126 below
• Note that, a total of six forces were mentioned in (3) and (4)
• There are indeed six lines in fig.4.126

Fig.4.126

7. Now we have an idea about the various forces
• We have to consider the following magnitudes:
(i) Magnitude of the net attractive force
(ii) Magnitude of the net repulsive force
■ Experiments show that (i) is greater than (ii)
■ That means, net attraction is greater than net repulsion
8. So the two atoms will move closer and closer to each other
• As they move closer, the energy of the system (the system consists of the two atoms) decreases
• This is graphically shown in the fig.4.127 below:

Fig.4.127

• Distance between the two nuclei is plotted along the x-axis
• Energy of the system is plotted along the y-axis
• Let us see the various features of the graph. The following steps from (9) to (14) will help us to clearly understand the features:
9. Usually, we read a graph from left to right. But in the present case, we have to read from right to left
(i) Put your finger tip on the extreme right end of the red curve
(ii) Move the finger tip slowly towards the left, through the curve
(iii) As the we proceed, the finger tip touches the various points on the curve
• Each of those points will have:
    ♦ a definite x coordinate
    ♦ a definite y coordinate
(iv) In the initial stage of the curve:
    ♦ The x coordinates decrease continuously
    ♦ But the y coordinates remain nearly the same
(v) This is so because, this stage of the curve is nearly horizontal
• This 'initial horizontal portion' is indicated in blue color (from A to B) in fig.4.128 below:

Fig.4.128

10. The reason for this horizontal nature of the curve can be explained in 3 steps:
(i) Initially, the two H atoms are at a large distance apart
    ♦ So there will be no interaction between them
(ii) Since there is no interaction, the energy of the system will be zero
    ♦ This 'zero energy condition' will continue up to the left end (point B) of the blue segment
    ♦ So the blue segment indicates that, actual interaction has not begun between the atoms
    ♦ The position 'I' shows a sample. At 'I', the atoms are far apart and hence there is no interaction
(iii) Note that, in addition to 'being nearly horizontal', the blue segment is very close to the x axis
    ♦ That means, all the points in the blue segment will have their y coordinates nearly zero
11. After the left end B of the blue segment, we see that, the curve dips
• This is indicated by the green segment BC
• We can write the details in 6 steps:
(i) As we move the finger tip through the green segment,
    ♦ The x coordinate decreases
    ♦ The y coordinate also decreases
(ii) The decrease in the x coordinate indicates that, the atoms are moving closer and closer to each other
    ♦ The position 'II' shows a sample. At 'II',. the atoms are closer to each other
    ♦ Hence there is some interaction
(iii) The decrease in y coordinate indicates that, the energy is decreasing

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When the distance decreases, the energy of the system also decreases. This is similar to an 'earth-stone' system that we see in the physics classes. When the stone is very high up, the potential energy of the earth-stone system is high. When the stone is at a lower height, the potential energy of the earth-stone system is low. We will see a more detailed explanation when we learn about 'interactions between charged particles' in physics classes

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(iv) At the bottom end of the green segment, the energy value is -435.8 kJ/mol

    ♦ This is the 'lowest energy' that the system can attain
(v) The ‘point of lowest energy’ is an important point
• We must note down the x coordinate at this point. It is equal to 74 pm
• So we can write:
    ♦ When the two hydrogen atoms approach each other, the energy of the system continuously decreases
    ♦ The energy reaches the least possible value when the distance between the atoms become 74 pm
(vi) The ‘least possible energy’ is an ideal condition for the ‘formation of a bond’
■ So the bond length between the two H atoms in a H₂ molecule will be 74 pm

12. Beyond the point C, we have the yellow segment CD. Here the situation changes dramatically. It can be written in 6 steps
(i) Beyond point C, the distance between the two H atoms is less than 74 pm
(ii) The atoms are so close to each other that,
    ♦ the two nuclei will begin to repel each other
    ♦ the two electrons will begin to repel each other
• So repulsive forces also come into play
(iii) Now the interactions are not so strong as in the green segment. This is because:
    ♦ The interaction is now the net of attraction and repulsion
    ♦ In the green segment, there was attraction only 
(iv) Since the interactions are not so strong, the energy of the system begins to increase
• This is indicated by the yellow segment
• As we move the finger tip through the yellow segment,
    ♦ The x coordinate decreases
    ♦ The y coordinate increases
(v) The decrease in the x coordinate indicates that, the atoms are moving closer and closer to each other
(vi) The increase in y coordinate indicates that, the energy is increasing
13. Note that, up to the left end D of the yellow segment, the energy is negative
• This is because, though there is repulsion, the force of attraction still has an upper hand
• But beyond this point D, repulsion takes over
• The energy due to ‘repulsive interactions’ is positive
• So the magenta portion DE is above the x axis

