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Calculation of the bond order of C2(2-) using the MO-diagram

The problem to solve is to formulate in which way two carbon atoms are described by using atomic orbitals.
First one has to find out the electronic structure of a carbon atom by locating this element in the periodic table.
Second step is then to fill the orbitals with the given number of electrons using Hund's rule.
The third step contains the formation of bonds between the atoms under respect of bonding and antibonding molecular orbitals.
With these given information the bond order is being calculated then.

Solution This solution is FREE courtesy of BrainMass!

Hi,

First, we have to find out how many electrons and atomic orbitals (AO) are contributed by every C - atom.
Carbon is an element of the second period and 14th group. That tells us the numbers of AO's and electrons.

Now let us have a look on the periodic table.
* The second period contains the following AO's : 1s , 2s , 2px , 2py , 2pz
* We can see that carbon has 6 electrons because of its group number.

Following Hund's rule we have to place the electrons from the lowest orbital up to the highest one. The Pauli rule tells us that every orbital can only get 2 electrons.
We find:

(1s)2 , (2s)2 , (2p x,y,z)2 ; ( In a free atom the p-orbitals have the same energy so that there is no difference between the x,y,z directions)

Combining the AO's of the 2 atoms we have to consider that :
* number of AO's is equal to the number of MO's :
* 6 AO's from every C result 12 MO's

The MO's are bonding (b) and antibonding (a) ones for every state.
Very important is the fact that NOW the p-orbitals are being split up in "sigma" and "pi" MO's.
Sigma MO's are bonds on the C - C axis (2p z) but pi MO's are formed by AO's perpendicular to it (2p x,y). The pi MO's are parallel to the sigma's.

In case of carbon the pi - MO's are at a lower energy level than the pz - sigma MO.
With 12 electrons ( 2 C - atoms ) plus 2 electrons of the ion we have 14 to fill the MO's.

So we get the MO diagram (the number of electrons are written in the brackets) with the lowest energy level on the left side:

1s(b,2) 1s(a,2) , 2s(b,2) , 2s(a,2) , 1pi (b,4) , 2p(b,2)

Bond order:

We have to subtract the number of electrons in antibonding MO's from the number in bonding ones. The obtained result we have to multiply with 1/2.

* We have in 10 electrons in bonding MO's :
1s(b,2), 2s(b,2), 1pi (b,4) , 2p(b,2)

and 4 electrons in antibonding MO's :
1s(a,2), 2s(a,2)

With the formula : B = 1/2 * { n(b) - n(a) } we calculate the bond order: B = 3

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