Question

Compare the relative stability of the following species and indicate their magnetic properties;

`"O"_2, "O"_2^+, "O"_2^-`(superoxide), `"O"_2^(2-)`(peroxide)

Answer 1

There are 16 electrons in a molecule of dioxygen, 8 from each oxygen atom. The electronic configuration of oxygen molecule can be written as:

`[sigma - (1"s")]^2[sigma^"*"(1"s")]^2[sigma(2"s")]^2[sigma(1"p"_"z")]^2[pi(2"p"_x)]^2[pi(2"p"_"y")]^2[pi^"*"(2"p"_x)]^1[pi^"*"(2"p"_"y")]^1`

Since the 1s orbital of each oxygen atom is not involved in boding, the number of bonding electrons = 8 = Nb and the number of anti-bonding orbitals = 4 = N_a.

Bond order  = `1/2("N"_"b" - "N"_"a")`

`= 1/2(8-4)`

= 2

Similarly, the electronic configuration of `"O"_2^+` can be written as:

`"KK"[sigma(2"s")]^2[sigma^"*"(2"s")]^2[sigma(2"p"_"z")]^2[pi(2"p"_x)]^2[pi(2"p"_"y")]^2[pi^"*"(2"p"_x)]^1`

`"N"_"b" = 8`

`"N"_"a" = 3`

Bond order of `"O"_2^+ = 1/2(8-3)`

= 2.5

Electronic configuration of `"O"_2^-` ion will be:

`KK[sigma(2"s")]^2[sigma^"*"("2s")]^2[sigma(2"p"_"z")]^2[pi(2"p"_x)]^2[pi(2"p"_"y")]^2[pi^"*"(2"p"_x)]^2[pi^"*"(2"p"_"y")]^1`

N_b = 8

N_a = 5

Bond order of `"O"_2^-` = `1/2 (8-5)`

= 1.5

Electronic configuration of `"O"_2^(2-)` ion will be:

`"KK"[sigma(2"s")]^2[sigma^"*"("2s")]^2[sigma("2p"_"z")]^2[pi("2p"_x)]^2[pi("2p"_"y")]^2[pi^"*"("2p"_x)]^2[pi^"*"(2"p"_"y")]^2`

N_b = 8

N_a = 6

Bond order of  `"O"_2^(2-) = 1/2(8 - 6)` =1

Bond dissociation energy is directly proportional to bond order. Thus, the higher the bond order, the greater will be the stability. On this basis, the order of stability is `"O"_2^+ > "O"_2 > "O"_2^- >"O"_2^(2-)`