• In the previous section 4.35, we were discussing the formation of the B2 molecule. We completed 5 steps
• We saw these:
♦ Merger between two 2pz orbitals resulting in σ2pz and σ*2pz
♦ Merger between two 2px orbitals resulting in 𝞹2px and 𝞹*2px
• In this section we will continue from step 7
7. In this step we will see the merger of 2py orbitals
• This step is very similar to the previous step 6
♦ Some minor differences are present only in the 'directions' and 'notations'
♦ However, we will write it again:
A. Bonding orbital
• This can be explained in 7 steps
(i) Assume that, px and pz are removed from the earlier fig.4.195
♦ Then we will be able to understand the 'merger of py' more easily
• The constructive addition is shown in fig.4.206 below:
(ii) The resulting molecular orbital has two 'U' shaped clouds:
♦ One on top side and the other on the bottom side of the z-axis
(iii) We have to give it a suitable name
♦ We see that, the two 2p orbitals add 'in a side-wise manner'
♦ So the resulting molecular orbital is denoted by the symbol '𝞹'
♦ Recall the ‘formation of 𝞹 bonds’ in the topic of hybridization (See fig.4.148 of section 4.26)
(iv) In our present case, the molecular orbital is formed by the merger of two 2py orbitals
♦ So we call the resulting molecular orbital as: 𝞹2py
(v) The above fig.4.201 is a 3D view. Such a view helps us to understand the actual shapes
• But in the present case, a 2D view will give us some extra details. It is shown in fig.4.207 below:
(vi) In fig.4.207, consider the shape of 𝞹2py
♦ Consider the region between the two nuclei
✰ This region is thicker
(vii) That means, the electrons will be spending more time in the middle portion between the two nuclei
♦ So both the nuclei has a strong influence on the electrons
♦ This gives rise to a strong bond between the two nuclei
(But it will not be as strong as a 'end-to-end' bond. 'Side-wise' bonding will be weaker than 'end-to-end' bonding)
• Since the addition is constructive, 𝞹2py is a bonding orbital
B. Anti-bonding orbital
• This can be explained in 8 steps
(i) The destructive addition resulting in the formation of anti-bonding orbital is shown in fig.4.208 below:
(ii) The resulting molecular orbital has two separate parts
• Each part has two equal lobes, one above and the other below the z-axis
♦ We have to give it a suitable name
♦ We see that, the two 2py orbitals add 'in a side-wise manner'
♦ So the resulting molecular orbital is denoted by the '𝞹' symbol
♦ Recall the ‘formation of 𝞹 bonds’ in the topic of hybridization (See fig.4.148 of section 4.26)
(iii) In or present case, the molecular orbital is formed by the merger of two 2py orbitals
♦ But the resulting molecular orbital is an anti-bonding orbital
♦ We have to distinguish it from the bonding orbital. For that, we give an extra '*' symbol
♦ So we call the resulting anti-bonding molecular orbital as: 𝞹*2py
(iv) The above fig.4.208 is a 3D view. Such a view helps us to understand the actual shapes
• But in the present case, a 2D view will give us some extra details. It is shown in fig.4.209 below:
(v) In fig.4.209, consider the shape of σ*2py
♦ Consider the region between the two nuclei
✰ This region is has very low density of electrons
(vi) Again in fig.4.209, consider the shape of σ*2py
♦ Consider the regions on the left and right sides of the two nuclei
✰ Those regions are thicker
✰ Those thicker regions are symmetrically distributed on either sides of the two nuclei
(In the 3D view, we will not recognize the fact that, the thicker portions are symmetrically distributed)
(vii) That means, the electrons will be spending more time away from the two nuclei
• It is clear that, in the σ*2py,
♦ The left side nucleus has no influence on the right side electron
♦ The right side nucleus has no effect on the left side electron
• The two 2py orbitals were independent before bond formation