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■ We see a sign convention:

• The energy due to attractive interaction is negative
• The energy due to repulsive interaction is positive
■ Why is there such a sign convention?
• We will see the answer in physics classes when we study the 'interactions between charged particles'

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14. Let us note an important inference from the above graph. It can be written in steps:

(i) When the two atoms are at a large distance apart, the energy of the system is zero
(ii) When the two atoms approach each other, the energy of the system decreases
    ♦ That means, the system gives off energy to the surroundings
■ When one mole of H atoms combine with another one mole of H atoms to form one mole of H₂ molecules, we will get 435.8 kJ of energy
    ♦ The energy that we receive is -ve
■ Conversely, when we supply 435 kJ of energy to one mole of H₂ molecules, we will get two mols of H atoms
    ♦ The energy that we supply is +ve

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We have seen the formation of a bond between two H atoms. Based on this, we will see valence bond theory in the next section

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Chapter 4.8 - Bond Order

In the previous section, we saw the bond enthalpies of diatomic molecules. We saw three examples related to homonuclear diatomic molecules. In this section, we will see the bond enthalpies of heteronuclear diatomic molecules

• H₂, O₂, and N₂ are diatomic molecules
    ♦ To be precise, they are homonuclear diatomic molecules
■ Diatomic molecules are molecules composed of only two atoms
• Consider a diatomic molecule
    ♦ If both the two atoms are of the same element, it is called a homonuclear diatomic molecule
    ♦ If the two atoms are of different elements, it is called a heteronuclear diatomic molecule

• Let us see the bond enthalpy in the case of heteronuclear diatomic molecules. We will explain it using an example:
Example 4:
1. Take one mole of HCl molecules
• We want all those molecules to be separated into 'individual H and Cl atoms'
• Let us see how it can be done:
2. For our present discussion, we will represent the HCl molecule as H-Cl
• Where '-' represents the single covalent bond between the H and Cl atoms
3. We have taken 1 mole of HCl molecules
• So there will be one mole (6.023 × 10²³ nos.) of 'H-Cl'
• So there will be 6.023 × 10²³ nos. of 'H-Cl single bonds'
4. We must break all those H-Cl bonds
• Then only we will be able to convert all of the '1 mole HCl' into individual H and Cl atoms
■ The energy required to break one mole (6.023 × 10²³ nos.) of 'H-Cl single bonds' is called 'bond enthalpy of H-Cl bond in hydrogen chloride molecule'
• We can write in any of the two ways:
    ♦ Bond enthalpy of H-Cl bond in hydrogen chloride molecule
       OR
    ♦ H-Cl bond enthalpy in hydrogen chloride molecule

• While breaking those bonds,

    ♦ The initial sample of HCl that we take, must be in gaseous state
    ♦ The resulting H atoms must also be in gaseous state
    ♦ The resulting Cl atoms must also be in gaseous state
• Scientists have determined this energy using experimental methods. It is equal to 431.0 kJ mol^-1
5. The process is represented as:
$\mathbf\small{\rm{HCl(g)\longrightarrow H(g)\,+Cl(g);\Delta_aH^\theta=431.0\,kJ\,mol^{-1}}}$
6. We can write the report as follows:
• Bond enthalpy of H-Cl bond in hydrogen chloride molecule is 431.0 kJ mol^-1
  OR
• H-Cl bond enthalpy in hydrogen chloride molecule is 431.0 kJ mol^-1

