What are the electron domain and molecular geometries of thionyl chloride, \( \mathrm{SOCl_2} \)?
| Electron domain geometry | Molecular geometry | |
|---|---|---|
| A. | Tetrahedral | Tetrahedral |
| B. | Tetrahedral | Trigonal pyramidal |
| C. | Trigonal planar | Trigonal planar |
| D. | Trigonal planar | Bent |
Correct Answer: \( \boxed{\mathrm{B}} \)
In \( \mathrm{SOCl_2} \), sulfur is the central atom. It has three bonding regions around it: one \( \mathrm{S=O} \) bond and two \( \mathrm{S-Cl} \) bonds, together with one lone pair.
The double bond counts as one electron domain, giving a total of \(4\) electron domains.
Four electron domains give a tetrahedral electron domain geometry. With one lone pair and three bonded atoms, the molecular geometry is trigonal pyramidal.
Therefore, the correct answer is (B).
Which molecule is most polar?
Correct Answer: \( \boxed{\mathrm{D}} \)
The polarity of a molecule depends on both the polarity of its bonds and its molecular geometry.
\( \mathrm{CF_4} \) is tetrahedral and symmetrical, so its bond dipoles cancel. \( \mathrm{CO_2} \) is linear and symmetrical, so its bond dipoles also cancel.
\( \mathrm{NBr_3} \) and \( \mathrm{NH_3} \) are both trigonal pyramidal because of the lone pair on nitrogen. However, the \( \mathrm{N-H} \) bonds in \( \mathrm{NH_3} \) are significantly more polar than the \( \mathrm{N-Br} \) bonds in \( \mathrm{NBr_3} \).
Therefore, \( \mathrm{NH_3} \) has the greatest overall dipole moment and is the most polar molecule. The correct answer is (D).
What is the hybridization of each circled atom I, II and III?
| I | II | III | |
|---|---|---|---|
| A. | \( \mathrm{sp^3} \) | \( \mathrm{sp} \) | \( \mathrm{sp^2} \) |
| B. | \( \mathrm{sp^3} \) | \( \mathrm{sp} \) | \( \mathrm{sp^3} \) |
| C. | \( \mathrm{sp^2} \) | \( \mathrm{sp^2} \) | \( \mathrm{sp^2} \) |
| D. | \( \mathrm{sp^2} \) | \( \mathrm{sp^2} \) | \( \mathrm{sp^3} \) |
Correct Answer: \( \boxed{\mathrm{D}} \)
Atom I: The carbon atom is involved in a \( \mathrm{C=C} \) double bond. It has three regions of electron density around it, giving \( \mathrm{sp^2} \) hybridization.
Atom II: The oxygen atom is involved in a \( \mathrm{C=O} \) double bond. It has \( \mathrm{sp^2} \) hybridization.
Atom III: The oxygen atom in the \( \mathrm{-OH} \) group has two sigma bonds and two lone pairs. It therefore has four electron domains and is \( \mathrm{sp^3} \) hybridized.
Thus:
\( \mathrm{I=sp^2,\quad II=sp^2,\quad III=sp^3} \)
Therefore, the correct answer is (D).
