In the previous section, we saw lattice enthalpy. In this section, we will see the factors on which the lattice enthalpy depends. First we will see bond length
1. Consider a molecule having two atoms A and B
• This is shown in fig.4.39 below:
• The nuclei of A and B are marked as '❌'
2. Let A and B be bonded by a covalent bond
■ Then we can write:
'Bond length' of the covalent bond between A and B = Distance between the nuclei of A and B
• In the fig.4.39, this distance is marked as R
3. We see that both A and B make contributions towards making up R
♦ The contribution made by A is denoted as rc,A
♦ The contribution made by B is denoted as rc,B
■ rc,A is the covalent radius of atom A
■ rc,B is the covalent radius of atom B
1. Consider two atoms of an element X
• This is shown in fig.4.40(a) below:
2. We know that, any atom will have two parts:
♦ The valence shell
✰ This is the outermost main-shell
♦ The core
✰ The core consists of the inner main-shells and the nucleus
• In the above fig.4.40(a), the valence shell is shown in blue color
• The core is shown in red color
3. Let the two atoms form a covalent bond with each other
• This is shown in fig.b
• In fig.b, we notice two points:
♦ The two valence shells overlap each other
♦ The two cores just touch each other
4. In this situation, we measure the 'radius of any one core'
• This radius is the covalent radius rc,X of the element X
• This is shown in fig.c
• Note that, radii of both the cores in fig.c are the same rc,X. So we need to measure any one only
5. So we can write the definition:
• The covalent radius of an atom is the radius of that atom’s core
♦ This core must be in contact with the core of an adjacent atom in a bonded situation
♦ The original atom and the adjacent atom must be in a bonded situation
♦ The original atom and the adjacent atom must belong to the same element
6. We notice another point also. It can be written in steps:
(i) The two cores are just touching each other
(ii) Consider the 'center to center' distance between the two cores
(iii) Take half of that distance
(iv) This 'half distance' is equal to rc,X
7. So we can write another method for measuring covalent radius:
• The covalent radius is half of the distance between two similar atoms
♦ Those two similar atoms should be part of a single molecule
♦ Those two similar atoms should be joined by a covalent bond
1. Consider fig.4.41(a) below:
• Note the two points in the fig.a:
♦ The center of the core
♦ The outer surface of the valence shell
2. Consider the distance between those two points
• This distance is marked as rvdw,X
■ This distance is the van der waals radius of the element X
3. So we can write the definition:
• The van der Waals radius represents the overall size of the atom
♦ This overall size includes the valence shell of the atom
♦ The atom should be in a non-bonded situation
4. Now consider fig.4.41(b)
• Two X2 molecules are just touching each other
• Note the two points in the fig:
♦ The center of an X in one molecule
♦ The center of an X in the other molecule
5. Consider the distance between those two points
• Take half of that distance
• This half distance will be equal to rvdw,X
6. So we can write another method for measuring van der waals radius:
• van der waals radius is half the distance between two similar atoms
♦ Those two atoms must be situated in two different molecules
♦ Those two molecules must be just touching each other
♦ The property we are talking about is: atomic radius (Details here)
• It is very difficult to directly measure the radius of an atom because, it is very small
• Also, the atom do not have a definite boundary. What we have is a cloud of electrons around the nucleus
• So an indirect method is used. It can be written in 2 steps:
(i) Find the distance between the nuclei of two atoms
♦ Those two atoms must be bonded together by a single covalent bond
(ii) Calculate half of that distance
♦ This 'half distance' is called 'covalent radius' of that atom
♦ This 'covalent radius' is taken as the 'atomic radius' of that atom
