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

##### 1 Answer

Apr 5, 2016

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

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

If **Paramagnetism** requires *unpaired* electrons.