What is the orientation of the two unhybridized p orbitals on be with respect to the two be−f bonds?
a) both p orbitals are parallel to the f−be−f bond axes.
b) both p orbitals are perpendicular to the f−be−f bond axes.
c) one p orbital is perpendicular to the f−be−f bond axis, the other is parallel.

b) Both p orbitals are perpendicular to the F−Be−F bond axes.

Explanation:

Be has 2 electrons in its valence shell, subshell s is fulfilled, so it has no unpaired electrons in its ground state to make bonds with F. So, it can promote the electrons to the 2p orbital and will having sp hybridization.

The bond between the orbitals sp and the p orbital of F are in opposite directions but the same ax. The two bonds are equivalent, and the molecule had a linear geometry. The two unhybridized p orbitals on Be are vacant, and so they are perpendicular to the F-Be-F bond axes.

specific gravity is the density of a substance divided by the density of water. since (at standard temperature and pressure) water has a density of 1 gram/cm3, and since all of the units cancel, specific gravity is usually very close to the same value as density

so "c" is the answer

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idk i’m struggling too