• In the previous section 4.37, we saw the details of the B2 and N2 molecules. In this section we will see the next molecule, which is O2
The details related to O2 molecule can be written in 9 steps:
1. The electronic configuration of O is 1s22s22p4
• 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 O2, 'type 1' is applicable
3. Now we can arrange the electrons in O2 into the various molecular orbitals
• The electronic configuration of O is 1s22s22p4
♦ So in an individual O atom, there are 8 electrons
• Then where will the 16 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 1' written in (2). So we can write:
• Out of the sixteen 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 σ2pz
♦ the next four will go to [𝞹2px, 𝞹2py]
♦ the last two will go to [𝞹*2px, 𝞹*2py]
✰ One will go to 𝞹*2px
✰ The other will go to 𝞹*2py
✰ (Recall 'Hund's rule of maximum multiplicity' that we applied in the case of B2)
■ Thus we get the diagram shown in fig.4.217 below:
• Details about this fig.4.217 can be written in 3 steps:
(i) The electronic configuration of O is 1s22s22p4
• We see that, the 1s orbital has 2 electrons
• So when two O 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 few previous molecules
♦ 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 O:1s22s22p4
• We see that, the 2s orbital has two electrons
• So when two O atoms combine, the 2s orbitals will contribute a total of four electrons
♦ In the fig.4.217 above,
✰ The two arrows in the left 2s box indicates those two electrons of the first O atom
✰ The two arrows in the right 2s box indicates those two electrons of the other O 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 O:1s22s22p4
• We see that, the 2p orbital has four electrons
• So when two O atoms combine, the 2p orbitals will contribute a total of eight electrons
♦ In the fig.4.217 above,
✰ The 2px, 2py and 2pz boxes on the left holds the four electrons of the first O atom
✰ The 2px, 2py and 2pz boxes on the right holds the four electrons of the other O atom
♦ Out of the above eight,
✰ The first two are filled up into the σ2pz
✰ The next four are filled up into the 𝞹2px and 𝞹2py
✰ The last two are filled up into the 𝞹*2px and 𝞹*2py
✰ (For the last two, Hund's rule is to be applied)
4. Based on fig.4.217 above, we can determine five items:
(i) Bond order in O2 molecule
(ii) Stability of O2 molecule
(iii) Nature of bond in O2 molecule
(iv) Electronic configuration of O2 molecule
(v) Magnetic nature of O2 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 O2, we have:
Nb = 8 and Na = 4
• Note:
♦ Nb is equal to eight because:
✰ The bonding orbital σ2s contains two electrons
✰ The bonding orbital σ2pz contains two electrons
✰ The bonding orbital 𝞹2px contains two electron
✰ The bonding orbital 𝞹2py contains two electron
♦ Na is equal to four because:
✰ The anti-bonding orbital σ*2s contains two electrons
✰ The anti-bonding orbital 𝞹*2px contains one electron
✰ The anti-bonding orbital 𝞹*2py contains one electron
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{8-4}{2}=2}}$
6. Next we determine the stability
• From (5) above, we have:
b.o of O2 = 2
• This is greater than zero. So the molecule is stable
7. Next we determine the nature of bond in O2 molecule
• From (5) above, we have:
b.o of O2 = 2
• So there will be a double bond between the two O atoms
8. Next we determine the electronic configuration of O2. This can be written in 2 steps:
(i) In O2, 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 σ2pz
♦ two electrons present in 𝞹2px
♦ two electrons present in 𝞹2py
♦ one electron present in 𝞹*2px
♦ one electron present in 𝞹*2py
(ii) So the electronic configuration is
(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(𝞹2px)2 (𝞹2py)2(𝞹*2px)1(𝞹*2py)1
♦ This is same as: KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2px)2 (𝞹2py)2(𝞹*2px)1(𝞹*2py)1
♦ Again, the two 𝞹2p orbitals can be combined. We get:
✰ KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2p)4 (𝞹*2px)1(𝞹*2py)1
♦ Again, the two 𝞹*2p orbitals can be combined. We get:
✰ KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2p)4 (𝞹*2p)2
9. Next we determine the magnetic nature of O2
• In fig.4.217, we see that two boxes contain unpaired electrons
✰ So O2 is paramagnetic
