What is meant by the term bond order? Calculate the bond order of: N2, O2, O2+ and O2.

Asked by Pragya Singh | 1 year ago |  232

##### Solution :-

Bond Order: It is defined as 0.5 times the difference between the “No. of electrons present in bonding orbitals and No. of electrons present in anti-bonding orbitals” of a molecule.

Bond Order = 0.5(Nb – Na);

N: No. of anti-bonding electrons

N: No. of bonding electrons

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

Electronic configuration of O2 is:

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

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

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

- 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 $$O_{2}^{-}$$​ (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 $$O_{2}^{+}$$​ 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

Electronic configuration of N2 is:

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

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

No. of bonding electrons = Nb = 10

No. of anti-bonding electrons = Na = 4

Now,

Bond order = 0.5(10 – 4) = 3

Answered by Abhisek | 1 year ago

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