• We have already seen the 'covalent radius rc,X of any element X' in fig.4.40 above
♦ We used figs.4.40 and 4.39 to define bond length
✰ We first defined 'bond length'
✰ Based on that, we defined 'covalent radius'
♦ But such a method is adopted for theoretical purposes only
• In the real world, we first measure the bond length
♦ This is done using spectroscopic techniques or X-ray diffraction techniques
• Half of that bond length will give the radius of the atom (covalent radius)
■ An example:
(i) Suppose that, the two atoms in the fig.4.40 are chlorine atoms
(ii) Scientists have measured the distance between the two '❌' in fig.4.40(c)
♦ The measurement is done using spectroscopic techniques or X-ray diffraction techniques
♦ The distance is found to be approximately equal to 198 pm
(iii) So the covalent radius of a chlorine atom = 198⁄2 = 99 pm
• In this way, we can determine the covalent radius of different atoms
1. Bond angle is simply, the angle between two bonds
2. The angle that we see in geometry classes, have a vertex and two legs
• In our present case,
♦ The vertex corresponds to a central atom
♦ The two legs correspond to two bonds originating from that atom
3. Let us see an example:
Fig.4.42(a) below shows the Lewis dot structure of H2O
(i) We see two O-H bonds
(ii) Both the bonds originate from the central atom O
♦ So O is the vertex
♦ The two bonds are the legs
(iii) The angle between the two bonds is the bond angle
4. Bond angle and structure of molecule:
• An orbital of O is involved in the left side O-H bond
• Another orbital of O is involved in the right side O-H bond
• So the bond angle gives us the angle between those two orbitals
• Thus bond angle is useful in understanding the distribution of 'orbitals of central atom' in space. This gives an idea about the shape of the molecule
• We will see more details about this in later sections
• We will explain bond enthalpy using some examples
Example 1:
1. Take one mole of H2 molecules
• We want all those molecules to be separated into 'individual H atoms'
• Let us see how it can be done:
2. For our present discussion, we will represent the H2 molecule as H-H
♦ See fig.4.13 in section 4.2
• Where '-' represents the single covalent bond between two hydrogen atoms
3. We have taken 1 mole of hydrogen molecules
• So there will be one mole (6.023 × 1023 nos.) of 'H-H'
• So there will be 6.023 × 1023 nos. of 'H-H single bonds'
4. We must break all those H-H bonds
• Then only we will be able to convert all of the '1 mole H2' into individual H atoms
■ The energy required to break one mole (6.023 × 1023 nos.) of 'H-H single bonds' is called 'bond enthalpy of H-H bond in hydrogen molecule'
• We can write in any of the two ways:
♦ Bond enthalpy of H-H bond in hydrogen molecule
OR
♦ H-H bond enthalpy in hydrogen molecule
♦ The resulting H atoms must also be in gaseous state
• Scientists have determined this energy using experimental methods. It is equal to 435.8 kJ mol-1
5. The process is represented as:
$\mathbf\small{\rm{H_2(g)\longrightarrow H(g)\,+H(g);\Delta_aH^\theta=435.8\,kJ\,mol^{-1}}}$
• $\mathbf\small{\rm{\Delta_aH^\theta}}$ is the symbol for bond enthalpy
6. We can write the report as follows:
• Bond enthalpy of H-H bond in hydrogen molecule is 435.8 kJ mol-1
OR
• H-H bond enthalpy in hydrogen molecule is 435.8 kJ mol-1
Example 2:
1. Take one mole of O2 molecules
• We want all those molecules to be separated into 'individual O atoms'
• Let us see how it can be done:
2. For our present discussion, we will represent the O2 molecule as O=O
♦ See fig.4.16 in section 4.2
• Where '=' represents the double bond between two oxygen atoms
3. We have taken 1 mole of oxygen molecules
• So there will be one mole (6.023 × 1023 nos.) of 'O=O'
• So there will be 6.023 × 1023 nos. of 'O=O double bonds'
4. We must break all those O=O bonds
• Then only we will be able to convert all of the '1 mole O2' into individual O atoms
■ The energy required to break one mole (6.023 × 1023 nos.) of 'O=O double bonds' is called 'bond enthalpy of O=O bond in oxygen molecule'