The details related to O2 molecule can be written in 9 steps:
1. The electronic configuration of O is 1s22s22p4
• 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 O2 into the various molecular orbitals
• The electronic configuration of O is 1s22s22p4
♦ So in an individual O atom, there are 8 electrons
• So when two individual O atoms combine to form a O2 molecule, there will be a total of 16 electrons
• In a O2 molecule, there are no 1s, 2s or 2p orbitals
• In a O2 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 1' written in (2). So we can write:
• Out of the sixteen 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 σ2pz
♦ the next four will go to [𝞹2px, 𝞹2py]
♦ the last two will go to [𝞹*2px, 𝞹*2py]
✰ One will go to 𝞹*2px
✰ The other will go to 𝞹*2py
✰ (Recall 'Hund's rule of maximum multiplicity' that we applied in the case of B2)
■ Thus we get the diagram shown in fig.4.217 below:
 |
| Fig.4.217 |
(i) The electronic configuration of O is 1s22s22p4
• We see that, the 1s orbital has 2 electrons
• So when two O 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 few previous molecules
♦ 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 O:1s22s22p4
• We see that, the 2s orbital has two electrons
• So when two O atoms combine, the 2s orbitals will contribute a total of four electrons
♦ In the fig.4.217 above,
✰ The two arrows in the left 2s box indicates those two electrons of the first O atom
✰ The two arrows in the right 2s box indicates those two electrons of the other O 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 O:1s22s22p4
• We see that, the 2p orbital has four electrons
• So when two O atoms combine, the 2p orbitals will contribute a total of eight electrons
♦ In the fig.4.217 above,
✰ The 2px, 2py and 2pz boxes on the left holds the four electrons of the first O atom
✰ The 2px, 2py and 2pz boxes on the right holds the four electrons of the other O atom
♦ Out of the above eight,
✰ The first two are filled up into the σ2pz
✰ The next four are filled up into the 𝞹2px and 𝞹2py
✰ The last two are filled up into the 𝞹*2px and 𝞹*2py
✰ (For the last two, Hund's rule is to be applied)
4. Based on fig.4.217 above, we can determine five items:
(i) Bond order in O2 molecule
(ii) Stability of O2 molecule
(iii) Nature of bond in O2 molecule
(iv) Electronic configuration of O2 molecule
(v) Magnetic nature of O2 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 O2, we have:
Nb = 8 and Na = 4
• Note:
♦ Nb is equal to eight because:
✰ The bonding orbital σ2s contains two electrons
✰ The bonding orbital σ2pz contains two electrons
✰ The bonding orbital 𝞹2px contains two electron
✰ The bonding orbital 𝞹2py contains two electron
♦ Na is equal to four because:
✰ The anti-bonding orbital σ*2s contains two electrons
✰ The anti-bonding orbital 𝞹*2px contains one electron
✰ The anti-bonding orbital 𝞹*2py contains one electron
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{8-4}{2}=2}}$
6. Next we determine the stability
• From (5) above, we have:
b.o of O2 = 2
• This is greater than zero. So the molecule is stable
7. Next we determine the nature of bond in O2 molecule
• From (5) above, we have:
b.o of O2 = 2
• So there will be a double bond between the two O atoms
8. Next we determine the electronic configuration of O2. This can be written in 2 steps:
(i) In O2, 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 σ2pz
♦ two electrons present in 𝞹2px
♦ two electrons present in 𝞹2py
♦ one electron present in 𝞹*2px
♦ one electron present in 𝞹*2py
(ii) So the electronic configuration is
(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(𝞹2px)2 (𝞹2py)2(𝞹*2px)1(𝞹*2py)1
♦ This is same as: KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2px)2 (𝞹2py)2(𝞹*2px)1(𝞹*2py)1
♦ Again, the two 𝞹2p orbitals can be combined. We get:
✰ KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2p)4 (𝞹*2px)1(𝞹*2py)1
♦ Again, the two 𝞹*2p orbitals can be combined. We get:
✰ KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2p)4 (𝞹*2p)2
9. Next we determine the magnetic nature of O2
• In fig.4.217, we see that two boxes contain unpaired electrons
✰ So O2 is paramagnetic
So we have written the details of O2 molecule. It is now easy to write the details about O2+ ion, O2- ion and O22- ion.