• Now, even after the formation of the anti-bond, they are trying to separate away from each other
• This is because of the repulsion between the two nuclei
(viii) In fact, there is a definite plane between the two nuclei in the σ*2py
• This plane can be clearly marked in the 3D view. This is shown in the fig.4.210 below:
• In this plane, we will never find any electrons. It is a nodal plane
8. So we have seen all the six molecular orbitals that result from the merger of p-orbitals
• Let us write a summary. It can be written in 3 steps:
(i) Two 2pz orbitals merge to give σ2pz and σ*2pz
♦ σ2pz is the bonding orbital
✰ It is the result of constructive addition
♦ σ*2pz is the anti-bonding orbital
✰ It is the result of destructive addition
■ This is a 'end-to-end' merger
♦ (Figs.4.196, 197, 198, 199 and 200 of the previous section)
(ii) Two 2px orbitals merge to give 𝞹2px and 𝞹*2px
♦ 𝞹2px is the bonding orbital
✰ It is the result of constructive addition
♦ 𝞹*2px is the anti-bonding orbital
✰ It is the result of destructive addition
■ This is a 'side-wise' merger
♦ (Figs.4.201, 202, 203, 204 and 205 of the previous section)
(iii) Two 2py orbitals merge to give 𝞹2py and 𝞹*2py
• 𝞹2py is the bonding orbital
✰ It is the result of constructive addition
• 𝞹*2py is the anti-bonding orbital
✰ It is the result of destructive addition
■ This is a 'side-wise' merger
♦ (Figs.4.206, 207, 208, 209 and 210)
9. So now we know the various molecular orbitals present in B2
• The electrons will enter those orbitals in the increasing order of energies
♦ So first we need to arrange the orbitals in the increasing order of energies
• There are two arrangements possible. They are shown in fig.4.211 below:
■ There is only a minor difference between the two types. This can be explained in 4 steps:
(i) Consider type 1
• Let us separate the various orbitals into groups. This is shown in fig.4.212 below:
(ii) The groups are separated by vertical red lines
♦ The first group consists of both bonding and anti-bonding orbitals related to 1s
♦ The second group consists of both bonding and anti-bonding orbitals related to 2s
♦ The third group consists of bonding orbitals related to 2p
♦ The fourth group consists of anti-bonding orbitals related to 2p
(iii) The only change occurs in the third group
♦ The sigma orbital takes the position of the pi orbitals
✰ This is indicated by the upper curved arrow
♦ The pi orbitals takes the position of the sigma orbital
✰ This is indicated by the lower curved arrow
(iv) When this interchanging takes place, we get the type 2
10. The 'molecular orbitals related to the atomic p-orbitals' can be graphically represented as shown in fig.4.213 below:
• The details can be written in 2 steps:
(i) Consider type 1
♦ The blue sloping lines indicate that:
✰ 2px and 2px undergo constructive addition to give 𝞹2px
✰ 2py and 2py undergo constructive addition to give 𝞹2py
♦ The green sloping lines indicate that:
✰ 2pz and 2pz undergo constructive addition to give σ2pz
♦ The white sloping lines indicate that:
✰ 2px and 2px undergo destructive addition to give 𝞹*2px
✰ 2py and 2py undergo destructive addition to give 𝞹*2py
♦ The magenta sloping lines indicate that:
✰ 2pz and 2pz undergo destructive addition to give σ*2pz
(ii) Consider type 2
• The four respective colors indicate the same type of addition and resulting orbitals as in type 1
♦ The only difference is this:
✰ σ2pz takes the position of [𝞹2px, 𝞹2py]
✰ [𝞹2px, 𝞹2py] take the position of σ2pz
• This 'interchanging of positions' is what we saw by the curved arrows in the previous fig.4.212
11. Now we can arrange the electrons in B2 into the various molecular orbitals
• In B2, the 'increasing order of type 2' is applicable
(We will see the reason in higher classes)
• Altogether in B2, there are ten molecular orbitals:
σ1s , σ*1s , σ2s , σ*2s , [𝞹2px, 𝞹2py], σ2pz , [𝞹*2px, 𝞹*2py] and σ*2pz
12. In an individual B atom, there are 5 electrons
• Then where will the 10 electrons reside?
