Question

Is `C_2 ^-`paramagnetic or diamagnetic?

Answer 1

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?

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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?

Answer 2

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.