First we will see O2+ ion
• The details can be written in 8 steps:
1. We know that, O2 has 16 electrons. So O2+ ion will have only 15 electrons
We can write:
• Out of the 15 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 σ2pz
♦ the next four will go to [𝞹2px, 𝞹2py]
♦ the last one will go to [𝞹*2px, 𝞹*2py]
✰ It will go to 𝞹*2px
■ Thus we get the diagram shown in fig.4.218 below:
2. Details about this fig.4.218 can be written in 2 steps:
(i) Electronic configurations:
♦ The electronic configuration of O is 1s22s22p4
♦ The electronic configuration of O+ is 1s22s22p3
• The arrangements related to 1s and 2s are same as in the earlier fig.4.217
• We need to discuss about 2p only
(ii) The two 2p orbitals together has seven electrons
♦ In the fig.4.218 above,
✰ The 2px, 2py and 2pz boxes on the left holds the four electrons of the first O atom
✰ The 2px, 2py and 2pz boxes on the right holds the three electrons of the O+ ion
♦ Out of the above seven,
✰ The first two are filled up into the σ2pz
✰ The next four are filled up into the 𝞹2px and 𝞹2py
✰ The last one is filled up into the 𝞹*2px
3. Based on fig.4.218 above, we can determine five items:
(i) Bond order in O2+ ion
(ii) Stability of O2+ ion
(iii) Nature of bond in O2+ ion
(iv) Electronic configuration of O2+ ion
(v) Magnetic nature of O2+ ion
• The following steps from (4) to (8) shows the calculations related to the five items:
4. First we determine the bond order (b.o). It can be written in 2 steps:
(i) In O2+ ion, we have:
Nb = 8 and Na = 3
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{8-3}{2}=2.5}}$
5. Next we determine the stability
• From (4) above, we have:
b.o of O2+ ion = 2.5
• This is greater than zero. So the ion is stable
6. Next we determine the nature of bond in O2+ ion
• From (4) above, we have:
b.o of O2+ ion = 2.5
• This is a fractional quantity. So we cannot write whether it is a single double or triple bond using 'bond order'
7. Next we determine the electronic configuration of O2+ ion. This can be written in 2 steps:
(i) In O2+ ion, 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 σ2pz
♦ two electrons present in 𝞹2px
♦ two electrons present in 𝞹2py
♦ one electron present in 𝞹*2px
(ii) So the electronic configuration is
(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(𝞹2px)2 (𝞹2py)2(𝞹*2px)1
♦ This is same as: KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2px)2 (𝞹2py)2(𝞹*2px)1
♦ Again, the two 𝞹2p orbitals can be combined. We get:
✰ KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2p)4 (𝞹*2px)1
8. Next we determine the magnetic nature of O2+ ion
• In fig.4.218, we see that one box contains unpaired electrons
✰ So O2+ ion is paramagnetic
The details can be written in 8 steps:
1. We know that, O2 has 16 electrons. So O2- ion will have 17 electrons
We can write:
• Out of the 17 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 σ2pz
♦ the next four will go to [𝞹2px, 𝞹2py]
♦ the last three will go to [𝞹*2px, 𝞹*2py]
✰ Two will go to 𝞹*2px
✰ The third will go to 𝞹*2py
■ Thus we get the diagram shown in fig.4.219 below:
2. Details about this fig.4.219 can be written in 2 steps:
(i) Electronic configurations:
♦ The electronic configuration of O is 1s22s22p4
♦ The electronic configuration of O- is 1s22s22p5
• The arrangements related to 1s and 2s are same as in the earlier fig.4.217
• We need to discuss about 2p only
(ii) The two 2p orbitals together has nine electrons
♦ In the fig.4.219 above,
✰ The 2px, 2py and 2pz boxes on the left holds the four electrons of the first O atom
✰ The 2px, 2py and 2pz boxes on the right holds the five electrons of the O- ion
♦ Out of the above nine,
✰ The first two are filled up into the σ2pz
✰ The next four are filled up into the 𝞹2px and 𝞹2py
✰ The last three are filled up into the 𝞹*2px and 𝞹*2py
3. Based on fig.4.218 above, we can determine five items:
(i) Bond order in O2- ion
(ii) Stability of O2- ion
(iii) Nature of bond in O2- ion
(iv) Electronic configuration of O2- ion
(v) Magnetic nature of O2- ion
• The following steps from (4) to (8) shows the calculations related to the five items:
4. First we determine the bond order (b.o). It can be written in 2 steps:
(i) In O2- ion, we have:
Nb = 8 and Na = 5
