Q4. Find the bond orders of N2+ and N2−. Can equal bond orders alone rank their stability? What unpaired-electron count does the simple model give?
N2+ (13 electrons) loses one σ2pz electron: Nb = 9, Na = 4, bond order 2.5. N2− (15 electrons) adds one π* electron: Nb = 10, Na = 5, bond order 2.5. Equal formal bond orders do not rank stability or bond length by themselves; comparing the raw number of antibonding electrons is insufficient. The elementary configurations each have one unpaired electron. N2+ is a bound molecular ion, whereas low-energy N2− is a temporary electron-scattering resonance that can lose its added electron.
Q5. NO has 15 electrons. Using the N2-type order, find its bond order and magnetic nature.
σ1s2 σ*1s2 σ2s2 σ*2s2 (π2px2 = π2py2) σ2pz2 π*2px1: Nb = 10, Na = 5, bond order 2.5. One unpaired π* electron: paramagnetic.
Q6. How many nodal planes do σ*1s, π2px and π*2px have?
σ*1s: one, perpendicular to the axis between the nuclei. π2px: one, the plane that contains the axis (from the p orbitals). π*2px: two, the plane containing the axis and the plane between the nuclei.
Q7. Taking z as the bond axis, which pairs can combine, and do they give σ or π orbitals: (a) 2px and 2px; (b) 2s and 2s; (c) 2py and 2pz; (d) 2px and 2py?
(a) π (π2px, π*2px); (b) σ (σ2s, σ*2s); (c) and (d) cannot combine: they differ in symmetry about the z-axis, though equal in energy.
Q8. Find the bond order of the cyanide ion, CN−, and name a neutral molecule with the same configuration.
CN− has 6 + 7 + 1 = 14 electrons, like N2: Nb = 10, Na = 4, bond order 3, diamagnetic. N2 (and CO) have corresponding elementary occupancy counts; their orbitals and orbital energies are not identical.