Name: All About Molecular Orbital Theory
Uploaded: 2019-07-19
Duration: 9 min 36 s
Description: The molecular orbital theory revolutionised the study of chemical bonding by approximating the states of bonded electrons. Let's learn more about this theory.

Question

Use the

MO

theory to predict the bond order and the number of unpaired electrons in

O_{2}^{2 -}, O_{2}^{-}, O_{2}, O_{2}^{+}, NO

and

CO

.

Accepted Answer

$O_{2}^{2 -}$ has $18$ electrons ( $O$ configuration $1 s^{2} 2 s^{2} 2 p^{4}$ )

According to molecular orbital theory, $18$ electrons are distributed as

$σ_{1 s^{2}} < σ_{1 s^{2}}^{*} < σ_{2 s^{2}} < σ_{2 s^{2}}^{*} < σ_{2 p_{z}^{2}} < (π_{2 p_{x}^{2}} = π_{2 p_{y}^{2}}) < (π_{2 p_{x}^{2}}^{*} = π_{2 p_{y}^{2}}^{*})$

bond order $= 1 / 2$ (number of electrons in bonding molecular orbitals-number of electrons of anti bonding molecular orbitals)

$= \frac{N_{b} - N_{a}}{2}$

$= (10 - 8) / 2 = 1$

All the electrons are paired up there are no unpaired electrons in $O_{2}^{2 -}$ .

$O_{2}^{-}$ has $17$ electrons these are distributed as

$σ_{1 s^{2}} < σ_{2 s^{2}}^{*} < σ_{2 s^{2}} < σ_{2 s^{2}}^{*} < σ_{2 p_{z}^{2}} < (π_{2 p_{x}^{2}} < π_{2 p_{y}^{2}}) < (π_{2 p_{x}^{2}}^{*} = π_{2 p_{y}^{1}}^{*})$

bond order $= \frac{N_{b} - N_{a}}{2}$

$= (10 - 7) / 2$

$= 1.5$

It has one unpaired electron in $π_{2 p_{y}}^{*}$ .

$O_{2}$ has $16$ electrons which are distributed as

$σ_{1 s^{2}} < σ_{2 s^{2}}^{*} < σ_{2 s^{2}} < σ_{2 s^{2}}^{*} < σ_{2 p_{z}^{2}} < (π_{2 p_{x}^{2}} = π_{2 p_{y}^{2}}) < (π_{2 p_{x}^{1}}^{*} = π_{2 p_{y}^{1}}^{*})$

bond order $= \frac{N_{b} - N_{a}}{2}$

$= (10 - 6) / 2$

$= 2$

It has two unpaired electrons in $π_{2 p_{x}}^{*}, π_{2 p_{y}}^{*}$ .

$O_{2}^{+}$ has $15$ electrons they are distributed as

$σ_{1 s^{2}} < σ_{1 s^{2}}^{*} < σ_{2 s^{2}} < σ_{2 s^{2}}^{*} < σ_{2 p_{z}^{2}} < (π_{2 p_{x}^{2}} = π_{2 p_{y}^{2}}) < (π_{2 p_{x}^{1}}^{*} = π_{2 p_{y}^{1}}^{*})$

bond order $= (10 - 5) / 2$

$= 2.5$

It has one unpaired electron in $π_{2 p_{x}}^{*}$ .

$NO$ has $15$ electrons ( $N$ electronic configuration is $1 s^{2} 2 s^{2} 2 p^{3}$ ) and ( $1 s^{2} 2 s^{2} 2 p^{4}$ is the electronic configuration of $O$ )

$σ_{1 s^{2}} < σ_{1 s^{2}}^{*} < σ_{2 s^{2}} < σ_{2 s^{2}}^{*} < σ_{2 p_{z}^{2}} < (π_{2 p_{x}^{2}} = π_{2 p_{y}^{2}}) < (π_{2 p_{x}^{1}}^{*} = π_{2 p_{x}^{2}}^{*})$

bond order $= (10 - 5) / 2$

$= 2.5$

It has one unpaired electron in $π_{2 p_{x}}^{*}$ .

$CO$ has $14$ electrons ( $C$ electronic configuration is $1 s^{2} 2 s^{2} 2 p^{2}$ and $1 s^{2} 2 s^{2} 2 p^{4}$ is the electronic configuration of $O$ )

$σ_{1 s^{2}} < σ_{1 s^{2}}^{*} < σ_{2 s^{2}} < σ_{2 s^{2}}^{*} < (π_{2 p_{x}^{2}} = π_{2 p_{y}^{2}}) < σ_{2 p_{z}^{2}}$

bond order $= (10 - 4) / 2$

$= 3$

It has no unpaired electrons.

Use the $MO$ theory to predict the bond order and the number of unpaired electrons in $O_{2}^{2 -}, O_{2}^{-}, O_{2}, O_{2}^{+}, NO$ and $CO$ .

Important Questions on The Covalent Bond

Learn and Prepare for any exam you want !

Use the MO theory to predict the bond order and the number of unpaired electrons in O22-, O2-, O2, O2+, NO and CO.

Important Questions on The Covalent Bond

Learn and Prepare for any exam you want !

Use the $MO$ theory to predict the bond order and the number of unpaired electrons in $O_{2}^{2 -}, O_{2}^{-}, O_{2}, O_{2}^{+}, NO$ and $CO$ .