Question #3aacd

1 Answer
Dec 11, 2016

Answer:

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

Explanation:

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

i.stack.imgur.com

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.

O2
(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

www.learnnext.com

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

O2-
(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.