• Next, let us see the bond enthalpy in the case of polyatomic molecules
• Polyatomic molecules are those molecules which have three or more atoms
Example 5:
1. Take one mole of H₂O molecules
• We want all those molecules to be separated into 'individual H and O atoms'
• Let us see how it can be done:
2. For our present discussion, we will represent the H₂O molecule as H-O-H
    ♦ See fig.4.15 in section 4.2
• Where '-' represents the single covalent bond between the H and O atoms
3. We have taken 1 mole of H₂O molecules
• So there will be one mole (6.023 × 10²³ nos.) of 'H-O-H'
4. In one molecule of H-O-H, there are two single bonds:
    ♦ The O-H bond on the left side
    ♦ The O-H bond on the right side
5. So in one mole of H₂O, there will be:
    ♦ 6.023 × 10²³ nos. of 'left side O-H bonds'
    ♦ 6.023 × 10²³ nos. of 'right side O-H bonds'
6. We must break all those O-H bonds
• Then only we will be able to convert all of the '1 mole H₂O' into individual H and O atoms
7. Now, there is a problem:
• The energy required to break the left side O-H bond
   is not equal to
• The energy required to break the right side O-H bond
8. So we must split up the process. It can be explained in 3 steps:
(i) Consider 6.023 × 10²³ nos. of H-O-H molecules
(ii) First we break the left side bond in each of them
We get two items:
    ♦ 6.023 × 10²³ nos. of H
    ♦ 6.023 × 10²³ nos. of O-H
• 502 kJ energy is required for this process
• So we can write:
$\mathbf\small{\rm{H_2O(g)\longrightarrow H(g)\,+OH(g);\Delta_aH^\theta=502.0\,kJ\,mol^{-1}}}$
(iii) Next we break the bond in 6.023 × 10²³ nos. of O-H obtained in (ii)
• 427 kJ energy is required for this process
• So we can write:
$\mathbf\small{\rm{OH(g)\longrightarrow H(g)\,+O(g);\Delta_aH^\theta=427.0\,kJ\,mol^{-1}}}$
9. The difference in the two energies can be explained in 3 steps:
(i) First we break the left side O-H bond
• This gives H and O-H
• Thus one H is removed from the original H-O-H
(ii) Consider the two environments:
• The first 'O-H bond' is in a chemical environment in which, 'that bond is a part of H-O-H'
• The second 'O-H bond' is in a chemical environment in which, 'that bond is a part of O-H'   
(iii) The two environments are clearly different
■ So the energies required will also be different
10. We want all 6.023 × 10²³ nos. of H-O-H molecules to be separated into 'individual H and O atoms' 
• Clearly, we will need '502 plus 427 kJ energy' for that
• These energies are called bond dissociation enthalpies
• These enthalpies are different from bond enthalpies that we were discussing about
11. So what is bond enthalpy in the case of H-O-H?
The answer can be written in 3 steps:
(i) Calculate the average (mean) of the two values: 502 and 427
• It is equal to 464.5
(ii) Since it is an average, we cannot simply write: 'bond enthalpy'
• We must write it as: 'average bond enthalpy'
(iii) So the final report is:
• The average bond enthalpy of the O-H bonds in water molecule is 464.5 kJ mol^-1
 OR
• Average O-H bond enthalpy in water molecule is 464.5 kJ mol^-1
12. Now we can write an interesting point:
■ For a homonuclear diatomic molecule, the 'bond dissociation enthalpy' will be same as 'bond enthalpy'
• The reader may write the reason in his/her own notebooks
13. While breaking the bonds in H-O-H,
    ♦ The initial sample of H₂O that we take, must be in gaseous state
    ♦ The resulting H atoms must also be in gaseous state
    ♦ The resulting O atoms must also be in gaseous state
14. So we have calculated the 'average bond enthalpy of the O-H bonds in water molecule'
• It is equal to 464.5 kJ mol^-1
■ Can we use this value for O-H bonds in other molecules?
• The answer can be written by taking an example. It can be written in 3 steps:
(i) Consider the C₂H₅OH (ethanol) molecule
• It contains one O-H bond
(ii) Suppose that we have 6.023 × 10²³ nos. of C₂H₅OH molecules
• There will be 6.023 × 10²³ nos. of O-H bonds
(iii) The energy required to break all those bonds will not be equal to 464.5 kJ
• This is because, the two environments mentioned below are different:
    ♦ The chemical environment in which 'O-H bond is a part of ethanol'
    ♦ The chemical environment in which 'O-H bond is a part of water'

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• We have completed a discussion on the basics of bond enthalpy. It is important to remember two points:

1. Strength of the bond
■ We must keep in mind that, bond enthalpy gives an idea about the 'strength of a bond'
• If the bond enthalpy of a bond is high, it is obvious that, the strength of that bond is high (a strong bond)
• If the bond enthalpy of a bond is low, it is obvious that, the strength of that bond is low (a weak bond)
2. Method of reporting:
• While reporting a bond enthalpy value, we must always include the following 3 items:
(i) The name of the two atoms between which the 'bond under consideration' is present
(ii) Whether the 'bond under consideration' is a single, double or triple bond
(iii) The molecule in which the 'bond under consideration' is present

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Next we will see bond order. It is the 'number of bonds'. We will explain it using some examples:
Example 1:
1. Consider one H₂ molecule
• It can be represented as H-H
    ♦ See fig.4.13 in section 4.2
• Where '-' represents the single covalent bond between the H atoms
2. Note down 2 points:
    ♦ It is a H₂ molecule
    ♦ There is one covalent bond: H-H 
3. We report the bond order using the 2 points written in (2):
■ In H₂ molecule, the H-H bond order is 1

Example 2:
1. Consider one O₂ molecule
• It can be represented as O=O
    ♦ See fig.4.16 in section 4.2
• Where '=' represents the double bond between the O atoms
2. Note down 2 points:
    ♦ It is a O₂ molecule
    ♦ There are two covalent bonds: O=O 
3. We report the bond order using the 2 points written in (2):
■ In O₂ molecule, the O=O bond order is 2

