Question 3aacd

Dec 11, 2016

$\text{O"_2^"+}$ has the strongest $\text{O—O}$ bond.

Explanation:

The blank molecular orbital diagram for $\text{O, F}$, and $\text{Ne}$ is Molecular orbitals for ${\text{O}}_{2}$

Each $\text{O}$ atom has 6 valence electrons, so the ${\text{O}}_{2}$ 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 ${\text{O}}_{2}^{+}$

We remove one electron from the highest energy orbital of ${\text{O}}_{2}$ to get In this diagram,

$\text{BO} = \frac{1}{2} \left(10 - 5\right) = \frac{5}{2} = 2.5$

Molecular orbitals for $\text{O"_2^"-}$

We add an electron to the molecular orbital diagram for ${\text{O}}_{2}$ and get (Adapted from Chemistry - Stack Exchange)

$\text{BO} = \frac{1}{2} \left(8 - 5\right) = \frac{3}{2} = 1.5$

$\text{O"_2^"+}$ has the strongest $\text{O-O}$ bond because it has the highest bond order.