Place the following substances in order of decreasing vapor pressure at a given temperature: BeF2, CH3OH, OF2? (a) OF2 > BeF2 > CH3OH
(b) BeF2 > OF2 > CH3OH
(c) BeF2 > CH3OH > OF2
(d) CH3OH > OF2 > BeF2
(e) OF2 > CH3OH > BeF2

(b) BeF2 > OF2 > CH3OH

Explanation:

The degree and type of intermolecular forces present in a substance influences its vapour pressure considerably. The greater the magnitude and strength of intermolecular forces in the substance, the lower the vapour pressure of the substance.

BeF2 molecules are held together by weak vanderwaals forces hence BeF2 will exhibit the least degree of intermolecular interaction and have the highest vapour pressure. OF2 molecules are bound together by dipole interactions hence it will exhibit a lower vapour pressure compared to BeF2. CH3OH molecules form hydrogen bonds with water molecules hence it will exhibit the least vapour pressure among the trio.

i'm assuming that the correct answer is indeed the first one/ "a"..

the objects have the same charge.