What molecular geometry would be expected for BF3 and NH3?

Explanation:

There are three valence electrons in a boron atom.

In boron trifluoride, the central boron atom did not achieve an octet with eight valence electrons. Rather, that boron atom would be electron deficient with only six valence electrons.

Each of the three fluoride atoms would have shared one valence electron with that boron atom, with a total of three boron-fluorine single bonds. On the other hand, all three of the valence electrons of that boron atom would be involved in bonding. Hence, there would be no extra valence electrons to act as lone pairs on that boron atom.

Hence, the central boron atom would have three electron domains (one for each boron-fluorine bond) with none of the electron domains coming from lone pairs. By the VSEPR theory, the geometry of the molecule would be trigonal planar. All four atoms in this molecule would be in the same plane.

There are five valence electrons in a nitrogen atom.

In ammonia, the central nitrogen atom is indeed able to achieve an octet (with eight valence electrons in total.) Three of the five valence electrons of nitrogen would form a total of three hydrogen-nitrogen bonds. The other two valence electrons would form a lone pair.

Hence the central nitrogen atom would have four electron domains (one for each of the three hydrogen-nitrogen bond, and one for the lone pair.) Hence, by the VSEPR theory, the geometry of this molecule would be trigonal pyramidal.

the product of the pressure and volume is also constant

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