• In the previous section 4.33, we saw the properties of He2, H2+ and He2+. In this section we will see Li2
• The details related to Li2 can be written in 11 steps:
1. Li has the electronic configuration: 1s22s1
• So when two Li atoms combine to form the Li2 molecule, the following mergers will take place:
♦ The 1s orbital of one Li atom merges with the 1s orbital of the other Li atom
✰ This results in the formation of bonding and anti-bonding orbitals: σ1s and σ*1s
♦ The 2s orbital of one Li atom merges with the 2s orbital of the other Li atom
✰ This results in the formation of bonding and anti-bonding orbitals: σ2s and σ*2s
2. The process of merger of the 2s orbitals is very similar to that of the 1s orbitals
• So we can use:
♦ Figs.4.185 and 186 in the section 4.32 to demonstrate the formation of the bonding orbital σ2s
♦ Figs.4.187, 188 and 189 in the section 4.32 to demonstrate the formation of the anti-bonding orbital σ*2s
• The only two points that we have to keep in mind are:
(i) The σ2s is larger than σ1s
♦ The two nuclei of Li atoms are the centers of σ1s
♦ The same two nuclei are the centers of σ2s also
(ii) The σ*2s is larger than σ*1s
♦ The two nuclei of Li atoms are the centers of σ*1s
♦ The same two nuclei are the centers of σ*2s also
3. So altogether, there are four molecular orbitals: σ1s , σ*1s , σ2s and σ*2s
• We can represent them in our usual graphical form. This is shown in fig.4.193(a) below:
• We need to note four points about the fig.4.193(a):
(i) The bottom pair of sloping green lines indicate that:
♦ The two 1s orbitals merge together to form σ1s
✰ We have already seen this in the cases of H2 and He2
(ii) The top pair of sloping green lines indicate that:
♦ The two 2s orbitals merge together to form σ2s
(iii) The bottom pair of sloping magenta lines indicate that:
♦ The two 1s orbitals merge together to form σ*1s
✰ We have already seen this in the cases of H2 and He2
(iv) The top pair of sloping magenta lines indicate that:
♦ The two 2s orbitals merge together to form σ*2s
4. Why is σ2s and σ*2s shown at a higher level than σ1s and σ*1s ?
Answer can be written in 3 steps:
(i) Experimental results shows that σ2s and σ*2s have higher energies than σ1s and σ*1s
(ii) The energies can be written in an increasing order as shown below:
σ1s < σ*1s < σ2s < σ*2s
(Note also that, among σ2s and σ*2s, the anti-bonding σ*2s has higher energy)
(iii) So, σ2s and σ*2s are shown at a higher level than σ1s and σ*1s
♦ The two levels are separated by a cyan dashed line
• Then where will the six electrons reside?
• The answer can be written in 2 steps:
(i) Obviously, they have no where else to go but the σ1s , σ*1s , σ2s and σ*2s
♦ The six electrons will go to those new molecular orbitals
(ii) The new molecular orbitals will be filled up according to the order written in 4(ii). So we can write:
• Out of the six electrons,
♦ the first two will go to σ1s
♦ the next two will go to σ*1s
♦ the last two will go to σ2s
• So the three squares corresponding to σ1s , σ*1s and σ2s will contain two arrows each
• The square corresponding to σ*2s will be vacant
• This is shown in fig.4.193(b) above
6. Based on fig.4.193(b) above, we can determine five items:
(i) Bond order in Li2 molecule
(ii) Stability of Li2 molecule
(iii) Nature of bond in Li2 molecule
(iv) Electronic configuration of Li2 molecule
