Question

Question #3aacd

Answer 1

`"O"_2^"+"` has the strongest `"O—O"` bond.

Explanation:

The blank molecular orbital diagram for `"O, F"`, and `"Ne"` is

Molecular orbitals for `"O"_2`

Each `"O"` atom has 6 valence electrons, so the `"O"_2` molecule has 12 valence electrons.

We use the Aufbau Principle and Hund's rule to place these electrons in the atomic and molecular orbitals.

(From www.grandinetti.org)

`"Bond order" = 1/2("bonding electrons -antibonding electrons")`

or

`"BO" = 1/2("B -A") = 1/2(8-4) = 4/2 = 2`

Molecular orbitals for `"O"_2^+`

We remove one electron from the highest energy orbital of `"O"_2` to get

In this diagram,

`"BO" = 1/2(10 - 5) = 5/2 = 2.5`

Molecular orbitals for `"O"_2^"-"`

We add an electron to the molecular orbital diagram for `"O"_2` and get

(Adapted from Chemistry - Stack Exchange)

`"BO" = 1/2(8-5) = 3/2 = 1.5`

`"O"_2^"+"` has the strongest `"O-O"` bond because it has the highest bond order.