For this us will start at the atomic orbitals and construct a molecular orbital (MO) diagram to it is in sure.

You are watching: C2 2+ paramagnetic or diamagnetic

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

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

How many an ext electrons go #"C"_2^(-)# have actually than #"C"_2#? where does that go?Is the unpaired?

My strategy begins favor this:

Carbon has accessibility to that is one #\mathbf(1s)#, one #\mathbf(2s)#, and three #\mathbf(2p)# orbitals (with the #1s# orbital much lower in energy than the #2s# and also #2p#"s).

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We don"t need to care around the #1s# electrons; they deserve to be omitted from the MO diagram due to the fact that they"re so short in energy.The #2s# orbital of every carbon combine head-on to kind a #\mathbf(sigma_"2s")# bonding and #\mathbf(sigma_"2s"^"*")# antibonding molecule orbital.The #2p_x# orbit of every carbon combine sidelong to form a #\mathbf(pi_(2p_x))# bonding and also #\mathbf(pi_(2p_x)^"*")# antibonding molecular orbital.The #2p_y# orbit of each 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 every carbon combine head-on to form 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 #"N"_2#, measures 4 and 5 give:

*

For #"O"_2# and #"F"_2#, procedures 6, 7, and 8 basically give:

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But... because that #"Li"_2#, #"Be"_2#, #"B"_2#, #\mathbf("C"_2)# and #"N"_2#, actions 6, 7, and 8 basically give:

*

Therefore, combine steps 4-8 to attain the MO diagram because that #"C"_2#:

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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, the is diamagnetic.

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