Is #C_2 ^-#paramagnetic or diamagnetic?

2 Answers
Jun 29, 2016

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

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?

  1. How many more electrons does #"C"_2^(-)# have than #"C"_2#?
  2. Where does it go?
  3. Is it unpaired?

My approach begins like this:

  1. 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.
  2. Each carbon has four valence electrons: two occupy the same #2s# orbital, and two singly occupy two of the three #2p# orbitals.
  3. Since one carbon has four valence electrons, two carbons bonded together must have a total of eight. This gives the number of valence electrons in #"C"_2#, but not #"C"_2^(-)#.
  4. The #1s# orbital of each carbon combine head-on to form a #\mathbf(sigma_"1s")# bonding and #\mathbf(sigma_"1s"^"*")# antibonding molecular orbital.
  5. The #2s# orbital of each carbon combine head-on to form a #\mathbf(sigma_"2s")# bonding and #\mathbf(sigma_"2s"^"*")# antibonding molecular orbital.
  6. 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.
  7. 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.
  8. 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?

Jun 29, 2016

Answer:

This is probably a paramagnet.

Explanation:

A Lewis structure of the acetylide ion is:

#""^(-):C-=C:^-#. Adn thus for the monoanion, whatever you call it:

#""^(-):C-=C*#, which is a formal paramagnet. Of course, the electron and the negative charge is delocalized over two carbon centres, nevertheless there is a formal lone electron.