Is #C_2^+# paramagnetic or diamagnetic? Is #C_2# paramagnetic or diamagnetic?

1 Answer
Apr 5, 2016

#"C"_2# has a similar MO ordering to #"N"_2#.

From bottom to top, we have the following valence MOs:

  1. #\mathbf(sigma_(2s))#, generated from the linear combination #2s_A + 2s_B# (head-on, in-phase)
  2. #\mathbf(sigma_(2s)^"*")#, generated from the linear combination #2s_A - 2s_B# (head-on, out-of-phase)
  3. #\mathbf(pi_(2p_x))#, generated from the linear combination #2p_(x,A) + 2p_(x,B)# (sidelong, in-phase)---degenerate with (4)
  4. #\mathbf(pi_(2p_y))#, generated from the linear combination #2p_(y,A) + 2p_(y,B)# (sidelong, in-phase)---degenerate with (3)
  5. #\mathbf(sigma_(2p_z))#, generated from the linear combination #2p_(z,A) + 2p_(z,B)# (head-on, in-phase)
  6. #\mathbf(pi_(2p_x)^"*")#, generated from the linear combination #2p_(x,A) - 2p_(x,B)# (sidelong, out-of-phase)---degenerate with (7)
  7. #\mathbf(pi_(2p_y)^"*")#, generated from the linear combination #2p_(y,A) - 2p_(y,B)# (sidelong, out-of-phase)---degenerate with (6)
  8. #\mathbf(sigma_(2p_z)^"*")#, generated from the linear combination #2p_(z,A) - 2p_(z,B)# (head-on, out-of-phase)

You can see the compatible bonding orbital combinations here (reverse one of the orbitals' signs to achieve the corresponding antibonding orbital combination):

As a result, we have the following MO diagram (with atomic orbital energies from Inorganic Chemistry, Miessler et al., pg. 134):

Thus, with no unpaired electrons, neutral #"C"_2# is diamagnetic.

If #"C"_2 -> "C"_2^(+) + e^(-)#, then is #"C"_2^(+)# paramagnetic or diamagnetic? Paramagnetism requires unpaired electrons.