The molecule BH3 is trigonal planar, with B in the center and H in the three vertices. Ther are no free electrons. All the valence electrons are paired in and forming bonds.
There are four kind of intermolecular attractions: ionic, hydrogen bonds, polar and dispersion forces.
B and H have very similar electronegativities, Boron's electronegativity is 2.0 and Hydrogen's electronegativity is 2.0.
The basis of ionic compounds are ions and the basis of polar compounds are dipoles.
The very similar electronegativities means that B and H will not form either ions or dipoles. So, that discards the possibility of finding ionic or polar interactions.
Regarding, hydrogen bonds, that only happens when hydrogen bonds to O, N or F atoms. This is not the case, so you are sure that there are not hydrogen bonds.
When this is the case, the only intermolecular force is dispersion interaction, which present in all molecules.
Then, the answer is dispersion interaction.
Question is incomplete, the complete question is as follows:
A student wants to examine a substance by altering the bonds within its molecules. Which of the following properties of the substance should the student examine?
A. Toxicity, because it can be observed by altering the state of the substance
B. Boiling point, because it can be observed by altering the state of the substance
C. Toxicity, because it can be observed by replacing the atoms of the substance with new atoms
D. Boiling point, because it can be observed by replacing the atoms of the substance with new atoms
Answer:
B.
Explanation:
A student can examine a substance without altering the bonds within the molecules by examining its boiling point.
The boiling point is the property of a substance, at which the substance changes its state, which is from solid to liquid, liquid to gas and others. So, examining the boiling point will alter the bonds within the molecules as the state of substance will change.
Hence, the correct answer is "B".