• The answer can be written in 2 steps:
(i) Obviously, they have no where else to go but the ten molecular orbitals mentioned in (11) above
(ii) The new molecular orbitals will be filled up according to the 'increasing order type 2' written in (9). So we can write:
• Out of the ten electrons,
♦ the first two will go to σ1s
♦ the next two will go to σ*1s
♦ the next two will go to σ2s
♦ the next two will go to σ*2s
♦ the last two will go to [𝞹2px, 𝞹2py]
13. Consider the last two electrons. They go to [𝞹2px, 𝞹2py]
• There are two boxes available: 𝞹2px and 𝞹2py
• We are inclined to put both the electrons in any one of those boxes
• But we must be careful at this stage
♦ The two orbitals are degenerate orbitals. They both have the same energy
♦ So we must apply Hund's rule of maximum multiplicity
♦ Based on that rule, we can put only one electron in one box
• So we can write:
♦ 𝞹2px will carry an unpaired electron
♦ 𝞹2py will carry an unpaired electron
■ Thus we get the arrangement shown in fig.4.214 below:
• Details about this fig.4.214 can be written in 3 steps:
(i) The electronic configuration of B is 1s22s22p1
• We see that, the 1s orbital has 2 electrons
• So when two B atoms combine, the 1s orbitals will contribute a total of four electrons
• Those four electrons will go to the σ1s and σ*1s orbitals
(ii) But we do not need to show the σ1s and σ*1s orbitals because, they are already filled up in the just previous molecule Be2
♦ Recall that, the electronic configuration of Be2 is:
✰ KK(σ2s)2(σ*2s)2
✰ which is same as
✰ (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2.
♦ The ‘KK’ indicates that, the first main-shell (the K main-shell) is completely used up
(ii) Again consider the electronic configuration of B:1s22s22p1
• We see that, the 2s orbital has two electrons
♦ In the fig.4.214 above,
✰ The two arrows in the left 2s box indicates those two electrons of the first B atom
✰ The two arrows in the right 2s box indicates those two electrons of the other B atom
♦ Those total four electrons are filled up into the σ2s and σ*2s
(iii) Next we take up the arrangement above the cyan dashed line
• Again consider the electronic configuration of B:1s22s22p1
• We see that, the 2p orbital has one electron
♦ In the fig.4.214 above,
✰ The one arrow in the left 2px box indicates the one electron of the first B atom
✰ The one arrow in the right 2px box indicates the one electron of the other B atom
♦ Those total two electrons are filled up into the 𝞹2px and 𝞹2py
14. Based on fig.4.214 above, we can determine five items:
(i) Bond order in B2 molecule
(ii) Stability of B2 molecule
(iii) Nature of bond in B2 molecule
(iv) Electronic configuration of B2 molecule
(v) Magnetic nature of B2 molecule
• The following steps from (15) to (19) shows the calculations related to the five items:
15. First we determine the bond order (b.o). It can be written in 2 steps:
(i) In B2, we have:
Nb = 4 and Na = 2
• Note:
♦ Nb is equal to four because:
✰ The bonding orbital σ2s contains two electrons
✰ The bonding orbital 𝞹2px contains one electron
✰ The bonding orbital 𝞹2py contains one electron
♦ Na is equal to two because:
✰ The anti-bonding orbital σ*2s contains two electrons
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{4-2}{2}=1}}$
16. Next we determine the stability
• From (15) above, we have:
b.o of B2 = 1
• This is greater than zero. So the molecule is stable
17. Next we determine the nature of bond in B2 molecule
• From (15) above, we have:
b.o of B2 = 1
• So there will be a single bond between the two B atoms
18. Next we determine the electronic configuration of B2. This can be written in 2 steps:
(i) In B2, we have:
♦ two electrons present in σ1s
♦ two electrons present in σ*1s
♦ two electrons present in σ2s
♦ two electrons present in σ*2s
♦ one electron present in 𝞹2px
♦ one electron present in 𝞹2py
(ii) So the electronic configuration is (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(𝞹2px)1(𝞹2py)1
♦ This is same as: KK(σ2s)2(σ*2s)2(𝞹2px)1(𝞹2py)1
♦ Again, the two 2p orbitals can be combined. We get: KK(σ2s)2(σ*2s)2(𝞹2p)2