• We can write in any of the two ways:
♦ Bond enthalpy of O=O bond in oxygen molecule
OR
♦ O=O bond enthalpy in oxygen molecule
♦ The resulting O atoms must also be in gaseous state
• Scientists have determined this energy using experimental methods. It is equal to 498 kJ mol-1
5. The process is represented as:
$\mathbf\small{\rm{O_2(g)\longrightarrow O(g)\,+O(g);\Delta_aH^\theta=435.8\,kJ\,mol^{-1}}}$
6. We can write the report as follows:
• Bond enthalpy of O=O bond in oxygen molecule is 498.0 kJ mol-1
OR
• O=O bond enthalpy in oxygen molecule is 498.0 kJ mol-1
Example 3:
1. Take one mole of N2 molecules
• We want all those molecules to be separated into 'individual N atoms'
• Let us see how it can be done:
2. For our present discussion, we will represent the N2 molecule as N≡N
♦ See fig.4.17 in section 4.2
• Where '≡' represents the triple bond between two nitrogen atoms
3. We have taken 1 mole of nitrogen molecules
• So there will be one mole (6.023 × 1023 nos.) of 'N≡N'
• So there will be 6.023 × 1023 nos. of 'N≡N triple bonds'
4. We must break all those N≡N bonds
• Then only we will be able to convert all of the '1 mole N2' into individual N atoms
■ The energy required to break one mole (6.023 × 1023 nos.) of 'N≡N single bonds' is called 'bond enthalpy of N≡N bond in nitrogen molecule'
• We can write in any of the two ways:
♦ Bond enthalpy of N≡N bond in nitrogen molecule
OR
♦ N≡N bond enthalpy in nitrogen molecule
♦ The resulting N atoms must also be in gaseous state
• Scientists have determined this energy using experimental methods. It is equal to 946.0 kJ mol-1
5. The process is represented as:
$\mathbf\small{\rm{N_2(g)\longrightarrow N(g)\,+N(g);\Delta_aH^\theta=946.0\,kJ\,mol^{-1}}}$
6. We can write the report as follows:
• Bond enthalpy of N≡N bond in nitrogen molecule is 946.0 kJ mol-1
OR
• N≡N bond enthalpy in nitrogen molecule is 946.0 kJ mol-1
1. Consider a molecule having two atoms A and B
• This is shown in fig.4.39 below:
 |
| Fig.4.39 |
2. Let A and B be bonded by a covalent bond
■ Then we can write:
'Bond length' of the covalent bond between A and B = Distance between the nuclei of A and B
• In the fig.4.39, this distance is marked as R
3. We see that both A and B make contributions towards making up R
♦ The contribution made by A is denoted as rc,A
♦ The contribution made by B is denoted as rc,B
■ rc,A is the covalent radius of atom A
■ rc,B is the covalent radius of atom B
• So our next aim is to define 'covalent radius' of an atom
• This is shown in fig.4.40(a) below:
 |
| Fig.4.40 |
♦ The valence shell
✰ This is the outermost main-shell
♦ The core
✰ The core consists of the inner main-shells and the nucleus
• In the above fig.4.40(a), the valence shell is shown in blue color
• The core is shown in red color
3. Let the two atoms form a covalent bond with each other
• This is shown in fig.b
• In fig.b, we notice two points:
♦ The two valence shells overlap each other
♦ The two cores just touch each other
4. In this situation, we measure the 'radius of any one core'
• This radius is the covalent radius rc,X of the element X
• This is shown in fig.c
• Note that, radii of both the cores in fig.c are the same rc,X. So we need to measure any one only
5. So we can write the definition:
• The covalent radius of an atom is the radius of that atom’s core
♦ This core must be in contact with the core of an adjacent atom in a bonded situation
♦ The original atom and the adjacent atom must be in a bonded situation
♦ The original atom and the adjacent atom must belong to the same element
6. We notice another point also. It can be written in steps:
(i) The two cores are just touching each other
(ii) Consider the 'center to center' distance between the two cores
(iii) Take half of that distance
(iv) This 'half distance' is equal to rc,X
7. So we can write another method for measuring covalent radius:
• The covalent radius is half of the distance between two similar atoms
♦ Those two similar atoms should be part of a single molecule
♦ Those two similar atoms should be joined by a covalent bond
