Predict the shapes of the following molecules, and then predict which would have resultant dipolemoments: (a) \(\mathrm{SO}_{2} ;\) (b) \(\mathrm{NH}_{3} ;\) (c) \(\mathrm{H}_{2} \mathrm{S} ;\) (d) \(\mathrm{C}_{2} \mathrm{H}_{4} ;\) (e) \(\mathrm{SF}_{6}\); (f) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\).

Based on the VSEPR theory, the shapes of molecules can be predicted. For \( \mathrm{SO}_{2} \), there are 2 bond pairs and 1 lone pair around the central atom, so its shape is bent or V-shaped. For \( \mathrm{NH}_{3} \), there are 3 bond pairs and 1 lone pair around the central atom, giving it a trigonal pyramidal shape. \( \mathrm{H}_{2} \mathrm{S} \) has 2 bond pairs and 2 lone pairs around the central atom, also providing a bent or V-shaped form. \( \mathrm{C}_{2} \mathrm{H}_{4} \) is a planar molecule since both carbon atoms are sp2 hybridized, forming a trigonal planar geometry. For \( \mathrm{SF}_{6} \), there are 6 bond pairs around the central atom, giving an octahedral shape. Lastly, \( \mathrm{CH}_{2} \mathrm{Cl}_{2} \) has a tetrahedral shape since there are 4 bond pairs around the central carbon atom.

A molecule will have a net dipole moment if its structure is unsymmetrical, which creates an imbalance of electron charge. Hence, \( \mathrm{SO}_{2} \), \( \mathrm{NH}_{3} \) and \( \mathrm{H}_{2} \mathrm{S} \) being V-shaped and trigonal pyramidal, will have a net dipole moment. \( \mathrm{C}_{2} \mathrm{H}_{4} \), being a symmetrical planar molecule, will not have a dipole moment. \( \mathrm{SF}_{6} \), being a symmetrical octahedral molecule, will not have a dipole moment. \( \mathrm{CH}_{2} \mathrm{Cl}_{2} \), although is a tetrahedral, it will have a net dipole moment due to the different atoms (H and Cl) connected to the central atom leading to a difference in electronegativities.