19. Next we determine the magnetic nature of B2
• In fig.4.214, we see that two boxes contain unpaired electrons
♦ Even if one unpaired box is present, the molecule will be paramagnetic
✰ So B2 is paramagnetic
• We saw these:
♦ Merger between two 2pz orbitals resulting in σ2pz and σ*2pz
♦ Merger between two 2px orbitals resulting in 𝞹2px and 𝞹*2px
• In this section we will continue from step 7
7. In this step we will see the merger of 2py orbitals
• This step is very similar to the previous step 6
♦ Some minor differences are present only in the 'directions' and 'notations'
♦ However, we will write it again:
A. Bonding orbital
• This can be explained in 7 steps
(i) Assume that, px and pz are removed from the earlier fig.4.195
♦ Then we will be able to understand the 'merger of py' more easily
• The constructive addition is shown in fig.4.206 below:
 |
| Fig.4.206 |
♦ One on top side and the other on the bottom side of the z-axis
(iii) We have to give it a suitable name
♦ We see that, the two 2p orbitals add 'in a side-wise manner'
♦ So the resulting molecular orbital is denoted by the symbol '𝞹'
♦ Recall the ‘formation of 𝞹 bonds’ in the topic of hybridization (See fig.4.148 of section 4.26)
(iv) In our present case, the molecular orbital is formed by the merger of two 2py orbitals
♦ So we call the resulting molecular orbital as: 𝞹2py
(v) The above fig.4.201 is a 3D view. Such a view helps us to understand the actual shapes
• But in the present case, a 2D view will give us some extra details. It is shown in fig.4.207 below:
 |
| Fig.4.207 |
♦ Consider the region between the two nuclei
✰ This region is thicker
(vii) That means, the electrons will be spending more time in the middle portion between the two nuclei
♦ So both the nuclei has a strong influence on the electrons
♦ This gives rise to a strong bond between the two nuclei
(But it will not be as strong as a 'end-to-end' bond. 'Side-wise' bonding will be weaker than 'end-to-end' bonding)
• Since the addition is constructive, 𝞹2py is a bonding orbital
B. Anti-bonding orbital
• This can be explained in 8 steps
(i) The destructive addition resulting in the formation of anti-bonding orbital is shown in fig.4.208 below:
 |
| Fig.4.208 |
• Each part has two equal lobes, one above and the other below the z-axis
♦ We have to give it a suitable name
♦ We see that, the two 2py orbitals add 'in a side-wise manner'
♦ So the resulting molecular orbital is denoted by the '𝞹' symbol
♦ Recall the ‘formation of 𝞹 bonds’ in the topic of hybridization (See fig.4.148 of section 4.26)
(iii) In or present case, the molecular orbital is formed by the merger of two 2py orbitals
♦ But the resulting molecular orbital is an anti-bonding orbital
♦ We have to distinguish it from the bonding orbital. For that, we give an extra '*' symbol
♦ So we call the resulting anti-bonding molecular orbital as: 𝞹*2py
(iv) The above fig.4.208 is a 3D view. Such a view helps us to understand the actual shapes
• But in the present case, a 2D view will give us some extra details. It is shown in fig.4.209 below:
 |
| Fig.4.209 |
♦ Consider the region between the two nuclei
✰ This region is has very low density of electrons
(vi) Again in fig.4.209, consider the shape of σ*2py
♦ Consider the regions on the left and right sides of the two nuclei
✰ Those regions are thicker
✰ Those thicker regions are symmetrically distributed on either sides of the two nuclei
(In the 3D view, we will not recognize the fact that, the thicker portions are symmetrically distributed)
(vii) That means, the electrons will be spending more time away from the two nuclei
• It is clear that, in the σ*2py,
♦ The left side nucleus has no influence on the right side electron
♦ The right side nucleus has no effect on the left side electron
• The two 2py orbitals were independent before bond formation
• Now, even after the formation of the anti-bond, they are trying to separate away from each other
• This is because of the repulsion between the two nuclei
(viii) In fact, there is a definite plane between the two nuclei in the σ*2py