Next we will see van der waals radius
 |
| Fig.4.41 |
♦ The center of the core
♦ The outer surface of the valence shell
2. Consider the distance between those two points
• This distance is marked as rvdw,X
■ This distance is the van der waals radius of the element X
3. So we can write the definition:
• The van der Waals radius represents the overall size of the atom
♦ This overall size includes the valence shell of the atom
♦ The atom should be in a non-bonded situation
4. Now consider fig.4.41(b)
• Two X2 molecules are just touching each other
• Note the two points in the fig:
♦ The center of an X in one molecule
♦ The center of an X in the other molecule
5. Consider the distance between those two points
• Take half of that distance
• This half distance will be equal to rvdw,X
6. So we can write another method for measuring van der waals radius:
• van der waals radius is half the distance between two similar atoms
♦ Those two atoms must be situated in two different molecules
♦ Those two molecules must be just touching each other
• Now we can revisit a property that we saw in the previous chapter
• It is very difficult to directly measure the radius of an atom because, it is very small
• Also, the atom do not have a definite boundary. What we have is a cloud of electrons around the nucleus
• So an indirect method is used. It can be written in 2 steps:
(i) Find the distance between the nuclei of two atoms
♦ Those two atoms must be bonded together by a single covalent bond
(ii) Calculate half of that distance
♦ This 'half distance' is called 'covalent radius' of that atom
♦ This 'covalent radius' is taken as the 'atomic radius' of that atom
• We have already seen the 'covalent radius rc,X of any element X' in fig.4.40 above
♦ We used figs.4.40 and 4.39 to define bond length
✰ We first defined 'bond length'
✰ Based on that, we defined 'covalent radius'
♦ But such a method is adopted for theoretical purposes only
• In the real world, we first measure the bond length
♦ This is done using spectroscopic techniques or X-ray diffraction techniques
• Half of that bond length will give the radius of the atom (covalent radius)
■ An example:
(i) Suppose that, the two atoms in the fig.4.40 are chlorine atoms
(ii) Scientists have measured the distance between the two '❌' in fig.4.40(c)
♦ The measurement is done using spectroscopic techniques or X-ray diffraction techniques
♦ The distance is found to be approximately equal to 198 pm
(iii) So the covalent radius of a chlorine atom = 198⁄2 = 99 pm
• In this way, we can determine the covalent radius of different atoms
Next we will see bond angle
2. The angle that we see in geometry classes, have a vertex and two legs
• In our present case,
♦ The vertex corresponds to a central atom
♦ The two legs correspond to two bonds originating from that atom
3. Let us see an example:
Fig.4.42(a) below shows the Lewis dot structure of H2O
 |
| Fig.4.42 |
(ii) Both the bonds originate from the central atom O
♦ So O is the vertex
♦ The two bonds are the legs
(iii) The angle between the two bonds is the bond angle
4. Bond angle and structure of molecule:
• An orbital of O is involved in the left side O-H bond
• Another orbital of O is involved in the right side O-H bond
• So the bond angle gives us the angle between those two orbitals
• Thus bond angle is useful in understanding the distribution of 'orbitals of central atom' in space. This gives an idea about the shape of the molecule
• We will see more details about this in later sections
Next we will see bond enthalpy
Example 1:
1. Take one mole of H2 molecules
• We want all those molecules to be separated into 'individual H atoms'
• Let us see how it can be done:
2. For our present discussion, we will represent the H2 molecule as H-H
♦ See fig.4.13 in section 4.2
• Where '-' represents the single covalent bond between two hydrogen atoms
3. We have taken 1 mole of hydrogen molecules
• So there will be one mole (6.023 × 1023 nos.) of 'H-H'
• So there will be 6.023 × 1023 nos. of 'H-H single bonds'
4. We must break all those H-H bonds