Example 3:
1. Consider one N₂ molecule
• It can be represented as N≡N
    ♦ See fig.4.17 in section 4.2
• Where '≡' represents the triple bond between the N atoms
2. Note down 2 points:
    ♦ It is a N₂ molecule
    ♦ There are three covalent bonds: N≡N
3. We report the bond order using the 2 points written in (2):
■ In N₂ molecule, the N≡N bond order is 3

Example 4:
1. Consider one CO molecule
• It can be represented as C≡O
    ♦ See fig.4.18 in section 4.2
• Where '≡' represents the triple bond between the C and O atoms
2. Note down 2 points:
    ♦ It is a CO molecule
    ♦ There are three covalent bonds: C≡O
3. We report the bond order using the 2 points written in (2):
■ In CO molecule, the C≡O bond order is 3

Example 5:
1. Consider one F₂ molecule
• It can be represented as F-F
• Where '-' represents the single covalent bond between the F atoms
2. Note down 2 points:
    ♦ It is a F₂ molecule
    ♦ There is one covalent bond: F-F 
3. We report the bond order using the 2 points written in (2):
■ In F₂ molecule, the F-F bond order is 1

Example 6:
1. Consider one O₂^2- ion
• It's Lewis dot structure is shown in fig.4.43(a) below:

Fig.4.43

(Note that, in the previous examples, there was not much need to draw Lewis dot structures. This is because, they were simple molecules and we are familiar with their structures. However, whenever any doubt arises, the reader may draw the Lewis dot structure, or refer already drawn ones)
• From the Lewis dot structure, it is clear that, there is a single bond ('-') between the two O atoms
2. Note down 2 points:
    ♦ It is a O₂^2- ion
    ♦ There is one covalent bond: O-O 
3. We report the bond order using the 2 points written in (2):
■ In O₂^2- ion, the O-O bond order is 1

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Let us compare examples 5 and 6:

• In example 5, F₂ molecule has (7+7) = 14 valence electrons
    ♦ The bond order is 1
• In example 6, O₂^2- ion has (6+6+2) = 14 valence electrons
    ♦ The bond order is 1
■ In general, isoelectronic species have the same bond orders

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Example 7:

1. Consider one NO⁺ ion
• It's Lewis dot structure is shown in fig.4.43(b) above
• From the Lewis dot structure, it is clear that, there is a triple bond ('≡') between the N and O atoms
2. Note down 2 points:
    ♦ It is a NO⁺ ion
    ♦ There are three covalent bonds: N≡O 
3. We report the bond order using the 2 points written in (2):
■ In NO⁺ ion, the N≡O bond order is 3

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Let us compare example 7 with N₂ and CO:

• In example 7, NO⁺ ion has (5+6-1) = 10 valence electrons
    ♦ The bond order is 3
• N2 molecule has (5+5) = 10 valence electrons
    ♦ The bond order is 3
• CO molecule has (4+6) = 10 valence electrons
    ♦ The bond order is 3
■ In general, isoelectronic species have the same bond orders

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Example 8:

1. Consider one C₂H₂ molecule
• It can be represented as H-C≡C-H
• Where:
    ♦ '-' represents the single covalent bond between the C and H atoms
    ♦ '≡' represents the triple covalent bond between the C atoms
2. Note down 2 points:
    ♦ It is a C₂H₂ molecule
    ♦ There is one covalent bond: C-H
    ♦ There is one covalent bond: C-H 
    ♦ There are three covalent bonds: C≡C
3. We report the bond order using the 4 points written in (2):
■ In C₂H₂ molecule:
    ♦ The C-H bond order is 1
    ♦ The C≡C bond order is 3

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• We have completed a discussion on the basics of bond order. It is important to remember two points:

1. Strength of the bond
■ We must keep in mind that, bond order gives an idea about the 'strength of a bond'
• If the bond order is high, bond enthalpy will also be high
    ♦ Thus it will be a strong bond
• If the bond order is low, bond enthalpy will also be low
    ♦ Thus it will be a weak bond
2. Method of specification:
• While reporting a bond order, we must always include the following 3 items:
(i) The name of the two atoms between which the 'bond under consideration' is present
(ii) Whether the 'bond under consideration' is a single, double or triple bond
(iii) The molecule in which the 'bond under consideration' is present

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Solved example 4.4

How do you express the bond strength in terms of bond order ?
Solution:
1. An increase in bond order indicates that, greater number of bonds are present
2. Greater number of bonds indicates that the bond is stronger
• For example:
    ♦ A double bond is stronger than a single bond
    ♦ A triple bond is stronger than a double bond
3. So we can write:
    ♦ When bond order increases, bond strength increases
    ♦ When bond order decreases, bond strength decreases

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In the next section, we will see resonance structures

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