10.2.6 Beyond Homonuclear Diatomics Flashcards Preview

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Flashcards in 10.2.6 Beyond Homonuclear Diatomics Deck (13)
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1
Q

Beyond Homonuclear Diatomics

A
  • Molecules containing atoms of similar electronegativities will have orbital energies that vary only slightly from homonuclear diatomic molecules.
  • Heteronuclear diatomics of very different electronegativities require the construction of a molecular orbital energy ladder appropriate to the placement of the valence electrons.
  • Electrons in delocalized bonds can move throughout the atoms of a given molecule, providing a low energy configuration with increased stability.
2
Q

note 1

A
  • Carbon monoxide is an example of a heteronuclear diatomic molecule. There are a total of 10 valence electrons in carbon monoxide.
  • Combining the electrons of the 2s and 2p atomic orbitals of carbon and oxygen to form molecular orbitals predicts a bond order of 3, indicating that there is a triple bond between carbon and oxygen.
  • The triple bond is formed from a sigma bond between two of the p orbitals and two pi bonds between the remaining p orbitals.
  • Since oxygen is more electronegative than carbon, the oxygen orbitals are slightly lower in energy than the carbon orbitals.
  • Hydrogen fluoride (HF) is a heteronuclear diatomic molecule with a very large difference in electronegativity between hydrogen and fluorine. This large difference in
    electronegativity causes the atomic orbitals of fluorine to be much lower in energy than the atomic orbitals of hydrogen.
  • The best bonding interaction that occurs between hydrogen and fluorine takes place between the 1s orbital of hydrogen and the 2p orbital of fluorine that is oriented towards hydrogen. These two atomic orbitals form a sigma bonding orbital ( ) as well as a sigma antibonding orbital ( *).
  • The 2s orbital of fluorine and the other two 2p orbitals of fluorine remain the same and may still hold electrons.
  • All eight valence electrons from HF are then used to fill the molecular orbital energy ladder. Most of the electrons (two 2s and four 2p electrons) are associated with F, while the two electrons in the sigma bond are associated with both H and F.
3
Q

note 2

A
  • The orbital diagram for butadiene shows the four possible orbital arrangements. The lowest energy arrangement (bottom of diagram) has no antibonding interactions. The next orbital has one antibonding interaction. The third orbital arrangement has two, and the highest energy arrangement has three antibonding interactions.
  • Electrons are placed in the orbitals from lowest energy to highest. Since there are four electrons in the system (one per p orbital) they fill only the bonding orbitals.
  • The orbital diagram predicts that the bond between the middle two carbons has some double bond character. It also predicts that the molecule is planar, allowing all four p orbitals to line up.
  • A molecule with overlapping pi bonds could have electrons delocalized across the entire length of the molecule.
4
Q

What is the most important extra step that has to be included in the first approach technique for assigning electrons in a heteronuclear diatomic molecule?

A

Account for the electronegativity (and orbital energy) differences between the two atoms.

5
Q

To account for the high electronegativity difference between hydrogen and fluorine, we reassign the orbital sites in HF in order to create an energy ladder that gives us a logical sequence for assigning electrons. Which statement about the reassigned orbital sites in HF is true?

A

There are two molecular orbital sites and three atomic orbital sites.

6
Q

Which of the following is not characteristic of HF?

A

HF involves just atomic orbitals because the hydrogen atom just has one valence electron.

7
Q

Which statement best explains the idea of delocalized electrons?

A

Delocalized electrons are electrons that are spread out in a molecule (instead of being confined to atomic orbitals). This helps explain some of the unique physical properties (color, magnetism, and so on) of molecules.

8
Q

Which statement best explains why the bonding between the hydrogen and fluorine atoms occurs where it does?

A

The fluorine atom has such lower bonding energy levels that its 2px orbital is the one that is best poised for bonding with the 1s orbital of hydrogen.

9
Q

Which statement about molecular bond theory is not true?

A

Molecular bond theory rejects valence bond theory and explains the bonding exclusively in terms of molecular (not atomic) bonding.

10
Q

We learned that four new molecular orbitals are created from the p orbitals for each carbon atom in the butadiene molecule (the axes perpendicular to the page). Which statement about these molecular orbitals is NOT true?

A

The four molecular p orbitals of the four carbon atoms can be lined up parallel in order to create four new molecular orbitals.

11
Q

In our first approach to assigning electrons to orbital sites in the CO molecule, we used the same technique that we used before with N2 and O2. Which of the following steps is not part of this first approach for assigning electrons in CO?

A

Assign the ten valence electrons to the bonding orbital, 2s and 2p sites.

12
Q

Which statement gives the best definition for isoelectronic molecules?

A

Isoelectronic molecules are molecules that have the same number of valence electrons and very similar orbitals that, as a result, give the molecules the same shape.

13
Q

Which statement about molecular orbital theory is not true?

A

Molecular orbital theory leads to more accurate bond order values.

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