Anonymous
Anonymous asked in Science & MathematicsChemistry · 1 month ago

# Nitrogen, N2 can ionize to form N2+ or add an electron to form N2-1. Using molecular orbital theory, compare these three species with these?

a. Magnetic character

b. Net number of π bonds

c. Bond order

d. Bond length

e. Bond strength

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• 1 month ago

Must use MO diagram with sp mixing: raises energy of σ3> π1; s,p labels changed to numerical labels. The exact bonding effect of σ*2 and σ3 are subject to debate, but I will use the more common description that σ*2 is antibonding and σ3 is bonding. (I prefer the view that both are nonbonding.)

N2 has ten valence electrons

N2: 10 e⁻ = σ1(2e⁻) σ*2 (2e⁻) π1(4e⁻) σ3(2e⁻) π2*(0e⁻) σ4*(0 e⁻)

Bond Order = ½[Σ (bonding e-) - Σ (antibonding e-)]

bo = ½[ Σ σ1(2e⁻)π1(4e⁻)σ3(2e⁻) - σ*2 (2e⁻)] = 3.0

All electrons are paired so N2 predicted (correctly) to be diamagnetic. There are two π bonds (π1(4e⁻)). One can’t predict bond length from MO diagram but lengths follow the order E-E> E=E> E≡ E. N2 has a triple bond so short; this is also the order for bond strength so N2 predicted to have to a high bond strength. N2 has the second highest bond strength of any diatomic (the bond strength of C≡O is higher)

[N2]^+ : 9 e⁻ = σ1(2e⁻) σ*2 (2e⁻) π1(4e⁻) σ3(1e⁻) π2*(0e⁻) σ4*(0 e⁻)

Bond Order = ½[Σ (bonding e-) - Σ (antibonding e-)]

bo = ½[Σ σ1(2e⁻)π1(4e⁻) σ3(1e⁻) - σ*2 (2e⁻)] = 2.5

There are also two π bonds (π1(4e⁻)). One unp e⁻ so paramagnetic, bo 2.5 so longer and weaker than that of N2.

[N2]^- 11 e⁻ = σ1(2e⁻) σ*2 (2e⁻) π1(4e⁻) σ3(2e⁻) π2*(1e⁻) σ4*(0 e⁻)

Bond Order = ½[Σ (bonding e-) - Σ (antibonding e-)]

bo = ½[Σ σ1(2e⁻)π1(4e⁻) - σ*2 (2e⁻) π2*(1e⁻)] = 2.5

Once again there are two π bonds (π1(4e⁻)).

Similar to N2^+ One unp e⁻ so paramagnetic, bo 2.5 so longer and weaker than those of N2.

This is the answer that is wanted but it is not quite correct as σ*2 is not completely antibonding.

So the bond orders/strengths are N2 (3)> N2^+~ N2^- (2.5).

If you use an MO diagram without sp mixing you will get the wrong number of π bonds.