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

You are watching: C2 2+ paramagnetic or diamagnetic

We find that because #"C"_2# has no unpaired electrons, the is *diamagnetic*.

So then, you"re 90% that the way there. Because *paramagnetism* needs an unpaired electron, is #"C"_2^(-)# paramagnetic or not?

My method begins choose this:

Carbon has accessibility to that**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).

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We don"t need to care about the #1s# electrons; they can be omitted from the MO diagram due to the fact that they"re so short in energy.The #2s# orbital of each carbon

**combine head-on**to kind a #mathbf(sigma_"2s")# bonding and #mathbf(sigma_"2s"^"*")# antibonding molecule orbital.The #2p_x# orbital of every carbon

**combine sidelong**to type a #mathbf(pi_(2p_x))# bonding and #mathbf(pi_(2p_x)^"*")# antibonding molecule orbital.The #2p_y# orbital of every carbon

**combine sidelong**to type a #mathbf(pi_(2p_y))# bonding and also #mathbf(pi_(2p_y)^"*")# antibonding molecule orbital.The #2p_z# orbital of each carbon

**combine head-on**to type a #mathbf(sigma_(2p_z))# bonding and #mathbf(sigma_(2p_z)^"*")# antibonding molecule orbital.

For #"Li"_2#, #"Be"_2#, #"B"_2#, #mathbf("C"_2)# and also #"N"_2#, measures 4 and 5 give:

For #"O"_2# and #"F"_2#, actions 6, 7, and also 8 usually give:

** But...** because that #"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 attain 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 actually no unpaired electrons, it is diamagnetic.

So then, because paramagnetism calls for an unpaired electron, is #"C"_2^(-)# paramagnetic or not?