# Is #C_2 ^-#paramagnetic or diamagnetic?

##### 2 Answers

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

We find that since *diamagnetic*.

So then, you're 90% of the way there. Since *paramagnetism* requires an unpaired electron, is

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

My approach begins like this:

- 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. - Each carbon has
**four**valence electrons:*two*occupy the same#2s# orbital, and*two*singly occupy two of the three#2p# orbitals. - 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^(-)# . - The
#1s# orbital of each carbon**combine head-on**to form a#\mathbf(sigma_"1s")# bonding and#\mathbf(sigma_"1s"^"*")# antibonding molecular orbital. - The
#2s# orbital of each carbon**combine head-on**to form a#\mathbf(sigma_"2s")# bonding and#\mathbf(sigma_"2s"^"*")# antibonding molecular orbital. - 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. - 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. - 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

For

** But...** for

Therefore, combine steps 4-8 to achieve the MO diagram for

and

#(sigma_(1s))^2(sigma_(1s)^"*")^2stackrel("valence electrons")overbrace((sigma_(2s))^2(sigma_(2s)^"*")^2(pi_(2p_x))^2(pi_(2p_y))^2)#

Since

So then, since paramagnetism requires an unpaired electron, is

This is probably a paramagnet.

#### Explanation:

A Lewis structure of the acetylide ion is: