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This is depicted in Fig. 5 in what is called a “molecular orbital energy diagram.” Each pair of atomic orbitals, one from each atom, is overlapped to form a bonding and an anti-bonding orbital. The three 2p orbitals from each atom form one σ size 12{σ} {} and σ * size 12{σ rSup { size 8{1} } } {} pair and two π size 12{π} {} and π * size 12{σ rSup { size 8{1} } } {} pairs. The lowering of the energies of the electrons in the σ size 12{σ} {} and π size 12{π} {} orbitals is apparent. The ten n=2 electrons from the nitrogen atoms are then placed pairwise, in order of increasing energy, into these molecular orbitals. Note that, in agreement with the Pauli Exclusion Principle, each pair in a single orbital consists of one spin up and one spin down electron.

Recall now that we began the discussion of bonding in N 2 size 12{N rSub { size 8{2} } } {} because of the curious result that the ionization energy of an electron in F 2 size 12{F rSub { size 8{2} } } {} is less than that of an electron in an F atom. By comparing the molecular orbital energy level diagrams for N 2 size 12{N rSub { size 8{2} } } {} and F 2 size 12{F rSub { size 8{2} } } {} we are now prepared to answer this puzzle. There are five p electrons in each fluorine atom. These ten electrons must be distributed over the molecular orbitals whose energies are shown in Fig. 6. (Note that the ordering of the bonding 2p orbitals differ between N 2 size 12{N rSub { size 8{2} } } {} and F 2 size 12{F rSub { size 8{2} } } {} .) We place two electrons in the σ size 12{σ} {} orbital, four more in the two π size 12{π rSup { size 8{ * } } } {} orbitals, and four more in the two π * size 12{σ rSup { size 8{1} } } {} orbitals. Overall, there are six electrons in bonding orbitals and four in anti-bonding orbitals. Since F 2 size 12{F rSub { size 8{2} } } {} is a stable molecule, we must conclude that the lowering of energy for the electrons in the bonding orbitals is greater than the raising of energy for the electrons in the antibonding orbitals. Overall, this distribution of electrons is, net, equivalent to having two electrons paired in a single bonding orbital.

This also explains why the ionization energy of F 2 size 12{F rSub { size 8{2} } } {} is less than that of an F atom. The electron with the highest energy requires the least energy to remove from the molecule or atom. The molecular orbital energy diagram in Fig. 6 clearly shows that the highest energy electrons in F 2 size 12{F rSub { size 8{2} } } {} are in anti-bonding orbitals. Therefore, one of these electrons is easier to remove than an electron in an atomic 2p orbital, because the energy of an anti-bonding orbital is higher than that of the atomic orbitals. (Recall that this is why an anti-bonding orbital is, indeed, anti-bonding.) Therefore, the ionization energy of molecular fluorine is less than that of atomic fluorine. This clearly demonstrates the physical reality and importance of the anti-bonding orbitals.

A particularly interesting case is the oxygen molecule, O 2 size 12{O rSub { size 8{2} } } {} . In completing the molecular orbital energy level diagram for oxygen, we discover that we must decide whether to pair the last two electrons in the same 2pπ * size 12{2pπ rSup { size 8{ * } } } {} orbital, or whether they should be separated into different 2pπ * size 12{2pπ rSup { size 8{ * } } } {} orbitals. To determine which, we note that oxygen molecules are paramagnetic, meaning that they are strongly attracted to a magnetic field. To account for this paramagnetism, we recall that electron spin is a magnetic property. In most molecules, all electrons are paired, so for each “spin up” electron there is a “spin down” electron and their magnetic fields cancel out. When all electrons are paired, the molecule is diamagnetic meaning that it responds only weakly to a magnetic field.

If the electrons are not paired, they can adopt the same spin in the presence of a magnetic field. This accounts for the attraction of the paramagnetic molecule to the magnetic field. Therefore, for a molecule to be paramagnetic, it must have unpaired electrons. The correct molecular orbital energy level diagram for an O 2 size 12{O rSub { size 8{2} } } {} molecule is shown in Fig. 7.

In comparing these three diatomic molecules, we recall that N 2 size 12{N rSub { size 8{2} } } {} has the strongest bond, followed by O 2 size 12{O rSub { size 8{2} } } {} and F 2 size 12{F rSub { size 8{2} } } {} . We have previously accounted for this comparison with Lewis structures, showing that N 2 size 12{N rSub { size 8{2} } } {} is a triple bond, O 2 size 12{O rSub { size 8{2} } } {} is a double bond, and F 2 size 12{F rSub { size 8{2} } } {} is a single bond. The molecular orbital energy level diagrams in Figs. 5 to 7 cast a new light on this analysis. Note that, in each case, the number of bonding electrons in these molecules is eight. The difference in bonding is entirely due to the number of antibonding electrons: 2 for N 2 size 12{N rSub { size 8{2} } } {} , 4 for O 2 size 12{O rSub { size 8{2} } } {} , and six for F 2 size 12{F rSub { size 8{2} } } {} . Thus, the strength of a bond must be related to the relative numbers of bonding and antibonding electrons in the molecule. Therefore, we now define the bond order as

Bond Order = 1 2 # bonding electrons # antibonding electrons size 12{ ital "Bond"` ital "Order"= { {1} over {2} } left ( ital "bonding"` ital "electrons" - ital "antibonding"` ital "electrons" right )} {}

Note that, defined this way, the bond order for N 2 size 12{N rSub { size 8{2} } } {} is 3, for O 2 size 12{O rSub { size 8{2} } } {} is 2, and for F 2 size 12{F rSub { size 8{2} } } {} is 1, which agrees with our conclusions from Lewis structures. We conclude that we can predict the relative strengths of bonds by comparing bond orders.

Review and discussion questions

  • Why does an electron shared by two nuclei have a lower potential energy than an electron on a single atom? Why does an electron shared by two nuclei have a lower kinetic energy than an electron on a single atom? How does this sharing result in a stable molecule? How can this affect be measured experimentally?
  • Explain why the bond in an H 2 size 12{H rSub { size 8{2} } } {} molecule is almost twice as strong as the bond in the H 2 + size 12{H rSub { size 8{2} } rSup { size 8{+{}} } } {} ion. Explain why the H 2 size 12{H rSub { size 8{2} } } {} bond is less than twice as strong as the H 2 + size 12{H rSub { size 8{2} } rSup { size 8{+{}} } } {} bond.
  • Be 2 size 12{H rSub { size 8{2} } } {} is not a stable molecule. What information can we determine from this observation about the energies of molecular orbitals?
  • Less energy is required to remove an electron from an F 2 size 12{F rSub { size 8{2} } } {} molecule than to remove an electron from an F atom. Therefore, the energy of that electron is higher in the molecule than in the atom. Explain why, nevertheless, F 2 size 12{F rSub { size 8{2} } } {} is a stable molecule, i.e., the energy of an F 2 size 12{F rSub { size 8{2} } } {} molecule is less than the energy of two F atoms.
  • Why do the orbitals of an atom "hybridize" when forming a bond?
  • Calculate the bond orders of the following molecules and predict which molecule in each pair has the stronger bond:
    • C 2 size 12{C rSub { size 8{2} } } {} or C 2 + size 12{H rSub { size 8{2} } rSup { size 8{+{}} } } {}
    • B 2 size 12{B rSub { size 8{2} } } {} or B 2 + size 12{H rSub { size 8{2} } rSup { size 8{+{}} } } {}
    • F 2 size 12{F rSub { size 8{2} } } {} or F 2 - size 12{H rSub { size 8{2} } rSup { size 8{+{}} } } {}
    • O 2 size 12{O rSub { size 8{2} } } {} or O 2 + size 12{H rSub { size 8{2} } rSup { size 8{+{}} } } {}
  • Which of the following diatomic molecules are paramagnetic: CO, Cl 2 size 12{ ital "Cl" rSub { size 8{2} } } {} , NO, N 2 size 12{N rSub { size 8{2} } } {} ?
  • B 2 size 12{B rSub { size 8{2} } } {} is observed to be paramagnetic. Using this information, draw an appropriate molecular orbital energy level diagram for B 2 size 12{B rSub { size 8{2} } } {} .

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, General chemistry i. OpenStax CNX. Jul 18, 2007 Download for free at http://cnx.org/content/col10263/1.3
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