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

$$O_2$$$$O_{2}^{+},$$ $$O_{2}^{-}​$$ (Superoxide), $$O_{2}^{2-}$$ (Peroxide)

Asked by Pragya Singh | 1 year ago |  144

##### Solution :-

O2 contain 16 electrons i.e. 8 electrons from each O- atom.

Electronic configuration of O2 is:

$$[\sigma – (1s)]^{2}[\sigma ^{*}(1s)]^{2}[\sigma (2s)]^{2}[\sigma ^{*}(2s)]^{2}[\sigma (1p_{z})]^{2}$$

$$[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi^{*} (2p_{x})]^{1}[\pi^{*} (2p_{y})]^{1}$$

As 1s- orbital of each O- atom does not involve in the bonding,

No. of bonding electrons = Nb = 8

No. of anti-bonding electrons = Na = 4

Now,

Bond order = 0.5(8 – 4) = 2

Electronic configuration of $$O_{2}^{+}$$​ is:

$$KK[\sigma(2s)]^{2}[\sigma ^{*}(2s)]^{2}[\sigma (2p_{z})]^{2}$$

$$[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi^{*} (2p_{x})]^{1}$$

No. of bonding electrons = Nb = 8

No. of anti-bonding electrons = Na = 3

Now,

Bond order = 0.5(8 – 3) = 2.5

As, bond dissociation energy $$\propto$$ bond order

Hence, higher the bond order, higher stability will be there.

Arrangement according to decreasing order of stability is given as:

$$O_{2}^{+} > O_{2} > O_{2}^{-} > O_{2}^{2-}$$

Electronic configuration of $$O_{2}^{-}$$​ (Superoxide) is:

$$KK[\sigma(2s)]^{2}[\sigma ^{*}(2s)]^{2}[\sigma (2p_{z})]^{2}$$

$$[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi^{*} (2p_{x})]^{2}[\pi^{*} (2p_{y})]^{1}$$

No. of bonding electrons = Nb = 8

No. of anti-bonding electrons = Na = 5

Now,

Bond order = 0.5(8 – 5) = 1.5

Electronic configuration of $$O_{2}^{2-}$$​ (peroxide) is:

$$KK[\sigma(2s)]^{2}[\sigma ^{*}(2s)]^{2}[\sigma (2p_{z})]^{2}$$

$$[\pi (2p_{x})]^{2}[\pi (2p_{y})]^{2}[\pi^{*} (2p_{x})]^{2}[\pi^{*} (2p_{y})]^{2}$$

No. of bonding electrons = Nb = 8

No. of anti-bonding electrons = Na = 6

Now,

Bond order = 0.5(8 – 6) = 1

Answered by Abhisek | 1 year ago

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