• In the previous section 4.36, we saw the details of the B2 molecule. In this section we will see the next molecule which is C2
The details related to C2 molecule can be written in 9 steps:
1. The electronic configuration of C is 1s22s22p2
• Since 2p orbitals are present, there will be a total of ten molecular orbitals in C2:
σ1s , σ*1s , σ2s , σ*2s , [𝞹2px, 𝞹2py], σ2pz , [𝞹*2px, 𝞹*2py] and σ*2pz
2. We want to arrange them in the increasing order of energies
• In the fig.4.211 of the previous section, we saw two types of increasing order
• For easy reference, that fig.4.211 is shown again below:
■ In the case of C2, 'type 2' is applicable
3. Now we can arrange the electrons in C2 into the various molecular orbitals
• The electronic configuration of C is 1s22s22p2
♦ So in an individual C atom, there are 6 electrons
• Then where will the 12 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 (1) above
(ii) The new molecular orbitals will be filled up according to the 'increasing order type 2' written in (2). So we can write:
• Out of the twelve 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 next two will go to [𝞹2px, 𝞹2py]
♦ the last two will also go to [𝞹2px, 𝞹2py]
✰ (Recall 'Hund's rule of maximum multiplicity' that we applied in the case of B2)
■ Thus we get the diagram shown in fig.4.215 below:
• Details about this fig.4.215 can be written in 3 steps:
(i) The electronic configuration of C is 1s22s22p2
• We see that, the 1s orbital has 2 electrons
• So when two C 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 B2
♦ Recall that, the electronic configuration of B2 is:
✰ KK(σ2s)2(σ*2s)2(𝞹2px)1(𝞹2py)1
♦ The ‘KK’ indicates that, the first main-shell (the K main-shell) is completely used up
(ii) Again consider the electronic configuration of C:1s22s22p2
• We see that, the 2s orbital has two electrons
• So when two C atoms combine, the 2s orbitals will contribute a total of four electrons
♦ In the fig.4.215 above,
✰ The two arrows in the left 2s box indicates those two electrons of the first C atom
✰ The two arrows in the right 2s box indicates those two electrons of the other C 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 C:1s22s22p2
• We see that, the 2p orbital has two electrons
• So when two B atoms combine, the 2p orbitals will contribute a total of four electrons
♦ In the fig.4.215 above,
✰ The one arrow in the left 2px box indicates the one electron of the first C atom
✰ The one arrow in the left 2py box indicates the second electron of the first C atom
✰ The one arrow in the right 2px box indicates the one electron of the other C atom
✰ The one arrow in the right 2py box indicates the second electron of the other C atom
♦ Those total four electrons are filled up into the 𝞹2px and 𝞹2py
4. Based on fig.4.215 above, we can determine five items:
(i) Bond order in C2 molecule
(ii) Stability of C2 molecule
(iii) Nature of bond in C2 molecule
(iv) Electronic configuration of C2 molecule
(v) Magnetic nature of C2 molecule
• The following steps from (5) to (9) shows the calculations related to the five items:
5. First we determine the bond order (b.o). It can be written in 2 steps:
(i) In C2, we have:
Nb = 6 and Na = 2
• Note:
♦ Nb is equal to six because:
✰ The bonding orbital σ2s contains two electrons
✰ The bonding orbital 𝞹2px contains two electron
✰ The bonding orbital 𝞹2py contains two electron
♦ Na is equal to two because:
✰ The anti-bonding orbital σ*2s contains two electrons
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{6-2}{2}=2}}$
6. Next we determine the stability
• From (5) above, we have:
b.o of C2 = 2
• This is greater than zero. So the molecule is stable
7. Next we determine the nature of bond in C2 molecule
• From (5) above, we have:
b.o of C2 = 2
• So there will be a double bond between the two C atoms
8. Next we determine the electronic configuration of C2. This can be written in 2 steps:
(i) In C2, we have:
♦ two electrons present in σ1s
♦ two electrons present in σ*1s
♦ two electrons present in σ2s
♦ two electrons present in σ*2s
♦ two electrons present in 𝞹2px
♦ two electrons present in 𝞹2py
(ii) So the electronic configuration is (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(𝞹2px)2(𝞹2py)2
♦ This is same as: KK(σ2s)2(σ*2s)2(𝞹2px)2(𝞹2py)2
♦ Again, the two 2p orbitals can be combined. We get: KK(σ2s)2(σ*2s)2(𝞹2p)4
9. Next we determine the magnetic nature of C2
• In fig.4.215, we see that no boxes contain unpaired electrons
✰ So C2 is diamagnetic
