• In the previous section 4.38, we saw the details of the O2 molecule and O2+, O2-, O22- ions. In this section we will see the next molecule which is F2
The details related to F2 molecule can be written in 9 steps:
1. The electronic configuration of F is 1s22s22p5
• Since 2p orbitals are present, there will be a total of ten molecular orbitals in F2:
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 a previous section, we saw two types of increasing order
• For easy reference, that fig.4.211 is shown again below:
■ In the case of F2, 'type 1' is applicable
3. Now we can arrange the electrons in F2 into the various molecular orbitals
• The electronic configuration of F is 1s22s22p5
♦ So in an individual F atom, there are 9 electrons
• Then where will the 18 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 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.4.221 below:
• Details about this fig.4.221 can be written in 3 steps:
(i) The electronic configuration of O is 1s22s22p5
• We see that, the 1s orbital has 2 electrons
• So when two F 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 F:1s22s22p4
• We see that, the 2s orbital has two electrons
• So when two F atoms combine, the 2s orbitals will contribute a total of four electrons
♦ In the fig.4.221 above,
✰ The two arrows in the left 2s box indicates those two electrons of the first F atom
✰ The two arrows in the right 2s box indicates those two electrons of the other F 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 F:1s22s22p5
• We see that, the 2p orbital has five electrons
• So when two F atoms combine, the 2p orbitals will contribute a total of ten electrons
♦ In the fig.4.221 above,
✰ The 2px, 2py and 2pz boxes on the left holds the five electrons of the first F atom
✰ The 2px, 2py and 2pz boxes on the right holds the five electrons of the other F atom
♦ 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
4. Based on fig.4.221 above, we can determine five items:
(i) Bond order in F2 molecule
(ii) Stability of F2 molecule
(iii) Nature of bond in F2 molecule
(iv) Electronic configuration of F2 molecule
(v) Magnetic nature of F2 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 F2, we have:
Nb = 8 and Na = 6
• 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 six because:
✰ The anti-bonding orbital σ*2s contains two electrons
✰ The anti-bonding orbital 𝞹*2px contains two electron
✰ The anti-bonding orbital 𝞹*2py contains two electron
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{8-6}{2}=1}}$
6. Next we determine the stability
• From (5) above, we have:
b.o of F2 = 1
• This is greater than zero. So the molecule is stable
7. Next we determine the nature of bond in F2 molecule
• From (5) above, we have:
b.o of F2 = 1
• So there will be a single bond between the two F atoms
8. Next we determine the electronic configuration of F2. This can be written in 2 steps:
(i) In F2, 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 electrons 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
9. Next we determine the magnetic nature of F2
• In fig.4.221, we see that no boxes contain unpaired electrons
✰ So F2 is diamagnetic
The details related to F2 molecule can be written in 9 steps:
1. The electronic configuration of F is 1s22s22p5
• Since 2p orbitals are present, there will be a total of ten molecular orbitals in F2:
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 a 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 F2 into the various molecular orbitals
• The electronic configuration of F is 1s22s22p5
♦ So in an individual F atom, there are 9 electrons
• So when two individual F atoms combine to form a F2 molecule, there will be a total of 18 electrons
• In a F2 molecule, there are no 1s, 2s or 2p orbitals
• In a F2 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 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.4.221 below:
 |
| Fig.4.221 |
(i) The electronic configuration of O is 1s22s22p5
• We see that, the 1s orbital has 2 electrons
• So when two F 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 F:1s22s22p4
• We see that, the 2s orbital has two electrons
• So when two F atoms combine, the 2s orbitals will contribute a total of four electrons
♦ In the fig.4.221 above,
✰ The two arrows in the left 2s box indicates those two electrons of the first F atom
✰ The two arrows in the right 2s box indicates those two electrons of the other F 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 F:1s22s22p5