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{8-5}{2}=1.5}}$
5. Next we determine the stability
• From (4) above, we have:
b.o of O2- ion = 1.5
• This is greater than zero. So the ion is stable
6. Next we determine the nature of bond in O2- ion
• From (4) above, we have:
b.o of O2- ion = 1.5
• This is a fractional quantity. So we cannot write whether it is a single, double or triple bond using 'bond order'
7. Next we determine the electronic configuration of O2- ion. This can be written in 2 steps:
(i) In O2- ion, 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 σ2pz
♦ two electrons present in 𝞹2px
♦ two electrons present in 𝞹2py
♦ two electrons present in 𝞹*2px
♦ one electron present in 𝞹*2py
(ii) So the electronic configuration is
(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(𝞹2px)2 (𝞹2py)2(𝞹*2px)2(𝞹*2py)1
♦ This is same as: KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2px)2 (𝞹2py)2(𝞹*2px)2(𝞹*2py)1
♦ Again, the two 𝞹2p orbitals can be combined. We get:
✰ KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2p)4 (𝞹*2px)2(𝞹*2py)1
♦ Again, the two 𝞹*2p orbitals can be combined. We get:
✰ KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2p)4 (𝞹*2p)3
8. Next we determine the magnetic nature of O2- ion
• In fig.4.217, we see that one box contains unpaired electrons
✰ So O2- ion is paramagnetic
The details can be written in 8 steps:
1. We know that, O2 has 16 electrons. So O22- ion will have 18 electrons
We can write:
• Out of the 18 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 σ2pz
♦ the next four will go to [𝞹2px, 𝞹2py]
♦ the last four will go to [𝞹*2px, 𝞹*2py]
■ Thus we get the diagram shown in fig.4.220 below:
2. Details about this fig.4.220 can be written in 2 steps:
(i) Electronic configurations:
♦ The electronic configuration of O- is 1s22s22p5
♦ The electronic configuration of O- is 1s22s22p5
• The arrangements related to 1s and 2s are same as in the earlier fig.4.217
• We need to discuss about 2p only
(ii) The two 2p orbitals together has ten electrons
♦ In the fig.4.220 above,
✰ The 2px, 2py and 2pz boxes on the left holds the five electrons of the first O- ion
✰ The 2px, 2py and 2pz boxes on the right holds the five electrons of the other O- ion
♦ Out of the above ten,
✰ The first two are filled up into the σ2pz
✰ The next four are filled up into the 𝞹2px and 𝞹2py
✰ The last four are filled up into the 𝞹*2px and 𝞹*2py
3. Based on fig.4.220 above, we can determine five items:
(i) Bond order in O22- ion
(ii) Stability of O22- ion
(iii) Nature of bond in O22- ion
(iv) Electronic configuration of O22- ion
(v) Magnetic nature of O22- ion
• The following steps from (4) to (8) shows the calculations related to the five items:
4. First we determine the bond order (b.o). It can be written in 2 steps:
(i) In O22- ion, we have:
Nb = 8 and Na = 6
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{8-6}{2}=1}}$
5. Next we determine the stability
• From (4) above, we have:
b.o of O22- ion = 1
• This is greater than zero. So the ion is stable
6. Next we determine the nature of bond in O22- ion
• From (4) above, we have:
b.o of O22- ion = 1
So it will be a single bond
7. Next we determine the electronic configuration of O22- ion. This can be written in 2 steps:
(i) In O22- ion, 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 σ2pz
♦ two electrons present in 𝞹2px
♦ two electrons present in 𝞹2py
♦ two electrons present in 𝞹*2px
♦ two electron present in 𝞹*2py
(ii) So the electronic configuration is
(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(𝞹2px)2 (𝞹2py)2(𝞹*2px)2(𝞹*2py)2
♦ This is same as: KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2px)2 (𝞹2py)2(𝞹*2px)2(𝞹*2py)2
♦ Again, the two 𝞹2p orbitals can be combined. We get:
✰ KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2p)4 (𝞹*2px)2(𝞹*2py)2
♦ Again, the two 𝞹*2p orbitals can be combined. We get:
✰ KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2p)4 (𝞹*2p)4
8. Next we determine the magnetic nature of O22- ion
• In fig.4.217, we see that no box contain unpaired electrons
✰ So O22- ion is diamagnetic
O2: 2, O2+: 2.5, O2-: 1.5, O22-: 1
• So we can arrange them in the increasing order of stability:
O22- < O2-< O2< O2+.
