# 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 ${\text{C}}_{2}$ has no unpaired electrons, it is diamagnetic.

So then, you're 90% of the way there. Since paramagnetism requires an unpaired electron, is ${\text{C}}_{2}^{-}$ paramagnetic or not?

1. How many more electrons does ${\text{C}}_{2}^{-}$ have than ${\text{C}}_{2}$?
2. Where does it go?
3. Is it unpaired?

My approach begins like this:

1. Carbon has access to its one $\setminus m a t h b f \left(1 s\right)$, one $\setminus m a t h b f \left(2 s\right)$, and three $\setminus m a t h b f \left(2 p\right)$ orbitals (with the $1 s$ orbital much lower in energy than the $2 s$ and $2 p$'s). We don't have to care about the $1 s$ 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 $2 s$ orbital, and two singly occupy two of the three $2 p$ 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 ${\text{C}}_{2}$, but not ${\text{C}}_{2}^{-}$.
4. The $1 s$ orbital of each carbon combine head-on to form a $\setminus m a t h b f \left({\sigma}_{\text{1s}}\right)$ bonding and $\setminus m a t h b f \left({\sigma}_{\text{1s"^"*}}\right)$ antibonding molecular orbital.
5. The $2 s$ orbital of each carbon combine head-on to form a $\setminus m a t h b f \left({\sigma}_{\text{2s}}\right)$ bonding and $\setminus m a t h b f \left({\sigma}_{\text{2s"^"*}}\right)$ antibonding molecular orbital.
6. The $2 {p}_{x}$ orbital of each carbon combine sidelong to form a $\setminus m a t h b f \left({\pi}_{2 {p}_{x}}\right)$ bonding and $\setminus m a t h b f \left({\pi}_{2 {p}_{x}}^{\text{*}}\right)$ antibonding molecular orbital.
7. The $2 {p}_{y}$ orbital of each carbon combine sidelong to form a $\setminus m a t h b f \left({\pi}_{2 {p}_{y}}\right)$ bonding and $\setminus m a t h b f \left({\pi}_{2 {p}_{y}}^{\text{*}}\right)$ antibonding molecular orbital.
8. The $2 {p}_{z}$ orbital of each carbon combine head-on to form a $\setminus m a t h b f \left({\sigma}_{2 {p}_{z}}\right)$ bonding and $\setminus m a t h b f \left({\sigma}_{2 {p}_{z}}^{\text{*}}\right)$ antibonding molecular orbital.

For ${\text{Li}}_{2}$, ${\text{Be}}_{2}$, ${\text{B}}_{2}$, $\setminus m a t h b f \left({\text{C}}_{2}\right)$ and ${\text{N}}_{2}$, steps 4 and 5 give:

For ${\text{O}}_{2}$ and ${\text{F}}_{2}$, steps 6, 7, and 8 basically give:

But... for ${\text{Li}}_{2}$, ${\text{Be}}_{2}$, ${\text{B}}_{2}$, $\setminus m a t h b f \left({\text{C}}_{2}\right)$ and ${\text{N}}_{2}$, steps 6, 7, and 8 basically give:

Therefore, combine steps 4-8 to achieve the MO diagram for ${\text{C}}_{2}$:

and ${\text{C}}_{2}$ 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 ${\text{C}}_{2}$ has no unpaired electrons, it is diamagnetic.

So then, since paramagnetism requires an unpaired electron, is ${\text{C}}_{2}^{-}$ paramagnetic or not?

Jun 29, 2016

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.