How bond energy varies from n2- to n2+
Answers
Foe the energy we need to determines the Bond order for which we need to do the electronic configuration of both the molecules by the MOT.
∴ For N₂
Total number of the electrons = 7 × 2 = 14
∴ E.C. = σ1s²σ*1s² σ2s² σ*2s² (π2p²x = π2p²y) σ2p²z
Therefore, Bond Order = (Nb - Na)/2
= (10 - 4)/2 = 3
For N₂⁺,
Total number of the electrons = 7 × 2 - 1 = 13
∴ E.C. = σ1s²σ*1s² σ2s² σ*2s² (π2p²x = π2p²y) σ2p¹z
Therefore,
Bond Order = (9 - 4)/2 = 5/2 = 2.5
We know the Molecule having the hind bond order will be having the high bond energy.
Therefore, Bond energy of the N₂ will be more than N₂⁺.
Hope it helps.
Answer:
We will need to think in terms of bond order and its relationship with molecular orbitals, thus I assume you are familiar with these topics.
In the construction of the molecular orbitals associated with the N2 molecule, we have two identical sets of atomic orbitals to work with, more specifically, orbitals 1s, 2s and three degenerated 2p orbitals for each atomic nucleus.
We then must categorise these 5 individual orbitals in each N atom by their symmetry element in accordance with the molecular symmetry group of the N2 molecule (D∞h). By doing this, we find three orbitals with σg (σ for simplicity) symmetry - 1s, 2s and 2pz -, and two degenerate orbitals of πu (π for short) symmetry - 2px and 2py. When combined, these set of orbitals will generate a new collection of molecular orbitals, which are properly labelled, in ascending order of orbital energy, 1σ, 2σ, 3σ, 4σ, π, 5σ, π* and 6σ.
Acknowledging that the neutral N2 molecule holds a total of 14 electrons, we will need to fill a total of 7 molecular orbitals, thus giving the configuration 1σ², 2σ², 3σ², 4σ², π⁴, 5σ².
For bonding energy purposes, we can ignore the low energy orbitals 1σ, 2σ, 3σ, 4σ, all containing electrons that do not contribute to the bond order in the N2 system, and focus on the remaining π, 5σ, π* and 6σ molecular orbitals.
Finally, we need to calculate the bond order of each of these systems, and remember that higher bond orders are typically associated with higher bonding energies in molecules - and it is certainly the case for examples with the same atomic nuclei, but different number of electrons. Let’s first obtain the electronic configuration of all three systems:
N2: 1σ², 2σ², 3σ², 4σ², π⁴, 5σ²
N2-: 1σ², 2σ², 3σ², 4σ², π⁴, 5σ², π*¹ (obtained by adding one electron to N2)
N2+: 1σ², 2σ², 3σ², 4σ², π⁴, 5σ¹ (obtained by removing one electron from N2)
Now we are ready to calculate the relevant bond orders by the formula:
B = (Nb - Na)/2, where B is the bond order, Nb is the number of electrons in bonding orbitals (π and 5σ in our case) and Na is the number of electrons in anti-bonding orbitals (π* in our example). Thus:
B = (Nb - Na)/2
B[N2] = (4 + 2 - 0)/2 = 6/2 = 3
B[N2-] = (4 + 2 - 1)/2 = 5/2
B[N2+] = (4 + 1 - 0)/2 = 5/2
This allows us to unequivocally classify N2 as the lowest energy species. For the N2- and N2+ pair, a more subtle effect must be taken into account.
More specifically, as we remove one electron from the neutral molecule N2, consequently yielding N2+, we arrive at a system with lower total electron-electron repulsion than that associated with the original N2 molecule. Conversely, adding an electron to neutral N2, thus affording N2-, results in a system with more total electronic repulsive energy.
Whilst just a small component of bond energy, the total repulsion amongst electrons participating in bonding interactions is expected to destabilise the bond between two atomic nuclei. As a consequence, we arrive at the following energetic ordering in terms of bond energies: N2 >> N2+ > N2-.