Chapter 4.29 - Structure of Trichloro boron and Beryllium chloride

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

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

Fig.4.164

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

Fig.4.165

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

5. So in fig.4.165, we have:

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

Fig.4.166

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

Fig.4.167

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

Fig.4.168

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

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

Fig.4.169

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

Fig.4.170

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

5. So in fig.4.170, we have:

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

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

Fig.4.171

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

Fig.4.172

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

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

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Chapter 4.26 - Structure of Ethene

In the previous section 4.25, we saw the structure of CH₄ and C₂H₆. We also discussed sigma bond. In this section, we will see the structure of C₂H₄ (ethene). We will also see the basic details about pi bonds

• Fig.4.144(a) below shows the Lewis dot structure of C₂H₄
• The details about the model of C₂H₄ can be written in 13 steps:
1. Both the C atoms in C₂H₄ is sp² hybridized
• Let us see how this sp² hybridization is achieved. It can be written in 4 steps:
(i) Fig.4.144(b) below shows the orbitals in the valence shell of C

Fig.4.144

(ii) When enough energy is given, one electron in the 2s orbital jumps to the 2p_z orbital
• Thus we get four half filled orbitals. This is shown in fig.c
(iii) The 2s orbital mixes together with 2p_x and 2p_y. This is shown in fig.d
(Note that, 2p_z does not participate in the mixing)
(iv) Since there is one s-orbital and two p-orbitals, it is a sp² hybridization
2. We know that:
    ♦ In sp² hybridization, there will be three hybrid orbitals
          ✰ Together, they form a triangular planar shape
          ✰ The larger lobes will be directed towards the corners of a triangle
• This is shown in fig.4.145(a) below:

Fig.4.145

• Let us write four 'important points to remember' about fig.a:
(i) The three sp² hybrid orbitals are at an angular distance of 120^o apart 
(ii) The nucleus of the C atom is shown as a small red sphere
    ♦ This sphere is situated at the origin of the coordinate axes
(iii) One of the orbitals lies exactly along the x-axis (red axis)
(iv) We know that, all the three sp² hybrid orbitals will lie on a plane
    ♦ So in fig.a, all the three orbitals are lying on the xy-plane
3. Remember that, the 2p_z orbital did not take part in the hybridization
    ♦ So it will 'remain as such' even after hybridization
    ♦ We have to show it in fig.a
• The modified model is shown in fig.b
• This model in fig.b represents the 'sp² hybridized C atom'
• So we can write:
    ♦ The 'sp² hybridized C atom' consists of three items:
          ✰ The three hybrid orbitals
          ✰ The 2p_z orbital
          ✰ The nucleus (small red sphere)
 • In the figs. a and b, the 'smaller lobes of hybrid orbitals' are not shown
    ♦ This is to reduce congestion and thus obtain greater clarity
4. Distribution of electrons in the hybrid orbitals:
• This can be written in 3 steps:
(i) We know that, the sp² hybrid orbitals are formed from 'one s orbital' and 'two p orbitals'
(ii) In our present case of C atom, they are: 'one 2s orbital' and 'two 2p orbitals'
(iii) Before the hybridization, the orbitals mentioned in (ii) carry a total of three electrons
    ♦ After hybridization, the orbitals mentioned in (ii) will no longer exist
    ♦ Then what will happen to the three electrons?
Answer:
• The three electrons will be distributed among the three sp² hybrid orbitals
    ♦ So each hybrid orbital will carry one electron
    ♦ This is indicated by the arrows in fig.4.145(a) above
5. Remember that, the 2p_z orbital also has one electron
• So an arrow is shown in the 2p_z orbital in fig.b also
6. So we have a 'sp² hybridized C atom' in fig.4.145(b)
• We want a mirror image of that C atom
• To obtain a 'required mirror image', we need to place the mirror in the 'correct position'
• In our present case, the 'correct position' can be specified using 2 steps:
(i) The x-axis should penetrate through the mirror
(ii) The the x-axis should be perpendicular to the 'plane of the mirror'
• Such a 'correct position' is shown in fig.4.146(a) below
• The 'resulting mirror image' is also shown in the same fig.a

Fig.4.146

7. So now we have two C atoms
• Remember that, each one of them is sp² hybridized
• We want the two C atoms to bond together
• For that we use the following four steps:
(i) We have two orbitals lying along the x-axis
    ♦ One belongs to the 'original C atom'
    ♦ The other belongs to the 'mirror image C atom'
(ii) Move the mirror image towards the original
    ♦ The movement should be exactly along the x-axis
(iii) When an optimum distance is reached, overlapping takes place between the 'two orbitals mentioned in (i)'
    ♦ This is shown in fig.4.146(b) above
• 'overlapping' means that, there will be a particular region, which will belong to both the orbitals
• This region is called 'overlapping region'
(iv) The two electrons (one from each orbital) will then lie in that 'overlapping region'
• Thus the two electrons will belong to both the C atoms
• In this way, a bond is formed between those two C atoms
• This is shown in fig.4.147(a) below:

Fig.4.147

8. So in fig.4.147(a) above, we have two C atoms which are bonded together
• Now bonding with H atoms can begin:
    ♦ Two H atoms will come and overlap with the free orbitals of the original C atom
    ♦ Another two H atoms will come and overlap with the free orbitals of the 'C atom which is the mirror image'
    ♦ This is shown in fig.4.147(b)
9. The structure in fig.4.147(b), is a single unit. It consists of:
    ♦ Two C atoms
    ♦ Four H atoms
• But the molecule of C₂H₄ is not yet formed. This is because, each of the two p_z orbitals is still half occupied
10. Each of the two p_z orbitals in fig.4.147(b) above needs one more electron
• So they share electrons among themselves
• Let us see how this sharing is done. It can be written in 3 steps
(i) We know that, all orbitals are electron clouds
    ♦ The two p_z orbitals are also electron clouds
(ii) The two p_z clouds overlap in a 'side-wise manner'
• The result is a cloud with 'U' shapes
    ♦ A 'inverted U' above the x-axis
    ♦ A 'upright U' below the x-axis
• This is shown in fig.4.148 below
(iii) When the overlapping is complete, the resulting cloud will contain two electrons

Fig.4.148

11. So the shape of the C₂H₄ molecule will be as shown in fig.4.149 below

Fig.4.149

 12. Consider the plane in fig.4.150 below
    ♦ The two C atoms lie on that plane
    ♦ The four H atoms also lie on that plane
(The plane is given a bit of transparency so that, the bonding below the plane, also becomes visible)

Fig.4.150

 • So the 2D representation of C₂H₄ will be as shown in the fig.4.150(b)
    ♦ It will not have any solid triangles
    ♦ It will not have any dashed triangles
    ♦ It will have only solid lines
13. Next we will see an interesting point. It can be written in 4 steps:
(i) Consider the final model in fig.4.149 above
• Let us consider it as two separate units
    ♦ The original C atom and it's two H atoms constitute the first unit
    ♦ The mirror image C atom and it's two H atoms constitute the second unit
(ii) Keep the first unit 'fixed'
    ♦ Rotate the second unit
    ♦ Nucleus of the second unit is the pivot of rotation
    ♦ x-axis is the axis of rotation
• This is indicated by the yellow curved arrow in fig.4.151 below
(iii) We can do this type of rotation, only by causing damage to the structure
• This is shown in fig.4.151 below:

Fig.4.151

(iv) We see that, the 'U' shaped clouds will break
    ♦ If they break, the two electrons in the cloud can no longer be together
    ♦ The bond is lost
• That is why, such a rotation is not possible

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• The structure of C₂H₄ gives us an opportunity to learn about pi bond

• It can be explained in 9 steps:
1. The fig.4.152(a) below shows the final structure of C₂H₄

Fig.4.152

2. We see that, all bonds between C and H are sigma bonds
• In other words: All C-H bonds are sigma bonds
3. We see that the C-C bond which lies along the x-axis is also a sigma bond
4. But there is a second bond between the two C atoms
• This can be explained using fig.4.152(b)
• This fig.b shows the situation before the overlapping of p_z orbitals
• We can write about the 'second bond' in 4 steps
(i) Initially, the first C atom has a p_z orbital
(ii) Initially, the second C atom also has a p_z orbital  
(iii) The above two p_z orbitals overlap in a 'side-wise manner'
    ♦ After this overlap, the the two p_z orbitals become a single orbital
    ♦ This is because, after the overlap, the new orbital has two electrons
(iv) When the two p_z orbitals become one, 'their parent C atoms' will naturally unite
    ♦ Thus a second bond is formed between the two C atoms
5. Consider the two yellow lines in the fig.b
• They are the axes of the p_z original orbitals
    ♦ Those two axes are parallel to each other
    ♦ Those two axes are perpendicular to the internuclear axis (the red axis)
• Even after the 'side-wise overlapping' (shown in fig.a), 
    ♦ Those two axes remain parallel to each other
    ♦ Those two axes remain perpendicular to the internuclear axis (the red axis)
■ Such a bond is called pi bond
6. So we can write the definition:
• If the axes of the atomic orbitals
    ♦ remain parallel to each other,
    ♦ and also remain perpendicular to the internuclear axis,
it is called a pi bond.
    ♦ In symbolic form, it is written as: 𝛑 bond
7. In the Lewis structure of C₂H₄, we see a double bond between the two C atoms
(See fig.4.144(a) at the beginning of this section)
• A double bond is represented by putting a '=' between atoms
• So now we know that:
    ♦ the upper '-' in the '=' is a 𝛔 bond
    ♦ the lower '-' in the '=' is a 𝛑 bond
• We will see more examples of 𝛑 bond in later sections

Strength of sigma and pi bonds

This can be written in 4 steps:
1. The strength of a bond depends on the 'extent of overlapping' between the orbitals
    ♦ If there is greater overlapping, the strength will be high
    ♦ If there is lesser overlapping, the strength will be low
2. In 𝛔 bond, there is end-to-end overlapping
    ♦ So the overlapping can take place to a greater extent
3. In 𝛑 bond, the overlapping is side-wise
    ♦ So the overlapping can take place only to a lower extent
4. So the 𝛔 bond will be stronger than the 𝛑 bond

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• In the next section, we will see the structure of C₂H₂ (ethyne)

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