C2 2- Diamagnetic Or Paramagnetic

For this we volition first at the atomic orbitals and construct a molecular orbital (MO) diagram to be certain.

We discover that since #"C"_2# has no unpaired electrons, it is diamagnetic.

So and so, you lot're xc% of the way there. Since paramagnetism requires an unpaired electron, is #"C"_2^(-)# paramagnetic or not?

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

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My arroyo 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 near the #1s# electrons; they can exist omitted from the MO diagram because they're so depression 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 ane carbon has four valence electrons, two carbons bonded together must have a total of viii. This gives the number of valence electrons in #"C"_2#, just non #"C"_2^(-)#.
4. The #1s# orbital of each carbon combine caput-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 course 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 #"Due north"_2#, steps 4 and 5 give:

For #"O"_2# and #"F"_2#, steps half dozen, 7, and eight basically requite:

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 accomplish the MO diagram for #"C"_2#:

and #"C"_2# has this configuration:

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

Since #"C"_2# has no unpaired electrons, information technology is diamagnetic.

So then, since paramagnetism requires an unpaired electron, is #"C"_2^(-)# paramagnetic or not?

C2 2- Diamagnetic Or Paramagnetic,

Source: https://socratic.org/questions/is-c-2-paramagnetic-or-diamagnetic-1

Posted by: holmesswuzzy1946.blogspot.com

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