10.2.5 Applications of the Molecular Orbital Theory Flashcards Preview

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Flashcards in 10.2.5 Applications of the Molecular Orbital Theory Deck (12)
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1
Q

Applications of the Molecular Orbital Theory

A
  • In molecular orbital theory (MO theory), atomic orbitals from different atoms are mixed to form molecular orbitals.
  • MO theory predicts that dinitrogen (N 2 ) is diamagnetic while dioxygen (O 2 ) is paramagnetic.
2
Q

note

A
  • 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 ( 2s ) as well as an antibonding orbital ( * 2s ). Sigma molecular orbitals (such as 2s ) have electron density between the atoms forming the bond, while pi molecular orbitals (such as 2py ) 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 (N 2 ), this results in four filled bonding orbitals (sigma 2s, sigma 2pz, pi 2px, and pi 2py ) and one filled antibonding orbital ( sigma* 2s ), with no unpaired electrons. Therefore, N 2 is diamagnetic (N 2 is not attracted to a magnetic field).
  • In dioxygen (O 2 ), this results in four filled bonding orbitals (sigma 2s, sigma 2pz, pi 2px, and pi 2py), one filled antibonding orbital (sigma* 2s ), and two unpair
3
Q

Which statement about electrons is not true?

A

When there is a minimum number of electrons between atoms, there is a high probability of bonding

4
Q

What is the main weakness of hybridization theory?

A

It does not acknowledge the fact that electrons do spend time away from their parent nuclei.

5
Q

Which statement about the nitrogen molecule is not correct?

A

The bond order is 4.

6
Q

Why is the energy ladder only helpful for elements that have numbers of 18 or less?

A

If the element had a higher atomic number, you would have to include orbital sites higher than 2p.

7
Q

Which statement about the bonding and anti-bonding sites in the 2s and 2p valence orbitals is not true?

A

The greatest bond order value is four.

8
Q

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?

A

The energy level for this site is very similar to the energy level for the σ 2s orbital site.

9
Q

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?

A

These sites are at lower energies than the π orbital sites in the 2p level.

10
Q

Which statement about the electron site assignments in the oxygen molecule is not correct?

A

The assignment of two lone electrons in the π* orbital does not have a great effect on any physical properties of the oxygen molecule.

11
Q

Which statement about the forming of orbitals (for n = 2) is not true?

A

A σ bond and a σ* (anti-) bond have similar energy levels.

12
Q

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)?

A

one 2s σ bond, one 2s σ* bond, one 2p σ bond, two 2p π bonds, two 2p π* bonds, one 2p σ* bond

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