The effects of London dispersion forces are best demonstrated by viewing the halogen group. The lighter halogens are gaseous at room temperature, but transition to liquid and then to solid when moving down the group (bromine is a liquid, and iodine is a solid). When moving down the group, the halogens become much larger. Larger atoms experience greater London dispersion forces due to increased surface contact and greater polarizability. These increased London dispersion forces are enough to raise the boiling points of the larger atoms, causing them to exist as liquids of gases at room temperature.

While it is true that the larger atoms are heavier and will lose their valence electrons more readily, these factors do not impact the phase of the compound.

The answer is resonance structures. When different Lewis structures can be drawn for a single molecule, the molecule exists as a composite of these structures, which are called resonance structures.

Isotopes refers to multiple nuclear compositions for a single element, based on varying numbers of neutrons. Isomers are different configurations of a given molecular formula, based on geometry and orientation. Epimers are a specific class of isomers involving a single stereocenter.

This is a straightforward question that has little to do with the passage, but it's a concept almost certainly to be seen on the MCAT.

Make sure to understand the concept of valence electrons and what that number may tell you about bonding properties. Questions like these are essentially "freebies," and should be answered without hesitation.

To determine the number of electrons an atom has, you must look at which column the atom is in on the periodic table. Boron is in column 3A, so it has three valence electrons. Nitrogen is in column 5A, so it has five valence electrons.

You should be familiar with common elements, like nitrogen, without looking at the periodic table. This will save you time on the exam.

The idea of VSEPR (valence shell electron pair repulsion) theory is that valence electron pairs repel each other, arranging themselves into positions that minimize their repulsions by maximizing the distance between them. The positions of these electron pairs then determine the overall geometry of the molecule. Molecular geometry is thus determined by the arrangement of electrons and nuclei such that the electrons are as far from one another as possible, while remaining as close to the positively charged nucleus as possible.

Of the available answer choices, only has a tetrahedral geometry. Tetrahedral molecules have four constituents bound to the central without any lone pairs.

has three bonds and a lone pair. It has a tetrahedral electronic geometry, but not molecular geometry.  has two double bonds, which give it a linear geometry.  has two lone pairs, giving the compound a square planar geometry. 

Sulfur hexaflouride (SF₆) is an example of octahedral geometry, as it follows the skeleton of AX₆E₀ format. A refers to sulfur, X to fluorine, and E to the lone pair electrons.

Square planar has an AX₄E₂ format, while tetrahedral and trigonal bipyramidal follow AX₄ and AX₅ formats, respectively. 

Recall the following relationships between geometry and number of pairs of electrons on the central atom.

2: linear

3: trigonal planar

4: tetrahedral

5: trigonal bipyriamidal

6: octahedral

To visualize the geometry, we need to think of how many electron pairs are on the central atom. Drawing Lewis dot diagrams may be helpful here. None of the answer choices has lone central electron pairs, with the exception of water, so the number of atoms bound to the central atom is the same as the number of central electron pairs.

The only one that does not match up with the correct geometry is SF₆, which is actually octahedral since it has six central electron pairs. In a water molecule, the central oxygen has six valence electrons, plus one from each bond with hydrogen, for a total of eight central electrons and four central electron pairs. So, this geometry is a variation on the tetrahedral form (bent), in which two central electron pairs are not bound.

In order to determine which molecule will have the smallest bond angle(s), make sure to factor in both the number of atoms around the central atom as well any lone pairs on the central atom.  has two lone pairs around the central sulfur atom, which pushes the two hydrogens closer together than the ones found in  and .

Octahedral geometry always corresponds to the  hybridization.  hybridization corresponds to a general formula of .  hybridization corresponds to a general formula of ,  hybridization corresponds to a general formula of , and  hybridization corresponds to a general formula of .