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# Is C2 Paramagnetic Or Diamagnetic? Molecular Orbitals: Problems And Solutions

For this we will start at the atomic orbitals and construct a molecular orbital (MO) diagram to be sure.

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We find that since #”C”_2# has no unpaired electrons, it is diamagnetic.

So then, you”re 90% of the way there.

Since paramagnetism requires an unpaired electron, is #”C”_2^(-)# paramagnetic or not?

How many more electrons does #”C”_2^(-)# have than #”C”_2#? Where does it go?Is it unpaired?

My approach begins like this:

Carbon has access to its one #mathbf(1s)#, one #mathbf(2s)#, and three #mathbf(2p)# orbitals (with the #1s# orbital much lower in energy than the #2s# and #2p#”s). We don”t have to care about the #1s# electrons; they can be omitted from the MO diagram because they”re so low in energy.The #2s# orbital of each carbon combine head-on to form a #mathbf(sigma_”2s”)# bonding and #mathbf(sigma_”2s”^”*”)# antibonding molecular orbital.The #2p_x# orbital of each carbon combine sidelong to form a #mathbf(pi_(2p_x))# bonding and #mathbf(pi_(2p_x)^”*”)# antibonding molecular orbital.The #2p_y# orbital of each carbon combine sidelong to form a #mathbf(pi_(2p_y))# bonding and #mathbf(pi_(2p_y)^”*”)# antibonding molecular orbital.The #2p_z# orbital of each carbon combine head-on to form a #mathbf(sigma_(2p_z))# bonding and #mathbf(sigma_(2p_z)^”*”)# antibonding molecular orbital.

For #”Li”_2#, #”Be”_2#, #”B”_2#, #mathbf(“C”_2)# and #”N”_2#, steps 4 and 5 give:

For #”O”_2# and #”F”_2#, steps 6, 7, and 8 basically give:

But… for #”Li”_2#, #”Be”_2#, #”B”_2#, #mathbf(“C”_2)# and #”N”_2#, steps 6, 7, and 8 basically give:

Therefore, combine steps 4-8 to achieve the MO diagram for #”C”_2#:

and #”C”_2# has this configuration:

#(sigma_(1s))^2(sigma_(1s)^”*”)^2stackrel(“valence electrons”)overbrace((sigma_(2s))^2(sigma_(2s)^”*”)^2(pi_(2p_x))^2(pi_(2p_y))^2)#

Since #”C”_2# has no unpaired electrons, it is diamagnetic.

So then, since paramagnetism requires an unpaired electron, is #”C”_2^(-)# paramagnetic or not?