• We see that, the 2p orbital has five electrons
• So when two F atoms combine, the 2p orbitals will contribute a total of ten electrons
♦ In the fig.4.221 above,
✰ The 2px, 2py and 2pz boxes on the left holds the five electrons of the first F atom
✰ The 2px, 2py and 2pz boxes on the right holds the five electrons of the other F atom
♦ 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
4. Based on fig.4.221 above, we can determine five items:
(i) Bond order in F2 molecule
(ii) Stability of F2 molecule
(iii) Nature of bond in F2 molecule
(iv) Electronic configuration of F2 molecule
(v) Magnetic nature of F2 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 F2, we have:
Nb = 8 and Na = 6
• 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 six because:
✰ The anti-bonding orbital σ*2s contains two electrons
✰ The anti-bonding orbital 𝞹*2px contains two electron
✰ The anti-bonding orbital 𝞹*2py contains two electron
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{8-6}{2}=1}}$
6. Next we determine the stability
• From (5) above, we have:
b.o of F2 = 1
• This is greater than zero. So the molecule is stable
7. Next we determine the nature of bond in F2 molecule
• From (5) above, we have:
b.o of F2 = 1
• So there will be a single bond between the two F atoms
8. Next we determine the electronic configuration of F2. This can be written in 2 steps:
(i) In F2, 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 electrons 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
9. Next we determine the magnetic nature of F2
• In fig.4.221, we see that no boxes contain unpaired electrons
✰ So F2 is diamagnetic
Next we will see the Ne2 molecule
The details related to Ne2 molecule can be written in 9 steps:
1. The electronic configuration of Ne is 1s22s22p6
• Since 2p orbitals are present, there will be a total of ten molecular orbitals in Ne2:
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 a previous section, we saw two types of increasing order
• For easy reference, that fig.4.211 is shown again below:
■ In the case of Ne2, 'type 1' is applicable
3. Now we can arrange the electrons in Ne2 into the various molecular orbitals
• The electronic configuration of Ne is 1s22s22p5
♦ So in an individual Ne atom, there are 10 electrons
• Then where will the 20 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 twenty 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 next four will go to [𝞹*2px, 𝞹*2py]
♦ The last two will go to σ*2pz
■ Thus we get the diagram shown in fig.4.222 below:
• Details about this fig.4.222 can be written in 3 steps:
(i) The electronic configuration of Ne is 1s22s22p6
• We see that, the 1s orbital has 2 electrons
• So when two Ne 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 Ne:1s22s22p6
• We see that, the 2s orbital has two electrons
• So when two Ne atoms combine, the 2s orbitals will contribute a total of four electrons
♦ In the fig.4.222 above,
✰ The two arrows in the left 2s box indicates those two electrons of the first Ne atom
✰ The two arrows in the right 2s box indicates those two electrons of the other Ne 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 Ne: 1s22s22p6
• We see that, the 2p orbital has six electrons
• So when two Ne atoms combine, the 2p orbitals will contribute a total of 12 electrons
♦ In the fig.4.221 above,
✰ The 2px, 2py and 2pz boxes on the left holds the six electrons of the first Ne atom
✰ The 2px, 2py and 2pz boxes on the right holds the six electrons of the other Ne atom
♦ Out of the above 12,
✰ The first two are filled up into the σ2pz
✰ The next four are filled up into the 𝞹2px and 𝞹2py
✰ The next four are filled up into the 𝞹*2px and 𝞹*2py
✰ The last two are filled up into the σ*2pz
4. Based on fig.4.222 above, we can determine five items:
(i) Bond order in Ne2 molecule
(ii) Stability of Ne2 molecule
(iii) Nature of bond in Ne2 molecule
(iv) Electronic configuration of Ne2 molecule
(v) Magnetic nature of Ne2 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 Ne2, we have:
Nb = 8 and Na = 8
• 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 six because:
✰ The anti-bonding orbital σ*2s contains two electrons
✰ The anti-bonding orbital 𝞹*2px contains two electron
✰ The anti-bonding orbital 𝞹*2py contains two electron
✰ The anti-bonding orbital σ*2pz contains two electrons
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{8-8}{2}=0}}$
6. Next we determine the stability
• From (5) above, we have:
b.o of Ne2 = 0
• So the Ne2 molecule does not exist
