Pregunta 5 · Complements de química URV Examen Primer Parcial. Complements de Química

a) Draw the molecular orbital energy diagram for the valence electrons (s-p mixing occurs)

Solución: Orden de energía de los orbitales moleculares: $$\sigma_{2s} < \sigma^*_{2s} < \pi_{2p_x} = \pi_{2p_y} < \sigma_{2p_z} < \pi^*_{2p_x} = \pi^*_{2p_y} < \sigma^*_{2p_z}$$ Carbono aporta 4 electrones de valencia ($2s^2 2p^2$) y Oxígeno 6 electrones de valencia ($2s^2 2p^4$), dando un total de 10 electrones de valencia colocados de menor a mayor energía.

b) Write the ground-state electron configuration of the molecule

Solución: $(\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p})^4 (\sigma_{2p_z})^2$

c) Calculate the bond order (B.O.)

Solución: 3

d) New bond order of CO+

Solución: 2.5

d) Magnetic property of CO+

Solución: paramagnetic

d) Explain how the bond order, bond dissociation energy, and magnetic properties are affected

Solución: The electron is removed from the HOMO ($\sigma_{2p_z}$, which has slight bonding character). New bond order $= 2.5$. Since the bond order decreases from 3 to 2.5, the bond weakens, so the bond dissociation energy decreases (easier to break). CO is diamagnetic, whereas $\mathrm{CO^+}$ has one unpaired electron and becomes paramagnetic.

e) Justify why the ionization potential of CO is higher than those of its constituent free atoms

Solución: The electrons residing in bonding molecular orbitals in CO are lower in energy than the atomic orbitals of isolated C (11.2 eV) and O (13.6 eV). Consequently, removing an electron from the stable CO molecular orbital requires more energy (14.0 eV).

f) Sketch the highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO)

Solución: HOMO: Corresponde a $\sigma_{2p_z}$, polarizado hacia el átomo de carbono, con un lóbulo grande apuntando en dirección opuesta al oxígeno (carácter de par no enlazante en C). LUMO: Corresponde a los orbitales antienlazantes degenerados $\pi^*_{2p}$, con un plano nodal entre el C y el O.