(v) Magnetic nature of Li2 molecule
• The following steps from (7) to (11) shows the calculations related to the five items:
7. First we determine the bond order (b.o). It can be written in 2 steps:
(i) In Li2, we have:
Nb = 4 and Na = 2
• Note:
♦ Nb is equal to four because:
✰ The bonding orbital σ1s contains two electrons
✰ The bonding orbital σ2s contains two electrons
♦ Na is equal to two because:
✰ The anti-bonding orbital σ*1s contains two electrons
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{4-2}{2}=1}}$
8. Next we determine the stability
• From (7) above, we have:
b.o of Li2 = 1
• This is greater than zero. So the molecule is stable
9. Next we determine the nature of bond in Li2 molecule
• From (7) above, we have:
b.o of Li2 = 1
• So there will be a single bond between the two Li atoms
10. Next we determine the electronic configuration of Li2. This can be written in 3 steps:
(i) In Li2, we have:
♦ two electrons present in σ1s
♦ two electrons present in σ*1s
♦ two electrons present in σ2s
(ii) So the electronic configuration is (σ1s)2(σ*1s)2(σ2s)2
(iii) In the above configuration, consider the first two components: (σ1s)2 and (σ*1s)2
• They indicate that
♦ The orbitals related to the 'first main-shell' are completely filled up
• We know that:
♦ The 'first main-shell' is the 'K main-shell'
• We want to indicate that:
♦ The orbitals related to the 'first main-shell' are completely filled up
• For that, we write: KK
♦ So 'KK' indicates: (σ1s)2 and (σ*1s)2
■ Thus the configuration written in (ii) becomes: KK(σ2s)2
11. Next we determine the magnetic nature of Li2
• In fig.4.193(b), we see that all the boxes contain paired electrons
♦ So Li2 is diamagnetic
• The details related to Be2 can be written in 11 steps:
1. Be has the electronic configuration: 1s22s2
• So when two Be atoms combine to form the Be2 molecule, the following mergers will take place:
♦ The 1s orbital of one Be atom merges with the 1s orbital of the other Be atom
✰ This results in the formation of bonding and anti-bonding orbitals: σ1s and σ*1s
♦ The 2s orbital of one Be atom merges with the 2s orbital of the other Be atom
✰ This results in the formation of bonding and anti-bonding orbitals: σ2s and σ*2s
2. The process of merger of the 2s orbitals is very similar to that of the 1s orbitals
We have seen the details in the 'step 2 of Li2' at the beginning of this section
3. So altogether, there are four molecular orbitals: σ1s , σ*1s , σ2s and σ*2s
• We can represent them in our usual graphical form. This is shown in fig.4.194(a) below:
• We need to note four points about the fig.a:
(i) The bottom pair of sloping green lines indicate that:
♦ The two 1s orbitals combine together to form σ1s
✰ We have already seen this in the cases of H2 and He2
(ii) The top pair of sloping green lines indicate that:
♦ The two 2s orbitals combine together to form σ2s
(iii) The bottom pair of sloping magenta lines indicate that:
♦ The two 1s orbitals combine together to form σ*1s
✰ We have already seen this in the cases of H2 and He2
(iv) The top pair of sloping magenta lines indicate that:
♦ The two 2s orbitals combine together to form σ*2s
4. Why is σ2s and σ*2s shown at a higher level than σ1s and σ*1s ?
We have already seen the answer in the 'step 4 of Li2' at the beginning of this section
• Then where will the 8 electrons reside?