• This plane can be clearly marked in the 3D view. This is shown in the fig.4.210 below:
 |
| Fig.4.210 |
8. So we have seen all the six molecular orbitals that result from the merger of p-orbitals
• Let us write a summary. It can be written in 3 steps:
(i) Two 2pz orbitals merge to give σ2pz and σ*2pz
♦ σ2pz is the bonding orbital
✰ It is the result of constructive addition
♦ σ*2pz is the anti-bonding orbital
✰ It is the result of destructive addition
■ This is a 'end-to-end' merger
♦ (Figs.4.196, 197, 198, 199 and 200 of the previous section)
(ii) Two 2px orbitals merge to give 𝞹2px and 𝞹*2px
♦ 𝞹2px is the bonding orbital
✰ It is the result of constructive addition
♦ 𝞹*2px is the anti-bonding orbital
✰ It is the result of destructive addition
■ This is a 'side-wise' merger
♦ (Figs.4.201, 202, 203, 204 and 205 of the previous section)
(iii) Two 2py orbitals merge to give 𝞹2py and 𝞹*2py
• 𝞹2py is the bonding orbital
✰ It is the result of constructive addition
• 𝞹*2py is the anti-bonding orbital
✰ It is the result of destructive addition
■ This is a 'side-wise' merger
♦ (Figs.4.206, 207, 208, 209 and 210)
9. So now we know the various molecular orbitals present in B2
• The electrons will enter those orbitals in the increasing order of energies
♦ So first we need to arrange the orbitals in the increasing order of energies
• There are two arrangements possible. They are shown in fig.4.211 below:
 |
| Fig.4.211 |
(i) Consider type 1
• Let us separate the various orbitals into groups. This is shown in fig.4.212 below:
 |
| Fig.4.212 |
♦ The first group consists of both bonding and anti-bonding orbitals related to 1s
♦ The second group consists of both bonding and anti-bonding orbitals related to 2s
♦ The third group consists of bonding orbitals related to 2p
♦ The fourth group consists of anti-bonding orbitals related to 2p
(iii) The only change occurs in the third group
♦ The sigma orbital takes the position of the pi orbitals
✰ This is indicated by the upper curved arrow
♦ The pi orbitals takes the position of the sigma orbital
✰ This is indicated by the lower curved arrow
(iv) When this interchanging takes place, we get the type 2
10. The 'molecular orbitals related to the atomic p-orbitals' can be graphically represented as shown in fig.4.213 below:
 |
| Fig.4.213 |
(i) Consider type 1
♦ The blue sloping lines indicate that:
✰ 2px and 2px undergo constructive addition to give 𝞹2px
✰ 2py and 2py undergo constructive addition to give 𝞹2py
♦ The green sloping lines indicate that:
✰ 2pz and 2pz undergo constructive addition to give σ2pz
♦ The white sloping lines indicate that:
✰ 2px and 2px undergo destructive addition to give 𝞹*2px
✰ 2py and 2py undergo destructive addition to give 𝞹*2py
♦ The magenta sloping lines indicate that:
✰ 2pz and 2pz undergo destructive addition to give σ*2pz
(ii) Consider type 2
• The four respective colors indicate the same type of addition and resulting orbitals as in type 1
♦ The only difference is this:
✰ σ2pz takes the position of [𝞹2px, 𝞹2py]
✰ [𝞹2px, 𝞹2py] take the position of σ2pz
• This 'interchanging of positions' is what we saw by the curved arrows in the previous fig.4.212
11. Now we can arrange the electrons in B2 into the various molecular orbitals
• In B2, the 'increasing order of type 2' is applicable
(We will see the reason in higher classes)
• Altogether in B2, there are ten molecular orbitals:
σ1s , σ*1s , σ2s , σ*2s , [𝞹2px, 𝞹2py], σ2pz , [𝞹*2px, 𝞹*2py] and σ*2pz
12. In an individual B atom, there are 5 electrons
• So when two individual B atoms combine to form a B2 molecule, there will be a total of 10 electrons
• In a B2 molecule, there are no 1s, 2s or 2p orbitals
• In a B2 molecule, there are no 1s, 2s or 2p orbitals
• The answer can be written in 2 steps:
(i) Obviously, they have no where else to go but the ten molecular orbitals mentioned in (11) above
(ii) The new molecular orbitals will be filled up according to the 'increasing order type 2' written in (9). So we can write:
• Out of the ten electrons,
♦ the first two will go to σ1s