• Then only we will be able to convert all of the '1 mole H2' into individual H atoms
■ The energy required to break one mole (6.023 × 1023 nos.) of 'H-H single bonds' is called 'bond enthalpy of H-H bond in hydrogen molecule'
• We can write in any of the two ways:
♦ Bond enthalpy of H-H bond in hydrogen molecule
OR
♦ H-H bond enthalpy in hydrogen molecule
• While breaking those bonds,
♦ The initial sample of H2 that we take must be in gaseous state♦ The resulting H atoms must also be in gaseous state
• Scientists have determined this energy using experimental methods. It is equal to 435.8 kJ mol-1
5. The process is represented as:
$\mathbf\small{\rm{H_2(g)\longrightarrow H(g)\,+H(g);\Delta_aH^\theta=435.8\,kJ\,mol^{-1}}}$
• $\mathbf\small{\rm{\Delta_aH^\theta}}$ is the symbol for bond enthalpy
6. We can write the report as follows:
• Bond enthalpy of H-H bond in hydrogen molecule is 435.8 kJ mol-1
OR
• H-H bond enthalpy in hydrogen molecule is 435.8 kJ mol-1
Example 2:
1. Take one mole of O2 molecules
• We want all those molecules to be separated into 'individual O atoms'
• Let us see how it can be done:
2. For our present discussion, we will represent the O2 molecule as O=O
♦ See fig.4.16 in section 4.2
• Where '=' represents the double bond between two oxygen atoms
3. We have taken 1 mole of oxygen molecules
• So there will be one mole (6.023 × 1023 nos.) of 'O=O'
• So there will be 6.023 × 1023 nos. of 'O=O double bonds'
4. We must break all those O=O bonds
• Then only we will be able to convert all of the '1 mole O2' into individual O atoms
■ The energy required to break one mole (6.023 × 1023 nos.) of 'O=O double bonds' is called 'bond enthalpy of O=O bond in oxygen molecule'
• We can write in any of the two ways:
♦ Bond enthalpy of O=O bond in oxygen molecule
OR
♦ O=O bond enthalpy in oxygen molecule
• While breaking those bonds,
♦ The initial sample of O2 that we take must be in gaseous state♦ The resulting O atoms must also be in gaseous state
• Scientists have determined this energy using experimental methods. It is equal to 498 kJ mol-1
5. The process is represented as:
$\mathbf\small{\rm{O_2(g)\longrightarrow O(g)\,+O(g);\Delta_aH^\theta=435.8\,kJ\,mol^{-1}}}$
6. We can write the report as follows:
• Bond enthalpy of O=O bond in oxygen molecule is 498.0 kJ mol-1
OR
• O=O bond enthalpy in oxygen molecule is 498.0 kJ mol-1
Example 3:
1. Take one mole of N2 molecules
• We want all those molecules to be separated into 'individual N atoms'
• Let us see how it can be done:
2. For our present discussion, we will represent the N2 molecule as N≡N
♦ See fig.4.17 in section 4.2
• Where '≡' represents the triple bond between two nitrogen atoms
3. We have taken 1 mole of nitrogen molecules
• So there will be one mole (6.023 × 1023 nos.) of 'N≡N'
• So there will be 6.023 × 1023 nos. of 'N≡N triple bonds'
4. We must break all those N≡N bonds
• Then only we will be able to convert all of the '1 mole N2' into individual N atoms
■ The energy required to break one mole (6.023 × 1023 nos.) of 'N≡N single bonds' is called 'bond enthalpy of N≡N bond in nitrogen molecule'
• We can write in any of the two ways:
♦ Bond enthalpy of N≡N bond in nitrogen molecule
OR
♦ N≡N bond enthalpy in nitrogen molecule
• While breaking those bonds,
♦ The initial sample of N2 that we take must be in gaseous state♦ The resulting N atoms must also be in gaseous state
• Scientists have determined this energy using experimental methods. It is equal to 946.0 kJ mol-1
5. The process is represented as:
$\mathbf\small{\rm{N_2(g)\longrightarrow N(g)\,+N(g);\Delta_aH^\theta=946.0\,kJ\,mol^{-1}}}$
6. We can write the report as follows:
• Bond enthalpy of N≡N bond in nitrogen molecule is 946.0 kJ mol-1
OR
• N≡N bond enthalpy in nitrogen molecule is 946.0 kJ mol-1
• H2, O2, and N2 are homonuclear diatomic molecules
• In the next section, we will see bond enthalpy in the case of heteronuclear diatomic molecules
• In the next section, we will see bond enthalpy in the case of heteronuclear diatomic molecules
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