Molecular Geometry, Hybridization, and Polarity

Consider the gaseous molecules of beryllium difluoride, $\mathrm{BeF_2}$, and boron trifluoride, $\mathrm{BF_3}$. The atomic numbers of the constituent elements are $Z(\mathrm{Be}) = 4$, $Z(\mathrm{B}) = 5$ and $Z(\mathrm{F}) = 9$.

1. a) Deduce the Lewis structure of each molecule and state the VSEPR formula of type $\mathrm{AX}_n\mathrm{E}_m$ around the central atom.
2. b) State the molecular geometry and the predicted bond angle in each case.

The molecules of methane, $\mathrm{CH_4}$, ammonia, $\mathrm{NH_3}$, and water, $\mathrm{H_2O}$, form a classic series of valence-isoelectronic species with a central atom from the second period. Experimental bond angle data: $\alpha(\mathrm{H-C-H}) = 109.5^\circ$, $\beta(\mathrm{H-N-H}) = 107.3^\circ$, $\gamma(\mathrm{H-O-H}) = 104.5^\circ$.

1. a) Write the VSEPR formula for each of the three molecules and determine their electron geometry and molecular geometry.
2. b) Justify the progressive decrease in the bond angle observed in the series, in terms of the relative repulsions between bonding pairs and lone pairs.

The carbon atom in methane, $\mathrm{CH_4}$, forms four identical bonds directed toward the vertices of a regular tetrahedron, with bond angles of $109.5^\circ$. Data: $Z(\mathrm{C}) = 6$ (ground-state configuration: $1s^2\,2s^2\,2p^2$), $Z(\mathrm{H}) = 1$.

1. a) Explain the valence bond and hybridization model used to justify the four equivalent bonds.
2. b) State the type of overlap that gives rise to the $\sigma$ bonds and briefly explain why hybridization is a working model and not an observable quantum state.

Consider the hydrocarbons ethene, $\mathrm{C_2H_4}$, and ethyne, $\mathrm{C_2H_2}$. The atomic numbers are $Z(\mathrm{C}) = 6$ and $Z(\mathrm{H}) = 1$.

1. a) Describe the type of hybridization shown by the carbon atom in each hydrocarbon, stating the geometric orientation of the hybrid orbitals and the number of pure $p$ atomic orbitals that remain unhybridized.
2. b) Determine, with justification, the total number of $\sigma$ bonds and $\pi$ bonds present in each molecule, stating which orbitals overlap to form each one.

The carbon tetrachloride molecule, $\mathrm{CCl_4}$, has strongly polarized bonds due to the difference in electronegativity between chlorine ($\chi = 3.16$) and carbon ($\chi = 2.55$). However, experimental measurements show that its overall dipole moment is zero ($\mu = 0\,\mathrm{D}$).

1. a) Justify the molecular geometry of $\mathrm{CCl_4}$ according to the VSEPR model from the valence configurations ($Z(\mathrm{C}) = 6$, $Z(\mathrm{Cl}) = 17$).
2. b) Explain using vectors why the molecule is nonpolar despite the polar nature of its individual bonds.

Carbon dioxide, $\mathrm{CO_2}$, and water, $\mathrm{H_2O}$, are triatomic molecules with polar bonds due to the high electronegativity of oxygen compared with that of carbon and hydrogen. However, their behavior in an electrostatic field is radically different. The following illustration shows the spatial orientations and bond dipole moments for both substances:

1. a) Justify the molecular geometry of both species using VSEPR theory (specify the $\mathrm{AX}_n\mathrm{E}_m$ notation).
2. b) Deduce, from the vector sum of the bond dipole moments, whether or not each molecule has a resultant molecular dipole moment.

Phosgene, $\mathrm{COCl_2}$, and sulfur dioxide, $\mathrm{SO_2}$, are two gaseous substances of great industrial importance. Data: $Z(\mathrm{C}) = 6$, $Z(\mathrm{O}) = 8$, $Z(\mathrm{Cl}) = 17$, $Z(\mathrm{S}) = 16$. Pauling electronegativities: $\chi(\mathrm{C}) = 2.55$; $\chi(\mathrm{S}) = 2.58$; $\chi(\mathrm{Cl}) = 3.16$; $\chi(\mathrm{O}) = 3.44$.

1. a) Draw the most plausible Lewis structures of both species and determine the molecular geometry and the hybridization model of the central atom.
2. b) In $\mathrm{COCl_2}$ the experimental bond angle $\mathrm{Cl-C-Cl}$ is $111.8^\circ$, clearly less than $120^\circ$. Explain this deviation and deduce, using a reasoned vector argument, whether both molecules are polar or nonpolar.

You can do this on paper and send a photo with your answer.

1.2-Dichloroethene, $\mathrm{C_2H_2Cl_2}$, occurs in two stereoisomeric forms that are stable at room temperature: the cis isomer and the trans isomer. Data: $Z(\mathrm{C}) = 6$, $Z(\mathrm{H}) = 1$, $Z(\mathrm{Cl}) = 17$. Experimental normal boiling points: substance A ($60.3\,^\circ\mathrm{C}$), substance B ($47.5\,^\circ\mathrm{C}$).

1. a) Describe the hybridization of the carbon atoms and the bonds that make up the double bond, and justify why both isomers can be separated at ordinary temperature.
2. b) Represent the bond dipole moments of each isomer as vectors, identify with justification which is the cis and which the trans isomer from their boiling points, and explain the cause of the observed physical difference.

The molecules of boron trifluoride, $\mathrm{BF_3}$, and nitrogen trifluoride, $\mathrm{NF_3}$, are compared. Electronegativities: $\chi(\mathrm{B}) = 2.04$; $\chi(\mathrm{N}) = 3.04$; $\chi(\mathrm{F}) = 3.98$. Measured dipole moments: one species has $\mu = 0.00\,\mathrm{D}$ and the other $\mu = 0.235\,\mathrm{D}$.

1. a) Deduce the molecular geometry and the estimated hybridization of the central atom for each species, stating the relative orientation of the bond dipoles.
2. b) Analyze the dipole values vectorially, assigning each to its compound. Justify why the dipole moment of $\mathrm{NF_3}$ is strikingly smaller than that of $\mathrm{NH_3}$ ($\mu = 1.47\,\mathrm{D}$), commenting on the contribution of the lone pair compared with the bonds, and recall why hybridization is not an exact quantum-mechanical physical description but a valence bond device.