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What is the bond order of (Fe2) and (Fe2)+ ?
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[QUOTE="TeethWhitener, post: 6180646, member: 511972"] 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 Fe[SUB]2[/SUB], 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 [B][I]much [/I][/B]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 [I]manganese does not form a covalent dimer at all![/I] 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. [/QUOTE]
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What is the bond order of (Fe2) and (Fe2)+ ?
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