What is the bond order of (Fe2) and (Fe2)+ ?

[CGandC]

Homework Statement

Hi ,
I was wondering what is the bond order of $Fe_2$ and $(Fe_2)^+$ ?
Is the molecule $Fe_2$ even able to be covalently bonded? if so , why not?

Relevant Equations

Bond order = ( # bonding electrons - # antibonding electrons ) /2

From almost writing the complete Molecular orbital electron configuration for $Fe_2$ and $(Fe_2)^+$ ( I said ' almost ' because I don't know how to write the electronic configuration for the d-bonds ) then I think the bond order for $Fe_2$ is 0 and the bond order for $(Fe_2)^+$ is 1/2

 

[mjc123]

Try this:
https://aip.scitation.org/doi/10.1063/1.466959The abstract doesn't answer your question (except that it's obviously not 0), but the article may.

 

[TeethWhitener]

Problem Statement: Hi ,
I was wondering what is the bond order of $Fe_2$ and $(Fe_2)^+$ ?
Is the molecule $Fe_2$ even able to be covalently bonded? if so , why not?
Relevant Equations: Bond order = ( # bonding electrons - # antibonding electrons ) /2

From almost writing the complete Molecular orbital electron configuration for $Fe_2$ and $(Fe_2)^+$ ( I said ' almost ' because I don't know how to write the electronic configuration for the d-bonds ) then I think the bond order for $Fe_2$ is 0 and the bond order for $(Fe_2)^+$ is 1/2

You might need a few more relevant pieces of information than you've given. One, if you have $N$ total atomic orbitals, how many molecular orbitals will you have? Two, it's important to note that the valence electrons of the d-block include $ns$ as well as $(n-1)d$ orbitals. So for iron, you'd need to take into account the $4s$ orbitals in addition to the $3d$ orbitals.

One easy way to approach bond order in the transition metals is to look at the analog in p-block elements. If you can figure out (e.g.,) why N₂ has a triple bond, but C₂ and O₂ only have double bonds, then you should be able to apply the information I gave you above to figure out what the bond order of Fe₂ should be.

NB--As the paper from @mjc123 points out, the formal bond order does not always correspond to the experimental bond strength, mainly because the LCAO approximation is just that: an approximation.

 

[CGandC]

Thanks.

I was given the answer by an instructor for what the bond order of $Fe_2$ and $(Fe_2)^+$ is:

The electron configuration of $Fe$ is $(1s)^2 (2s)^2 (2p)^6 (3s)^2 (3p)^6 (4s)^2 (3d)^6$

Therefore I have 6 electrons in the last orbital corresponding to the final highest energy level ( the $3d$ orbital ) for a single $Fe$ atom.

Therefore, for $Fe_2$ molecule I'll have at total 12 electrons which occupy the final highest energy level ( according to aufbau principle of course ).

Since there is a $Fe_2$ molecule , then in the $3d$ orbital there two kinds of orbital divisions : bonding orbitals and anti-bonding orbitals.
Since each orbital in the d-orbital can occupy 10 electrons, then according to aufbau principle, there will be 10 electrons that will occupy the bonding orbitals and then there will be 2 electrons that will occupy the anti-bonding orbitals.

Therefore , the bond order for $Fe_2$ is: ( 10 - 2)/2 = 4

And for $(Fe_2)^+$ there will be 10 electrons that will occupy the bonding orbitals and then there will be 1 electron that will occupy the anti-bonding orbitals ( still talking about the $3d$ orbital ) .

Therefore , the bond order for $(Fe_2)^+$ is: ( 10 - 1 )/2 = 4.5

 

[TeethWhitener]

Thanks.

I was given the answer by an instructor for what the bond order of $Fe_2$ and $(Fe_2)^+$ is:

The electron configuration of $Fe$ is $(1s)^2 (2s)^2 (2p)^6 (3s)^2 (3p)^6 (4s)^2 (3d)^6$

Therefore I have 6 electrons in the last orbital corresponding to the final highest energy level ( the $3d$ orbital ) for a single $Fe$ atom.

Therefore, for $Fe_2$ molecule I'll have at total 12 electrons which occupy the final highest energy level ( according to aufbau principle of course ).

Since there is a $Fe_2$ molecule , then in the $3d$ orbital there two kinds of orbital divisions : bonding orbitals and anti-bonding orbitals.
Since each orbital in the d-orbital can occupy 10 electrons, then according to aufbau principle, there will be 10 electrons that will occupy the bonding orbitals and then there will be 2 electrons that will occupy the anti-bonding orbitals.

Therefore , the bond order for $Fe_2$ is: ( 10 - 2)/2 = 4

And for $(Fe_2)^+$ there will be 10 electrons that will occupy the bonding orbitals and then there will be 1 electron that will occupy the anti-bonding orbitals ( still talking about the $3d$ orbital ) .

Therefore , the bond order for $(Fe_2)^+$ is: ( 10 - 1 )/2 = 4.5

This is the right answer, although I want to point out one thing. As I said previously, both the 4s and 3d orbitals contribute to the valence electron shell. This gives a total of 6 orbitals for each Fe atom. Since the total number of molecular orbitals must equal the total number of atomic orbitals in the system, we have 12 molecular orbitals in all. By symmetry, 6 of these will be bonding and 6 will be antibonding. For Fe₂, each Fe has 6 d electrons and 2 s electrons in the valence orbitals, meaning there is a total of 16 electrons to fill the 12 molecular orbitals via the aufbau principle. So all 6 of the bonding orbitals will be filled, and 2 of the antibonding orbitals will be filled, giving an overall bond order of 4. Likewise, removing one electron to get $(Fe_2)^+$ will give a bond order of 4.5.

As I said, the answer was right, but the s orbital plays an important role. In fact, it's the whole reason we say species like $Mo_2$ and $W_2$ have a formal sextuple bond. This actually has experimental ramifications. Ignoring the s orbitals leads to the prediction that the formal bond order (and ultimately the strength of the bond) is higher for $Mn_2$, $Tc_2$, and $Re_2$ (quintuple bond) than for $Cr_2$, $Mo_2$, and $W_2$ (sextuple bond including s orbitals, quadruple bond excluding s orbitals). Experimental measurements show that the latter are the stronger bonds.

One final point: there is reasonable disagreement over whether the 4s orbital really counts as a valence orbital for the first row of transition metals (it definitely does for the second and third rows), as the energy between 4s and 3d orbitals is proportionally larger than 5s/4d and 6s/5d. I will point out that $Cr_2$ has a much stronger bond than $Mn_2$. But, even that isn't the whole story. The manganese atom has a ground state configuration of $4s^2 3d^5$, a half-filled d-subshell. The first excited state has a configuration of $4s^1 3d^6$, but because the ground state is so stable, the excitation energy to this state is quite large (>2eV). But the total bond energy of the $Cr_2$ dimer, for comparison, is only about 1.5eV. So if we expect that the bond energy of $Mn_2$ is comparable, once we factor in the energy required to hybridize electrons in Mn atoms to form the covalent bond, we find that manganese does not form a covalent dimer at all! There simply isn't enough energy available. In fact, that's what we observe experimentally, with $Mn_2$ being a very weakly bound (<0.1eV) van der Waals dimer.