1. The electronic configuration of N is 1s22s22p3
Since 2p orbitals are present, there will be a total of ten molecular orbitals in N2:
σ1s , σ*1s , σ2s , σ*2s , [𝞹2px, 𝞹2py], σ2pz , [𝞹*2px, 𝞹*2py] and σ*2pz
2. We want to arrange them in the increasing order of energies
• In the fig.4.211 of the previous section, we saw two types of increasing order
• For easy reference, that fig.4.211 is shown again below:
■ In the case of N2, 'type 2' is applicable
3. Now we can arrange the electrons in N2 into the various molecular orbitals
• The electronic configuration of N is 1s22s22p3
♦ So in an individual N atom, there are 7 electrons
• Then where will the 14 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 (1) above
(ii) The new molecular orbitals will be filled up according to the 'increasing order type 2' written in (2). So we can write:
• Out of the fourteen 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 next four will go to [𝞹2px, 𝞹2py]
♦ the last two will go to σ2pz
■ Thus we get the diagram shown in fig.4.216 below:
• Details about this fig.4.216 can be written in 3 steps:
(i) The electronic configuration of N is 1s22s22p3
• 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 previous few molecules
♦ Recall that, the electronic configuration of C2 is:
✰ KK(σ2s)2(σ*2s)2(𝞹2px)2(𝞹2py)2
♦ The ‘KK’ indicates that, the first main-shell (the K main-shell) is completely used up
(ii) Again consider the electronic configuration of N:1s22s22p3
• We see that, the 2s orbital has two electrons
• So when two B atoms combine, the 2s orbitals will contribute a total of four electrons
♦ In the fig.4.215 above,
✰ The two arrows in the left 2s box indicates those two electrons of the first N atom
✰ The two arrows in the right 2s box indicates those two electrons of the other N 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 C:1s22s22p3
• We see that, the 2p orbital has three electrons
• So when two B atoms combine, the 2p orbitals will contribute a total of six electrons
♦ In the fig.4.216 above,
✰ The one arrow in the left 2px box indicates the one electron of the first N atom
✰ The one arrow in the left 2py box indicates the second electron of the first N atom
✰ The one arrow in the left 2pz box indicates the third electron of the first N atom
✰ The one arrow in the right 2px box indicates the one electron of the other N atom
✰ The one arrow in the right 2py box indicates the second electron of the other N atom
✰ The one arrow in the right 2py box indicates the third electron of the other N atom
♦ Out of the above six,
✰ The first four electrons are filled up into the 𝞹2px and 𝞹2py
✰ The last two are filled up into the σ2pz
4. Based on fig.4.216 above, we can determine five items:
(i) Bond order in N2 molecule
(ii) Stability of N2 molecule
(iii) Nature of bond in N2 molecule
(iv) Electronic configuration of N2 molecule
(v) Magnetic nature of N2 molecule
• The following steps from (5) to (9) shows the calculations related to the five items:
5. First we determine the bond order (b.o). It can be written in 2 steps:
(i) In N2, we have:
Nb = 8 and Na = 2
• Note:
♦ Nb is equal to four because:
✰ The bonding orbital σ2s contains two electrons
✰ The bonding orbital 𝞹2px contains two electron
✰ The bonding orbital 𝞹2py contains two electron
♦ Na is equal to two because:
✰ The anti-bonding orbital σ*2s contains two electrons
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{8-2}{2}=3}}$
6. Next we determine the stability
• From (5) above, we have:
b.o of N2 = 3
• This is greater than zero. So the molecule is stable
7. Next we determine the nature of bond in N2 molecule
• From (5) above, we have:
b.o of N2 = 3
• So there will be a triple bond between the two N atoms
8. Next we determine the electronic configuration of N2. This can be written in 2 steps:
(i) In N2, we have:
♦ two electrons present in σ1s
♦ two electrons present in σ*1s
♦ two electrons present in σ2s
♦ two electrons present in σ*2s
♦ two electrons present in 𝞹2px
♦ two electrons present in 𝞹2py
♦ two electrons present in σ2pz
(ii) So the electronic configuration is (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(𝞹2px)2(𝞹2py)2(σ2pz)2
♦ This is same as: KK(σ2s)2(σ*2s)2(𝞹2px)2(𝞹2py)2(σ2pz)2
♦ Again, the two 𝞹2p orbitals can be combined. We get: KK(σ2s)2(σ*2s)2(𝞹2p)4(σ2pz)2
9. Next we determine the magnetic nature of N2
• In fig.4.215, we see that no boxes contain unpaired electrons
✰ So N2 is diamagnetic
The details related to C2 molecule can be written in 9 steps:
1. The electronic configuration of C is 1s22s22p2
• Since 2p orbitals are present, there will be a total of ten molecular orbitals in C2:
σ1s , σ*1s , σ2s , σ*2s , [𝞹2px, 𝞹2py], σ2pz , [𝞹*2px, 𝞹*2py] and σ*2pz
2. We want to arrange them in the increasing order of energies
• In the fig.4.211 of the previous section, we saw two types of increasing order
• For easy reference, that fig.4.211 is shown again below:
 |
| Fig.4.211 |
3. Now we can arrange the electrons in C2 into the various molecular orbitals
• The electronic configuration of C is 1s22s22p2
♦ So in an individual C atom, there are 6 electrons
• So when two individual C atoms combine to form a C2 molecule, there will be a total of 12 electrons
• In a C2 molecule, there are no 1s, 2s or 2p orbitals
• In a C2 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 (1) above
(ii) The new molecular orbitals will be filled up according to the 'increasing order type 2' written in (2). So we can write:
• Out of the twelve 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 next two will go to [𝞹2px, 𝞹2py]
♦ the last two will also go to [𝞹2px, 𝞹2py]
✰ (Recall 'Hund's rule of maximum multiplicity' that we applied in the case of B2)
■ Thus we get the diagram shown in fig.4.215 below:
 |
| Fig.4.215 |
(i) The electronic configuration of C is 1s22s22p2
• We see that, the 1s orbital has 2 electrons
• So when two C 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 B2
♦ Recall that, the electronic configuration of B2 is:
✰ KK(σ2s)2(σ*2s)2(𝞹2px)1(𝞹2py)1
♦ The ‘KK’ indicates that, the first main-shell (the K main-shell) is completely used up
(ii) Again consider the electronic configuration of C:1s22s22p2
• We see that, the 2s orbital has two electrons
• So when two C atoms combine, the 2s orbitals will contribute a total of four electrons
♦ In the fig.4.215 above,
✰ The two arrows in the left 2s box indicates those two electrons of the first C atom
✰ The two arrows in the right 2s box indicates those two electrons of the other C 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 C:1s22s22p2
• We see that, the 2p orbital has two electrons
• So when two B atoms combine, the 2p orbitals will contribute a total of four electrons
♦ In the fig.4.215 above,
✰ The one arrow in the left 2px box indicates the one electron of the first C atom
✰ The one arrow in the left 2py box indicates the second electron of the first C atom
✰ The one arrow in the right 2px box indicates the one electron of the other C atom
✰ The one arrow in the right 2py box indicates the second electron of the other C atom
♦ Those total four electrons are filled up into the 𝞹2px and 𝞹2py
4. Based on fig.4.215 above, we can determine five items:
(i) Bond order in C2 molecule
(ii) Stability of C2 molecule
(iii) Nature of bond in C2 molecule
(iv) Electronic configuration of C2 molecule
(v) Magnetic nature of C2 molecule
• The following steps from (5) to (9) shows the calculations related to the five items:
5. First we determine the bond order (b.o). It can be written in 2 steps:
(i) In C2, we have:
Nb = 6 and Na = 2
• Note:
♦ Nb is equal to six because:
✰ The bonding orbital σ2s contains two electrons
✰ The bonding orbital 𝞹2px contains two electron
✰ The bonding orbital 𝞹2py contains two electron
♦ Na is equal to two because:
✰ The anti-bonding orbital σ*2s contains two electrons
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{6-2}{2}=2}}$
6. Next we determine the stability
• From (5) above, we have:
b.o of C2 = 2
• This is greater than zero. So the molecule is stable
7. Next we determine the nature of bond in C2 molecule
• From (5) above, we have:
b.o of C2 = 2
• So there will be a double bond between the two C atoms
8. Next we determine the electronic configuration of C2. This can be written in 2 steps:
(i) In C2, we have:
♦ two electrons present in σ1s
♦ two electrons present in σ*1s
♦ two electrons present in σ2s
♦ two electrons present in σ*2s
♦ two electrons present in 𝞹2px
♦ two electrons present in 𝞹2py
(ii) So the electronic configuration is (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(𝞹2px)2(𝞹2py)2
♦ This is same as: KK(σ2s)2(σ*2s)2(𝞹2px)2(𝞹2py)2
♦ Again, the two 2p orbitals can be combined. We get: KK(σ2s)2(σ*2s)2(𝞹2p)4
9. Next we determine the magnetic nature of C2
• In fig.4.215, we see that no boxes contain unpaired electrons
✰ So C2 is diamagnetic
The details related to N2 molecule can be written in 9 steps:
Since 2p orbitals are present, there will be a total of ten molecular orbitals in N2:
σ1s , σ*1s , σ2s , σ*2s , [𝞹2px, 𝞹2py], σ2pz , [𝞹*2px, 𝞹*2py] and σ*2pz
2. We want to arrange them in the increasing order of energies
• In the fig.4.211 of the previous section, we saw two types of increasing order
• For easy reference, that fig.4.211 is shown again below:
|
| Fig.4.211 |
3. Now we can arrange the electrons in N2 into the various molecular orbitals
• The electronic configuration of N is 1s22s22p3
♦ So in an individual N atom, there are 7 electrons
• So when two individual N atoms combine to form a N2 molecule, there will be a total of 14 electrons
• In a N2 molecule, there are no 1s, 2s or 2p orbitals
• In a N2 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 (1) above
(ii) The new molecular orbitals will be filled up according to the 'increasing order type 2' written in (2). So we can write:
• Out of the fourteen 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 next four will go to [𝞹2px, 𝞹2py]
♦ the last two will go to σ2pz
■ Thus we get the diagram shown in fig.4.216 below:
 |
| Fig.4.216 |
(i) The electronic configuration of N is 1s22s22p3
• 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 previous few molecules
♦ Recall that, the electronic configuration of C2 is:
✰ KK(σ2s)2(σ*2s)2(𝞹2px)2(𝞹2py)2
♦ The ‘KK’ indicates that, the first main-shell (the K main-shell) is completely used up
(ii) Again consider the electronic configuration of N:1s22s22p3
• We see that, the 2s orbital has two electrons
• So when two B atoms combine, the 2s orbitals will contribute a total of four electrons
♦ In the fig.4.215 above,
✰ The two arrows in the left 2s box indicates those two electrons of the first N atom
✰ The two arrows in the right 2s box indicates those two electrons of the other N 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 C:1s22s22p3
• We see that, the 2p orbital has three electrons
• So when two B atoms combine, the 2p orbitals will contribute a total of six electrons
♦ In the fig.4.216 above,
✰ The one arrow in the left 2px box indicates the one electron of the first N atom
✰ The one arrow in the left 2py box indicates the second electron of the first N atom
✰ The one arrow in the left 2pz box indicates the third electron of the first N atom
✰ The one arrow in the right 2px box indicates the one electron of the other N atom
✰ The one arrow in the right 2py box indicates the second electron of the other N atom
✰ The one arrow in the right 2py box indicates the third electron of the other N atom
♦ Out of the above six,
✰ The first four electrons are filled up into the 𝞹2px and 𝞹2py
✰ The last two are filled up into the σ2pz
4. Based on fig.4.216 above, we can determine five items:
(i) Bond order in N2 molecule
(ii) Stability of N2 molecule
(iii) Nature of bond in N2 molecule
(iv) Electronic configuration of N2 molecule
(v) Magnetic nature of N2 molecule
• The following steps from (5) to (9) shows the calculations related to the five items:
5. First we determine the bond order (b.o). It can be written in 2 steps:
(i) In N2, we have:
Nb = 8 and Na = 2
• Note:
♦ Nb is equal to four because:
✰ The bonding orbital σ2s contains two electrons
✰ The bonding orbital 𝞹2px contains two electron
✰ The bonding orbital 𝞹2py contains two electron
♦ Na is equal to two because:
✰ The anti-bonding orbital σ*2s contains two electrons
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{8-2}{2}=3}}$
6. Next we determine the stability
• From (5) above, we have:
b.o of N2 = 3
• This is greater than zero. So the molecule is stable
7. Next we determine the nature of bond in N2 molecule
• From (5) above, we have:
b.o of N2 = 3
• So there will be a triple bond between the two N atoms
8. Next we determine the electronic configuration of N2. This can be written in 2 steps:
(i) In N2, we have:
♦ two electrons present in σ1s
♦ two electrons present in σ*1s
♦ two electrons present in σ2s
♦ two electrons present in σ*2s
♦ two electrons present in 𝞹2px
♦ two electrons present in 𝞹2py
♦ two electrons present in σ2pz
(ii) So the electronic configuration is (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(𝞹2px)2(𝞹2py)2(σ2pz)2
♦ This is same as: KK(σ2s)2(σ*2s)2(𝞹2px)2(𝞹2py)2(σ2pz)2
♦ Again, the two 𝞹2p orbitals can be combined. We get: KK(σ2s)2(σ*2s)2(𝞹2p)4(σ2pz)2
9. Next we determine the magnetic nature of N2
• In fig.4.215, we see that no boxes contain unpaired electrons
✰ So N2 is diamagnetic
• In the next section, we will see the details related to O2 molecule
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