# 3.11 Molecular orbital theory  (Page 10/26)

 Page 10 / 26

## Key equations

• $\text{bond order}=\phantom{\rule{0.2em}{0ex}}\frac{\left(\text{number of bonding electron}\right)-\left(\text{number of antibonding electrons}\right)}{2}$

## Chemistry end of chapter exercises

Sketch the distribution of electron density in the bonding and antibonding molecular orbitals formed from two s orbitals and from two p orbitals.

How are the following similar, and how do they differ?

(a) σ molecular orbitals and π molecular orbitals

(b) ψ for an atomic orbital and ψ for a molecular orbital

(c) bonding orbitals and antibonding orbitals

(a) Similarities: Both are bonding orbitals that can contain a maximum of two electrons. Differences: σ orbitals are end-to-end combinations of atomic orbitals, whereas π orbitals are formed by side-by-side overlap of orbitals. (b) Similarities: Both are quantum-mechanical constructs that represent the probability of finding the electron about the atom or the molecule. Differences: ψ for an atomic orbital describes the behavior of only one electron at a time based on the atom. For a molecule, ψ represents a mathematical combination of atomic orbitals. (c) Similarities: Both are orbitals that can contain two electrons. Differences: Bonding orbitals result in holding two or more atoms together. Antibonding orbitals have the effect of destabilizing any bonding that has occurred.

If molecular orbitals are created by combining five atomic orbitals from atom A and five atomic orbitals from atom B combine, how many molecular orbitals will result?

Can a molecule with an odd number of electrons ever be diamagnetic? Explain why or why not.

An odd number of electrons can never be paired, regardless of the arrangement of the molecular orbitals. It will always be paramagnetic.

Can a molecule with an even number of electrons ever be paramagnetic? Explain why or why not.

Why are bonding molecular orbitals lower in energy than the parent atomic orbitals?

Bonding orbitals have electron density in close proximity to more than one nucleus. The interaction between the bonding positively charged nuclei and negatively charged electrons stabilizes the system.

Calculate the bond order for an ion with this configuration:

${\left({\text{σ}}_{2s}\right)}^{2}{\left({\text{σ}}_{2s}^{*}\right)}^{2}{\left({\text{σ}}_{2px}\right)}^{2}{\left({\text{π}}_{2py},\phantom{\rule{0.2em}{0ex}}{\text{π}}_{2pz}\right)}^{4}{\left({\text{π}}_{2py}^{*},\phantom{\rule{0.2em}{0ex}}{\text{π}}_{2pz}^{*}\right)}^{3}$

Explain why an electron in the bonding molecular orbital in the H 2 molecule has a lower energy than an electron in the 1 s atomic orbital of either of the separated hydrogen atoms.

The pairing of the two bonding electrons lowers the energy of the system relative to the energy of the nonbonded electrons.

Predict the valence electron molecular orbital configurations for the following, and state whether they will be stable or unstable ions.

(a) ${\text{Na}}_{2}{}^{\text{2+}}$

(b) ${\text{Mg}}_{2}{}^{\text{2+}}$

(c) ${\text{Al}}_{2}{}^{\text{2+}}$

(d) ${\text{Si}}_{2}{}^{\text{2+}}$

(e) ${\text{P}}_{2}{}^{\text{2+}}$

(f) ${\text{S}}_{2}{}^{\text{2+}}$

(g) ${\text{F}}_{2}{}^{\text{2+}}$

(h) ${\text{Ar}}_{2}{}^{\text{2+}}$

Determine the bond order of each member of the following groups, and determine which member of each group is predicted by the molecular orbital model to have the strongest bond.

(a) H 2 , ${\text{H}}_{2}{}^{\text{+}},$ ${\text{H}}_{2}{}^{\text{−}}$

(b) O 2 , ${\text{O}}_{2}{}^{\text{2+}},$ ${\text{O}}_{2}{}^{\text{2−}}$

(c) Li 2 , ${\text{Be}}_{2}{}^{\text{+}},$ Be 2

(d) F 2 , ${\text{F}}_{2}{}^{\text{+}},$ ${\text{F}}_{2}{}^{\text{−}}$

(e) N 2 , ${\text{N}}_{2}{}^{\text{+}},$ ${\text{N}}_{2}{}^{\text{−}}$

(a) H 2 bond order = 1, ${\text{H}}_{2}{}^{\text{+}}$ bond order = 0.5, ${\text{H}}_{2}{}^{\text{−}}$ bond order = 0.5, strongest bond is H 2 ; (b) O 2 bond order = 2, ${\text{O}}_{2}{}^{\text{2+}}$ bond order = 3; ${\text{O}}_{2}{}^{\text{2−}}$ bond order = 1, strongest bond is ${\text{O}}_{2}{}^{\text{2+}};$ (c) Li 2 bond order = 1, ${\text{Be}}_{2}{}^{\text{+}}$ bond order = 0.5, Be 2 bond order = 0, strongest bond is ${\text{Li}}_{2}$ ;(d) F 2 bond order = 1, ${\text{F}}_{2}{}^{\text{+}}$ bond order = 1.5, ${\text{F}}_{2}{}^{\text{−}}$ bond order = 0.5, strongest bond is ${\text{F}}_{2}{}^{\text{+}};$ (e) N 2 bond order = 3, ${\text{N}}_{2}{}^{\text{+}}$ bond order = 2.5, ${\text{N}}_{2}{}^{\text{−}}$ bond order = 2.5, strongest bond is N 2

For the first ionization energy for an N 2 molecule, what molecular orbital is the electron removed from?

Compare the atomic and molecular orbital diagrams to identify the member of each of the following pairs that has the highest first ionization energy (the most tightly bound electron) in the gas phase:

(a) H and H 2

(b) N and N 2

(c) O and O 2

(d) C and C 2

(e) B and B 2

(a) H 2 ; (b) N 2 ; (c) O; (d) C 2 ; (e) B 2

Which of the period 2 homonuclear diatomic molecules are predicted to be paramagnetic?

A friend tells you that the 2 s orbital for fluorine starts off at a much lower energy than the 2 s orbital for lithium, so the resulting σ 2 s molecular orbital in F 2 is more stable than in Li 2 . Do you agree?

Yes, fluorine is a smaller atom than Li, so atoms in the 2 s orbital are closer to the nucleus and more stable.

True or false: Boron contains 2 s 2 2 p 1 valence electrons, so only one p orbital is needed to form molecular orbitals.

What charge would be needed on F 2 to generate an ion with a bond order of 2?

2+

Predict whether the MO diagram for S 2 would show s-p mixing or not.

Explain why ${\text{N}}_{2}{}^{\text{2+}}$ is diamagnetic, while ${\text{O}}_{2}{}^{\text{4+}},$ which has the same number of valence electrons, is paramagnetic.

N 2 has s-p mixing, so the π orbitals are the last filled in ${\text{N}}_{2}{}^{\text{2+}}.$ O 2 does not have s-p mixing, so the σ p orbital fills before the π orbitals.

Using the MO diagrams, predict the bond order for the stronger bond in each pair:

(a) B 2 or ${\text{B}}_{2}{}^{\text{+}}$

(b) F 2 or ${\text{F}}_{2}{}^{\text{+}}$

(c) O 2 or ${\text{O}}_{2}{}^{\text{2+}}$

(d) ${\text{C}}_{2}{}^{\text{+}}$ or ${\text{C}}_{2}{}^{\text{−}}$

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Which of the following will increase the percent of HF that is converted to the fluoride ion in water? (a) addition of NaOH (b) addition of HCl (c) addition of NaF
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