# 0.7 Chemical bonding and molecular energy levels  (Page 7/7)

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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 $\sigma$ and ${\sigma }^{*}$ pair and two $\pi$ and ${\pi }^{*}$ pairs. The lowering of the energies of the electrons in the $\sigma$ and $\pi$ 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 ${N}_{2}$ because of the curious result that the ionization energy of an electron in ${F}_{2}$ is less than that of an electron in an F atom. By comparing the molecular orbital energy level diagrams for ${N}_{2}$ and ${F}_{2}$ 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 ${N}_{2}$ and ${F}_{2}$ .) We place two electrons in the $\sigma$ orbital, four more in the two ${\pi }^{}$ orbitals, and four more in the two ${\pi }^{*}$ orbitals. Overall, there are six electrons in bonding orbitals and four in anti-bonding orbitals. Since ${F}_{2}$ 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 ${F}_{2}$ 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 ${F}_{2}$ 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, ${O}_{2}$ . 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 ${2p\pi }^{*}$ orbital, or whether they should be separated into different ${2p\pi }^{*}$ 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 ${O}_{2}$ molecule is shown in Fig. 7.

In comparing these three diatomic molecules, we recall that ${N}_{2}$ has the strongest bond, followed by ${O}_{2}$ and ${F}_{2}$ . We have previously accounted for this comparison with Lewis structures, showing that ${N}_{2}$ is a triple bond, ${O}_{2}$ is a double bond, and ${F}_{2}$ 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 ${N}_{2}$ , 4 for ${O}_{2}$ , and six for ${F}_{2}$ . 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

$\text{Bond}\text{Order}=\frac{1}{2}\left(\text{# bonding electrons}-\text{# antibonding electrons}\right)$

Note that, defined this way, the bond order for ${N}_{2}$ is 3, for ${O}_{2}$ is 2, and for ${F}_{2}$ 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 ${H}_{2}$ molecule is almost twice as strong as the bond in the ${H}_{2}^{+}$ ion. Explain why the ${H}_{2}$ bond is less than twice as strong as the ${H}_{2}^{+}$ bond.
• ${\mathrm{Be}}_{2}$ 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 ${F}_{2}$ 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, ${F}_{2}$ is a stable molecule, i.e., the energy of an ${F}_{2}$ 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:
• ${C}_{2}$ or ${C}_{2}^{+}$
• ${B}_{2}$ or ${B}_{2}^{+}$
• ${F}_{2}$ or ${F}_{2}^{-}$
• ${O}_{2}$ or ${O}_{2}^{+}$
• Which of the following diatomic molecules are paramagnetic: CO, ${\mathrm{Cl}}_{2}$ , NO, ${N}_{2}$ ?
• ${B}_{2}$ is observed to be paramagnetic. Using this information, draw an appropriate molecular orbital energy level diagram for ${B}_{2}$ .

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
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
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
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.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
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
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
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.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
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
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.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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