• The answer can be written in 2 steps:
(i) Obviously, they have no where else to go but the σ1s , σ*1s , σ2s and σ*2s
♦ The 8 electrons will go to those new molecular orbitals
(ii) The new molecular orbitals will be filled up according to the order written in 4(ii) for Li2 molecule earlier
• Let us write it again: σ1s < σ*1s < σ2s < σ*2s
• So we can write:
• Out of the six electrons,
♦ the first two will go to σ1s
♦ the next two will go to σ*1s
♦ the next two will go to σ2s
♦ the last two will go to σ*2s
• So all the four squares will contain two arrows each
• No square is left vacant
• This is shown in fig.4.194(b) above
6. Based on fig.4.194(b) above, we can determine five items:
(i) Bond order in Be2 molecule
(ii) Stability of Be2 molecule
(iii) Nature of bond in Be2 molecule
(iv) Electronic configuration of Be2 molecule
(v) Magnetic nature of Be2 molecule
The following steps from (7) to (11) shows the calculations related to the five items:
7. First we determine the bond order (b.o). It can be written in 2 steps:
(i) In Be2, we have:
Nb = 4 and Na = 4
• Note:
♦ Nb is equal to four because:
✰ The bonding orbital σ1s contains two electrons
✰ The bonding orbital σ2s contains two electrons
♦ Na is equal to four because:
✰ The anti-bonding orbital σ*1s contains two electrons
✰ The anti-bonding orbital σ*2s contains two electrons
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{4-4}{2}=0}}$
8. Next we determine the stability
• From (7) above, we have:
b.o of Be2 = 0
• Since the b.o is zero, the molecule Be2 is unstable
9. Next we determine the nature of bond in Be2 molecule
• Since the molecule is unstable, there is no point in writing whether the bond is single, double or triple
10. Next we determine the electronic configuration of Be2. This can be written in 3 steps:
(i) In Be2, we have:
♦ two electrons present in σ1s
♦ two electrons present in σ*1s
♦ two electrons present in σ2s
♦ two electrons present in σ*2s
(ii) So the electronic configuration is (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2
(iii) In the above configuration, consider the first two components: (σ1s)2 and (σ*1s)2
• They indicate that
♦ The orbitals related to the 'first main-shell' are completely filled up
• We know that:
♦ The 'first main-shell' is the 'K main-shell'
• We want to indicate that:
♦ The orbitals related to the 'first main-shell' are completely filled up
• For that, we write: KK
♦ So 'KK' indicates: (σ1s)2 and (σ*1s)2
■ Thus the configuration written in (ii) becomes: KK(σ2s)2(σ*2s)2
11. Next we determine the magnetic nature of Be2
• In fig.4.194(b), we see that all the boxes contain paired electrons
♦ So Be2 is diamagnetic
• The details related to Li2 can be written in 11 steps:
1. Li has the electronic configuration: 1s22s1
• So when two Li atoms combine to form the Li2 molecule, the following mergers will take place:
♦ The 1s orbital of one Li atom merges with the 1s orbital of the other Li atom
✰ This results in the formation of bonding and anti-bonding orbitals: σ1s and σ*1s
♦ The 2s orbital of one Li atom merges with the 2s orbital of the other Li atom
✰ This results in the formation of bonding and anti-bonding orbitals: σ2s and σ*2s
2. The process of merger of the 2s orbitals is very similar to that of the 1s orbitals
• So we can use:
♦ Figs.4.185 and 186 in the section 4.32 to demonstrate the formation of the bonding orbital σ2s
♦ Figs.4.187, 188 and 189 in the section 4.32 to demonstrate the formation of the anti-bonding orbital σ*2s
• The only two points that we have to keep in mind are:
(i) The σ2s is larger than σ1s
♦ The two nuclei of Li atoms are the centers of σ1s
♦ The same two nuclei are the centers of σ2s also
(ii) The σ*2s is larger than σ*1s
♦ The two nuclei of Li atoms are the centers of σ*1s
♦ The same two nuclei are the centers of σ*2s also
3. So altogether, there are four molecular orbitals: σ1s , σ*1s , σ2s and σ*2s
• We can represent them in our usual graphical form. This is shown in fig.4.193(a) below:
 |
| Fig.4.193 |
(i) The bottom pair of sloping green lines indicate that:
♦ The two 1s orbitals merge together to form σ1s
✰ We have already seen this in the cases of H2 and He2
(ii) The top pair of sloping green lines indicate that:
♦ The two 2s orbitals merge together to form σ2s
(iii) The bottom pair of sloping magenta lines indicate that:
♦ The two 1s orbitals merge together to form σ*1s
✰ We have already seen this in the cases of H2 and He2
(iv) The top pair of sloping magenta lines indicate that:
♦ The two 2s orbitals merge together to form σ*2s
4. Why is σ2s and σ*2s shown at a higher level than σ1s and σ*1s ?
Answer can be written in 3 steps:
(i) Experimental results shows that σ2s and σ*2s have higher energies than σ1s and σ*1s
(ii) The energies can be written in an increasing order as shown below:
σ1s < σ*1s < σ2s < σ*2s
(Note also that, among σ2s and σ*2s, the anti-bonding σ*2s has higher energy)
(iii) So, σ2s and σ*2s are shown at a higher level than σ1s and σ*1s
♦ The two levels are separated by a cyan dashed line
5. In an individual Li atom, there are 3 electrons
• So when two individual Li atoms combine to form a Li2 molecule, there will be a total of 6 electrons
• In a Li2 molecule, there are no 1s or 2s orbitals
• So when two individual Li atoms combine to form a Li2 molecule, there will be a total of 6 electrons
• In a Li2 molecule, there are no 1s or 2s orbitals
• The answer can be written in 2 steps:
(i) Obviously, they have no where else to go but the σ1s , σ*1s , σ2s and σ*2s
♦ The six electrons will go to those new molecular orbitals
(ii) The new molecular orbitals will be filled up according to the order written in 4(ii). So we can write:
• Out of the six electrons,
♦ the first two will go to σ1s
♦ the next two will go to σ*1s
♦ the last two will go to σ2s
• So the three squares corresponding to σ1s , σ*1s and σ2s will contain two arrows each
• The square corresponding to σ*2s will be vacant
• This is shown in fig.4.193(b) above
6. Based on fig.4.193(b) above, we can determine five items:
(i) Bond order in Li2 molecule
(ii) Stability of Li2 molecule
(iii) Nature of bond in Li2 molecule
(iv) Electronic configuration of Li2 molecule
(v) Magnetic nature of Li2 molecule
• The following steps from (7) to (11) shows the calculations related to the five items:
7. First we determine the bond order (b.o). It can be written in 2 steps:
(i) In Li2, we have:
Nb = 4 and Na = 2
• Note:
♦ Nb is equal to four because:
✰ The bonding orbital σ1s contains two electrons
✰ The bonding orbital σ2s contains two electrons
♦ Na is equal to two because:
✰ The anti-bonding orbital σ*1s contains two electrons
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{4-2}{2}=1}}$
8. Next we determine the stability
• From (7) above, we have:
b.o of Li2 = 1
• This is greater than zero. So the molecule is stable
9. Next we determine the nature of bond in Li2 molecule
• From (7) above, we have:
b.o of Li2 = 1
• So there will be a single bond between the two Li atoms
10. Next we determine the electronic configuration of Li2. This can be written in 3 steps:
(i) In Li2, we have:
♦ two electrons present in σ1s
♦ two electrons present in σ*1s
♦ two electrons present in σ2s
(ii) So the electronic configuration is (σ1s)2(σ*1s)2(σ2s)2
(iii) In the above configuration, consider the first two components: (σ1s)2 and (σ*1s)2
• They indicate that
♦ The orbitals related to the 'first main-shell' are completely filled up
• We know that:
♦ The 'first main-shell' is the 'K main-shell'
• We want to indicate that:
♦ The orbitals related to the 'first main-shell' are completely filled up
• For that, we write: KK
♦ So 'KK' indicates: (σ1s)2 and (σ*1s)2
■ Thus the configuration written in (ii) becomes: KK(σ2s)2
11. Next we determine the magnetic nature of Li2
• In fig.4.193(b), we see that all the boxes contain paired electrons
♦ So Li2 is diamagnetic
Next we will see Be2
1. Be has the electronic configuration: 1s22s2
• So when two Be atoms combine to form the Be2 molecule, the following mergers will take place:
♦ The 1s orbital of one Be atom merges with the 1s orbital of the other Be atom
✰ This results in the formation of bonding and anti-bonding orbitals: σ1s and σ*1s
♦ The 2s orbital of one Be atom merges with the 2s orbital of the other Be atom
✰ This results in the formation of bonding and anti-bonding orbitals: σ2s and σ*2s
2. The process of merger of the 2s orbitals is very similar to that of the 1s orbitals
We have seen the details in the 'step 2 of Li2' at the beginning of this section
3. So altogether, there are four molecular orbitals: σ1s , σ*1s , σ2s and σ*2s
• We can represent them in our usual graphical form. This is shown in fig.4.194(a) below:
 |
| Fig.4.194 |
(i) The bottom pair of sloping green lines indicate that:
♦ The two 1s orbitals combine together to form σ1s
✰ We have already seen this in the cases of H2 and He2
(ii) The top pair of sloping green lines indicate that:
♦ The two 2s orbitals combine together to form σ2s
(iii) The bottom pair of sloping magenta lines indicate that:
♦ The two 1s orbitals combine together to form σ*1s
✰ We have already seen this in the cases of H2 and He2
(iv) The top pair of sloping magenta lines indicate that:
♦ The two 2s orbitals combine together to form σ*2s
4. Why is σ2s and σ*2s shown at a higher level than σ1s and σ*1s ?
We have already seen the answer in the 'step 4 of Li2' at the beginning of this section
5. In an individual Be atom, there are 4 electrons
• So when two individual Be atoms combine to form a Be2 molecule, there will be a total of 8 electrons
• In a Be2 molecule, there are no 1s or 2s orbitals
• So when two individual Be atoms combine to form a Be2 molecule, there will be a total of 8 electrons
• In a Be2 molecule, there are no 1s or 2s orbitals
• The answer can be written in 2 steps:
♦ The 8 electrons will go to those new molecular orbitals
(ii) The new molecular orbitals will be filled up according to the order written in 4(ii) for Li2 molecule earlier
• Let us write it again: σ1s < σ*1s < σ2s < σ*2s
• So we can write:
• Out of the six electrons,
♦ the first two will go to σ1s
♦ the next two will go to σ*1s
♦ the next two will go to σ2s
♦ the last two will go to σ*2s
• So all the four squares will contain two arrows each
• No square is left vacant
• This is shown in fig.4.194(b) above
6. Based on fig.4.194(b) above, we can determine five items:
(i) Bond order in Be2 molecule
(ii) Stability of Be2 molecule
(iii) Nature of bond in Be2 molecule
(iv) Electronic configuration of Be2 molecule
(v) Magnetic nature of Be2 molecule
The following steps from (7) to (11) shows the calculations related to the five items:
7. First we determine the bond order (b.o). It can be written in 2 steps:
(i) In Be2, we have:
Nb = 4 and Na = 4
• Note:
♦ Nb is equal to four because:
✰ The bonding orbital σ1s contains two electrons
✰ The bonding orbital σ2s contains two electrons
♦ Na is equal to four because:
✰ The anti-bonding orbital σ*1s contains two electrons
✰ The anti-bonding orbital σ*2s contains two electrons
(ii) So $\mathbf\small{\rm{b.o=\frac{N_b-Na}{2}=\frac{4-4}{2}=0}}$
8. Next we determine the stability
• From (7) above, we have:
b.o of Be2 = 0
• Since the b.o is zero, the molecule Be2 is unstable
9. Next we determine the nature of bond in Be2 molecule
• Since the molecule is unstable, there is no point in writing whether the bond is single, double or triple
10. Next we determine the electronic configuration of Be2. This can be written in 3 steps:
(i) In Be2, we have:
♦ two electrons present in σ1s
♦ two electrons present in σ*1s
♦ two electrons present in σ2s
♦ two electrons present in σ*2s
(ii) So the electronic configuration is (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2
(iii) In the above configuration, consider the first two components: (σ1s)2 and (σ*1s)2
• They indicate that
♦ The orbitals related to the 'first main-shell' are completely filled up
• We know that:
♦ The 'first main-shell' is the 'K main-shell'
• We want to indicate that:
♦ The orbitals related to the 'first main-shell' are completely filled up
• For that, we write: KK
♦ So 'KK' indicates: (σ1s)2 and (σ*1s)2
■ Thus the configuration written in (ii) becomes: KK(σ2s)2(σ*2s)2
11. Next we determine the magnetic nature of Be2
• In fig.4.194(b), we see that all the boxes contain paired electrons
♦ So Be2 is diamagnetic
• In the next section, we will see the details related to B2 molecule
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