O2 molecules there are 1 sigma bond and 1 pi bond and 2 lone pair of electron for each oxygen ( non bonding electrons ) ?? so how MOT diagram shows 1 sigma and 2 pi bond and there is no non bonding electrons as all orbital combine together ?

1 Answer
Truong-Son N.
Dec 12, 2017

It should all check out. You just have to reinterpret the MO diagram of #"O"_2# in the following way.

The MO diagram of #"O"_2# is:

https://s3mn.mnimgs.com/

for the Lewis structure:

#:ddot"O"=ddot"O":#

The nonbonding electrons come from filled bonding AND antibonding MOs at the same time. So, what you should be looking at to correlate these is:

  • #sigma_(2s)# and #sigma_(2s)^"*"# (both electrons in each orbital)
  • #pi_(2p_x)#, #pi_(2p_y)#, #pi_(2p_x)^"*"#, and #pi_(2p_y)^"*"# (only one electron in each orbital)

There isn't always a neat correlation like this, but this gives #2 cdot 2 + 4 cdot 1 = 8# nonbonding electrons:

#ul(color(blue)bb(uarr) color(white)(darr))" "ul(color(blue)bb(uarr) color(white)(darr))#
#pi_(2p_x)^"*"" "" "pi_(2p_y)^"*"#

#ul(color(blue)bb(uarr) darr)" "ul(color(blue)bb(uarr) darr)#
#pi_(2p_x)" "" "pi_(2p_y)#

#ul(uarr darr)#
#sigma_(2p_z)#

#ul(bbcolor(blue)(uarr darr))#
#sigma_(2s)^"*"#

#ul(bbcolor(blue)(uarr darr))#
#sigma_(2s)#

That leaves the two #sigma_(2p_z)# and the one #pi_(2p_x)# and one #pi_(2p_y)# electrons to make the #sigma# and #pi# bonds, respectively.

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