2. Next we write the magnetic properties:
O2: Paramagnetic, O2+: Paramagnetic, O2-: Paramagnetic, O22-: Diamagnetic
First we will see O2+ ion
• The details can be written in 8 steps:
1. We know that, O2 has 16 electrons. So O2+ ion will have only 15 electrons
We can write:
• Out of the 15 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 σ2pz
♦ the next four will go to [𝞹2px, 𝞹2py]
♦ the last one will go to [𝞹*2px, 𝞹*2py]
✰ It will go to 𝞹*2px
■ Thus we get the diagram shown in fig.4.218 below:
 |
| Fig.4.218 |
(i) Electronic configurations:
♦ The electronic configuration of O is 1s22s22p4
♦ The electronic configuration of O+ is 1s22s22p3
• The arrangements related to 1s and 2s are same as in the earlier fig.4.217
• We need to discuss about 2p only
(ii) The two 2p orbitals together has seven electrons
♦ In the fig.4.218 above,
✰ The 2px, 2py and 2pz boxes on the left holds the four electrons of the first O atom
✰ The 2px, 2py and 2pz boxes on the right holds the three electrons of the O+ ion
♦ Out of the above seven,
✰ The first two are filled up into the σ2pz
✰ The next four are filled up into the 𝞹2px and 𝞹2py
✰ The last one is filled up into the 𝞹*2px
3. Based on fig.4.218 above, we can determine five items:
(i) Bond order in O2+ ion
(ii) Stability of O2+ ion
(iii) Nature of bond in O2+ ion
(iv) Electronic configuration of O2+ ion
(v) Magnetic nature of O2+ ion
• The following steps from (4) to (8) shows the calculations related to the five items:
4. First we determine the bond order (b.o). It can be written in 2 steps:
(i) In O2+ ion, we have:
Nb = 8 and Na = 3
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{8-3}{2}=2.5}}$
5. Next we determine the stability
• From (4) above, we have:
b.o of O2+ ion = 2.5
• This is greater than zero. So the ion is stable
6. Next we determine the nature of bond in O2+ ion
• From (4) above, we have:
b.o of O2+ ion = 2.5
• This is a fractional quantity. So we cannot write whether it is a single double or triple bond using 'bond order'
7. Next we determine the electronic configuration of O2+ ion. This can be written in 2 steps:
(i) In O2+ ion, 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 σ2pz
♦ two electrons present in 𝞹2px
♦ two electrons present in 𝞹2py
♦ one electron present in 𝞹*2px
(ii) So the electronic configuration is
(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(𝞹2px)2 (𝞹2py)2(𝞹*2px)1
♦ This is same as: KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2px)2 (𝞹2py)2(𝞹*2px)1
♦ Again, the two 𝞹2p orbitals can be combined. We get:
✰ KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2p)4 (𝞹*2px)1
8. Next we determine the magnetic nature of O2+ ion
• In fig.4.218, we see that one box contains unpaired electrons
✰ So O2+ ion is paramagnetic
Next we will see O2- ion
1. We know that, O2 has 16 electrons. So O2- ion will have 17 electrons
We can write:
• Out of the 17 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 σ2pz
♦ the next four will go to [𝞹2px, 𝞹2py]
♦ the last three will go to [𝞹*2px, 𝞹*2py]
✰ Two will go to 𝞹*2px
✰ The third will go to 𝞹*2py
■ Thus we get the diagram shown in fig.4.219 below:
 |
| Fig.4.219 |
(i) Electronic configurations:
♦ The electronic configuration of O is 1s22s22p4
♦ The electronic configuration of O- is 1s22s22p5
• The arrangements related to 1s and 2s are same as in the earlier fig.4.217
• We need to discuss about 2p only
(ii) The two 2p orbitals together has nine electrons
♦ In the fig.4.219 above,
✰ The 2px, 2py and 2pz boxes on the left holds the four electrons of the first O atom
✰ The 2px, 2py and 2pz boxes on the right holds the five electrons of the O- ion
♦ Out of the above nine,
✰ The first two are filled up into the σ2pz
✰ The next four are filled up into the 𝞹2px and 𝞹2py
✰ The last three are filled up into the 𝞹*2px and 𝞹*2py
3. Based on fig.4.218 above, we can determine five items:
(i) Bond order in O2- ion
(ii) Stability of O2- ion
(iii) Nature of bond in O2- ion
(iv) Electronic configuration of O2- ion
(v) Magnetic nature of O2- ion
• The following steps from (4) to (8) shows the calculations related to the five items:
4. First we determine the bond order (b.o). It can be written in 2 steps:
(i) In O2- ion, we have:
Nb = 8 and Na = 5
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{8-5}{2}=1.5}}$
5. Next we determine the stability
• From (4) above, we have:
b.o of O2- ion = 1.5