7. Next we determine the nature of bond in Ne2 molecule
• From (5) above, we have:
b.o of Ne2 = 0
Since the molecule does not exist, there is no point in writing the nature of bond
8. Next we determine the electronic configuration of Ne2. This can be written in 2 steps:
(i) In Ne2, 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 electrons present in 𝞹*2py
♦ two electrons present in σ*2pz
(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(σ*2pz)2
♦ This is same as: KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2px)2 (𝞹2py)2(𝞹*2px)2(𝞹*2py)2(σ*2pz)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(σ*2pz)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(σ*2pz)2
9. Next we determine the magnetic nature of Ne2
• In fig.4.222, we see that no boxes contain unpaired electrons
✰ So Ne2 is diamagnetic
The details related to Ne2 molecule can be written in 9 steps:
1. The electronic configuration of Ne is 1s22s22p6
• Since 2p orbitals are present, there will be a total of ten molecular orbitals in Ne2:
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 a 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 Ne2 into the various molecular orbitals
• The electronic configuration of Ne is 1s22s22p5
♦ So in an individual Ne atom, there are 10 electrons
• So when two individual Ne atoms combine to form a Ne2 molecule, there will be a total of 20 electrons
• In a Ne2 molecule, there are no 1s, 2s or 2p orbitals
• In a Ne2 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 twenty 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 next four will go to [𝞹*2px, 𝞹*2py]
♦ The last two will go to σ*2pz
■ Thus we get the diagram shown in fig.4.222 below:
 |
| Fig.4.222 |
(i) The electronic configuration of Ne is 1s22s22p6
• We see that, the 1s orbital has 2 electrons
• So when two Ne 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 Ne:1s22s22p6
• We see that, the 2s orbital has two electrons
• So when two Ne atoms combine, the 2s orbitals will contribute a total of four electrons
♦ In the fig.4.222 above,
✰ The two arrows in the left 2s box indicates those two electrons of the first Ne atom
✰ The two arrows in the right 2s box indicates those two electrons of the other Ne 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 Ne: 1s22s22p6
• We see that, the 2p orbital has six electrons
• So when two Ne atoms combine, the 2p orbitals will contribute a total of 12 electrons
♦ In the fig.4.221 above,
✰ The 2px, 2py and 2pz boxes on the left holds the six electrons of the first Ne atom
✰ The 2px, 2py and 2pz boxes on the right holds the six electrons of the other Ne atom
♦ Out of the above 12,
✰ The first two are filled up into the σ2pz
✰ The next four are filled up into the 𝞹2px and 𝞹2py
✰ The next four are filled up into the 𝞹*2px and 𝞹*2py
✰ The last two are filled up into the σ*2pz
4. Based on fig.4.222 above, we can determine five items:
(i) Bond order in Ne2 molecule
(ii) Stability of Ne2 molecule
(iii) Nature of bond in Ne2 molecule
(iv) Electronic configuration of Ne2 molecule
(v) Magnetic nature of Ne2 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 Ne2, we have:
Nb = 8 and Na = 8
• 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 six because:
✰ The anti-bonding orbital σ*2s contains two electrons
✰ The anti-bonding orbital 𝞹*2px contains two electron
✰ The anti-bonding orbital 𝞹*2py contains two electron
✰ The anti-bonding orbital σ*2pz contains two electrons
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{8-8}{2}=0}}$
6. Next we determine the stability
• From (5) above, we have:
b.o of Ne2 = 0
• So the Ne2 molecule does not exist
7. Next we determine the nature of bond in Ne2 molecule
• From (5) above, we have:
b.o of Ne2 = 0
Since the molecule does not exist, there is no point in writing the nature of bond
8. Next we determine the electronic configuration of Ne2. This can be written in 2 steps:
(i) In Ne2, 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 electrons present in 𝞹*2py
♦ two electrons present in σ*2pz
(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(σ*2pz)2
♦ This is same as: KK(σ2s)2(σ*2s)2(σ2pz)2(𝞹2px)2 (𝞹2py)2(𝞹*2px)2(𝞹*2py)2(σ*2pz)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(σ*2pz)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(σ*2pz)2
9. Next we determine the magnetic nature of Ne2
• In fig.4.222, we see that no boxes contain unpaired electrons
✰ So Ne2 is diamagnetic
• In the next section, we will see hydrogen bonding
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