Applications of the Molecular Orbital Theory (10.2.5) Flashcards
• In molecular orbital theory (MO theory), atomic orbitals from different atoms
are mixed to form molecular orbitals.
• In molecular orbital theory (MO theory), atomic orbitals from different atoms
are mixed to form molecular orbitals.
• MO theory predicts that dinitrogen (N2) is diamagnetic while dioxygen (O2) is
paramagnetic.
• MO theory predicts that dinitrogen (N2) is diamagnetic while dioxygen (O2) is
paramagnetic.
In molecular orbital theory (MO theory), atomic
orbitals from different atoms are mixed to form
molecular orbitals. Whenever two orbitals are
mixed, a bonding orbital (with high electron
probability density between the nuclei) is formed,
as well as an antibonding orbital (with a node in
electron probability density between the nuclei).
For example, when two 2s orbitals from two nuclei
are mixed, a bonding orbital is formed (σ2s) as well
as an antibonding orbital (σ2s). Sigma molecular
orbitals (such as σ2s) have electron density
between the atoms forming the bond, while pi
molecular orbitals (such as π2py) have electron
density above and below the sigma molecular
orbital.
Molecular orbitals fill from lowest to highest energy
following Hund’s rule and the Pauli exclusion
principle.
In dinitrogen (N2), this results in four filled bonding
orbitals (σ2s, σ2pz, π2px, and π2py) and one filled
antibonding orbital (σ2s), with no unpaired
electrons. Therefore, N2 is diamagnetic (N2 is not
attracted to a magnetic field).
In dioxygen (O2), this results in four filled bonding
orbitals (σ2s, σ2pz, π2px, and π2py), one filled
antibonding orbital (σ2s), and two unpaired
electrons in antibonding orbitals (π2px and π*2py).
Therefore, O2 is paramagnetic (O2 is attracted to a
magnetic field). This property of O2 cannot be
predicted or explained using earlier bonding
theories (Lewis dot structures, VSEPR theory,
and valence bond theory).
In molecular orbital theory (MO theory), atomic
orbitals from different atoms are mixed to form
molecular orbitals. Whenever two orbitals are
mixed, a bonding orbital (with high electron
probability density between the nuclei) is formed,
as well as an antibonding orbital (with a node in
electron probability density between the nuclei).
For example, when two 2s orbitals from two nuclei
are mixed, a bonding orbital is formed (σ2s) as well
as an antibonding orbital (σ2s). Sigma molecular
orbitals (such as σ2s) have electron density
between the atoms forming the bond, while pi
molecular orbitals (such as π2py) have electron
density above and below the sigma molecular
orbital.
Molecular orbitals fill from lowest to highest energy
following Hund’s rule and the Pauli exclusion
principle.
In dinitrogen (N2), this results in four filled bonding
orbitals (σ2s, σ2pz, π2px, and π2py) and one filled
antibonding orbital (σ2s), with no unpaired
electrons. Therefore, N2 is diamagnetic (N2 is not
attracted to a magnetic field).
In dioxygen (O2), this results in four filled bonding
orbitals (σ2s, σ2pz, π2px, and π2py), one filled
antibonding orbital (σ2s), and two unpaired
electrons in antibonding orbitals (π2px and π*2py).
Therefore, O2 is paramagnetic (O2 is attracted to a
magnetic field). This property of O2 cannot be
predicted or explained using earlier bonding
theories (Lewis dot structures, VSEPR theory,
and valence bond theory).
When we looked at assigning the electrons in nitrogen to sites in the 2s and 2p orbitals, we used the 2s anti-bonding site. Which statement about this site is not true?
The energy level for this site is very similar to the energy level for the σ 2s orbital site. (D)
This is false. It is important to understand that, within any orbital, the σ (or π) sites have lower energy levels than their counterpart σ* (or π*) sites.
Which of the following shows the correct order—from least to greatest energy—for the bond types found in 2s and 2p orbitals for the O2 molecule (for n = 2)?
one 2s σ bond, one 2s σ* bond, one 2p σ bond, two 2p π bonds, two 2p π* bonds, one 2p σ* bond (B)
This shows the six orbital sites in the 2s and 2p orbitals in order from least energy to greatest energy. This is the order in which electrons are assigned orbital sites.
Why is the energy ladder only helpful for elements that have numbers of 18 or less?
If the element had a higher atomic number, you would have to include orbital sites higher than 2p. (A)
The 2p orbital can accommodate a total of twelve electrons, while the 2s orbital can accommodate four electrons and the 1s orbital can accommodate two. Therefore, 12 + 4 + 2 = 18, the maximum number of electrons. Because 18 is the atomic number of argon, argon is an appropriate model for elements of atomic numbers of 18 or less.
Which statement about electrons is not true?
When there is a minimum number of electrons between atoms, there is a high probability of bonding. (D)
The exact opposite is true. Where there is a maximum number of electrons between atoms, there is a high probability of bonding.
Which statement about the electron site assignments in the oxygen molecule is not correct?
The assignment of two lone electrons in the π* orbital does not have a great effect on any physical properties of the oxygen molecule. (B)
This is not true. The two lone pair electrons cause the oxygen molecule to show magnetic properties. That is why O2 is considered paramagnetic.
Which statement about the forming of orbitals (for n = 2) is not true?
A σ bond and a σ* (anti-) bond have similar energy levels. (C)
This is not true. They have very different energies. A σ* bond requires much more energy to form.
Which statement about the nitrogen molecule is not correct?
The bond order is 4. (C)
The bond order is three. Since there are eight electrons in bonding orbitals and two electrons in anti-bonding orbitals, there is a total of (8 − 2) six electrons (or three pairs) in bonding orbitals.
What is the main weakness of hybridization theory?
It does not acknowledge the fact that electrons do spend time away from their parent nuclei. (C)
This is the most significant weakness of hybridization theory. It does not allow for an electron to assume orbitals far from the home-base nucleus. In fact, this happens quite often and accounts for unique behaviors of molecules.
Which statement about the bonding and anti-bonding sites in the 2s and 2p valence orbitals is not true?
The greatest bond order value is four. (B)
This is not true. The maximum bond order value is three. This occurs when there are a total of ten valence electrons (in dinitrogen). In that situation, there are bonding pairs in the 2s σ site, the 2p σ site, and two of them in the 2p π site. In addition, there is one anti-bonding pair in the 2s σ* site. So there are four bonding pairs and one anti-bonding pair. Therefore, the maximum bond order = 4 − 1 = 3. (As you go to higher energy sites above the 2p π sites, all the sites are anti-bonding sites.)
Look at the lower energy, more stable sites in the 2p level (indicated by the arrows).
Which of the following statements is the weakest explanation for why there are two π orbital sites within the 2p level?
These sites are at lower energies than the π orbital sites in the 2p level. (D)
Although this is true, it does not explain much about the need for two π orbital sites.