• This is greater than zero. So the ion is stable
6. Next we determine the nature of bond in O2- ion
• From (4) above, we have:
b.o of O2- ion = 1.5
• This is a fractional quantity. So we cannot write whether it is a single, double or triple bond using 'bond order'
7. Next we determine the electronic configuration of O2- ion. This can be written in 2 steps:
(i) In O2- ion, 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 σ2pz
♦ two electrons present in 𝞹2px
♦ two electrons present in 𝞹2py
♦ two electrons present in 𝞹*2px
♦ one electron present in 𝞹*2py
(ii) So the electronic configuration is
(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(𝞹2px)2 (𝞹2py)2(𝞹*2px)2(𝞹*2py)1
♦ This is same as: KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2px)2 (𝞹2py)2(𝞹*2px)2(𝞹*2py)1
♦ Again, the two 𝞹2p orbitals can be combined. We get:
✰ KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2p)4 (𝞹*2px)2(𝞹*2py)1
♦ Again, the two 𝞹*2p orbitals can be combined. We get:
✰ KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2p)4 (𝞹*2p)3
8. Next we determine the magnetic nature of O2- ion
• In fig.4.217, we see that one box contains unpaired electrons
✰ So O2- ion is paramagnetic
Next we will see O22- ion
1. We know that, O2 has 16 electrons. So O22- ion will have 18 electrons
We can write:
• Out of the 18 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 σ2pz
♦ the next four will go to [𝞹2px, 𝞹2py]
♦ the last four will go to [𝞹*2px, 𝞹*2py]
■ Thus we get the diagram shown in fig.4.220 below:
 |
| Fig.4.220 |
(i) Electronic configurations:
♦ The electronic configuration of O- is 1s22s22p5
♦ The electronic configuration of O- is 1s22s22p5
• The arrangements related to 1s and 2s are same as in the earlier fig.4.217
• We need to discuss about 2p only
(ii) The two 2p orbitals together has ten electrons
♦ In the fig.4.220 above,
✰ The 2px, 2py and 2pz boxes on the left holds the five electrons of the first O- ion
✰ The 2px, 2py and 2pz boxes on the right holds the five electrons of the other O- ion
♦ Out of the above ten,
✰ The first two are filled up into the σ2pz
✰ The next four are filled up into the 𝞹2px and 𝞹2py
✰ The last four are filled up into the 𝞹*2px and 𝞹*2py
3. Based on fig.4.220 above, we can determine five items:
(i) Bond order in O22- ion
(ii) Stability of O22- ion
(iii) Nature of bond in O22- ion
(iv) Electronic configuration of O22- ion
(v) Magnetic nature of O22- ion
• The following steps from (4) to (8) shows the calculations related to the five items:
4. First we determine the bond order (b.o). It can be written in 2 steps:
(i) In O22- ion, we have:
Nb = 8 and Na = 6
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{8-6}{2}=1}}$
5. Next we determine the stability
• From (4) above, we have:
b.o of O22- ion = 1
• This is greater than zero. So the ion is stable
6. Next we determine the nature of bond in O22- ion
• From (4) above, we have:
b.o of O22- ion = 1
So it will be a single bond
7. Next we determine the electronic configuration of O22- ion. This can be written in 2 steps:
(i) In O22- ion, 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 σ2pz
♦ two electrons present in 𝞹2px
♦ two electrons present in 𝞹2py
♦ two electrons present in 𝞹*2px
♦ two electron present in 𝞹*2py
(ii) So the electronic configuration is
(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(𝞹2px)2 (𝞹2py)2(𝞹*2px)2(𝞹*2py)2
♦ This is same as: KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2px)2 (𝞹2py)2(𝞹*2px)2(𝞹*2py)2
♦ Again, the two 𝞹2p orbitals can be combined. We get:
✰ KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2p)4 (𝞹*2px)2(𝞹*2py)2
♦ Again, the two 𝞹*2p orbitals can be combined. We get:
✰ KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2p)4 (𝞹*2p)4
8. Next we determine the magnetic nature of O22- ion
• In fig.4.217, we see that no box contain unpaired electrons
✰ So O22- ion is diamagnetic
Now we can compare the stability and magnetic properties of the following:
O2 molecule, O2+ ion, O2- ion and O22- ion.
1. We will first write the b.o values:
O2 molecule, O2+ ion, O2- ion and O22- ion.
• So we can arrange them in the increasing order of stability:
O22- < O2-< O2< O2+.
2. Next we write the magnetic properties:
O2: Paramagnetic, O2+: Paramagnetic, O2-: Paramagnetic, O22-: Diamagnetic
• In the next section, we will see the details related to F2 and Ne2 molecules
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