♦ the next two will go to σ*1s
♦ the next two will go to σ2s
♦ the next two will go to σ*2s
♦ the last two will go to [𝞹2px, 𝞹2py]
13. Consider the last two electrons. They go to [𝞹2px, 𝞹2py]
• There are two boxes available: 𝞹2px and 𝞹2py
• We are inclined to put both the electrons in any one of those boxes
• But we must be careful at this stage
♦ The two orbitals are degenerate orbitals. They both have the same energy
♦ So we must apply Hund's rule of maximum multiplicity
♦ Based on that rule, we can put only one electron in one box
• So we can write:
♦ 𝞹2px will carry an unpaired electron
♦ 𝞹2py will carry an unpaired electron
■ Thus we get the arrangement shown in fig.4.214 below:
 |
| Fig.4.214 |
(i) The electronic configuration of B is 1s22s22p1
• We see that, the 1s orbital has 2 electrons
• So when two B atoms combine, the 1s orbitals will contribute a total of four electrons
• Those four electrons will go to the σ1s and σ*1s orbitals
(ii) But we do not need to show the σ1s and σ*1s orbitals because, they are already filled up in the just previous molecule Be2
♦ Recall that, the electronic configuration of Be2 is:
✰ KK(σ2s)2(σ*2s)2
✰ which is same as
✰ (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2.
♦ The ‘KK’ indicates that, the first main-shell (the K main-shell) is completely used up
(ii) Again consider the electronic configuration of B:1s22s22p1
• We see that, the 2s orbital has two electrons
♦ In the fig.4.214 above,
✰ The two arrows in the left 2s box indicates those two electrons of the first B atom
✰ The two arrows in the right 2s box indicates those two electrons of the other B atom
♦ Those total four electrons are filled up into the σ2s and σ*2s
(iii) Next we take up the arrangement above the cyan dashed line
• Again consider the electronic configuration of B:1s22s22p1
• We see that, the 2p orbital has one electron
♦ In the fig.4.214 above,
✰ The one arrow in the left 2px box indicates the one electron of the first B atom
✰ The one arrow in the right 2px box indicates the one electron of the other B atom
♦ Those total two electrons are filled up into the 𝞹2px and 𝞹2py
14. Based on fig.4.214 above, we can determine five items:
(i) Bond order in B2 molecule
(ii) Stability of B2 molecule
(iii) Nature of bond in B2 molecule
(iv) Electronic configuration of B2 molecule
(v) Magnetic nature of B2 molecule
• The following steps from (15) to (19) shows the calculations related to the five items:
15. First we determine the bond order (b.o). It can be written in 2 steps:
(i) In B2, we have:
Nb = 4 and Na = 2
• Note:
♦ Nb is equal to four because:
✰ The bonding orbital σ2s contains two electrons
✰ The bonding orbital 𝞹2px contains one electron
✰ The bonding orbital 𝞹2py contains one electron
♦ Na is equal to two because:
✰ The anti-bonding orbital σ*2s contains two electrons
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{4-2}{2}=1}}$
16. Next we determine the stability
• From (15) above, we have:
b.o of B2 = 1
• This is greater than zero. So the molecule is stable
17. Next we determine the nature of bond in B2 molecule
• From (15) above, we have:
b.o of B2 = 1
• So there will be a single bond between the two B atoms
18. Next we determine the electronic configuration of B2. This can be written in 2 steps:
(i) In B2, we have:
♦ two electrons present in σ1s
♦ two electrons present in σ*1s
♦ two electrons present in σ2s
♦ two electrons present in σ*2s
♦ one electron present in 𝞹2px
♦ one electron present in 𝞹2py
(ii) So the electronic configuration is (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(𝞹2px)1(𝞹2py)1
♦ This is same as: KK(σ2s)2(σ*2s)2(𝞹2px)1(𝞹2py)1
♦ Again, the two 2p orbitals can be combined. We get: KK(σ2s)2(σ*2s)2(𝞹2p)2
19. Next we determine the magnetic nature of B2
• In fig.4.214, we see that two boxes contain unpaired electrons
♦ Even if one unpaired box is present, the molecule will be paramagnetic
✰ So B2 is paramagnetic
• In the next section, we will see the details related to C2 molecule
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