Question

According to VSEPR theory , in which fashion will the bonds and lone pairs of electrons be arranged about the central atom in the following molecules or molecular ions? (The VSEPR theory link may take a moment or two to load.) PO43- ion (The central atom is P.) H3O+ ion (The central atom is O.) AsF5 molecule (The central atom is As.)

Answer 1

Answer:

Tetrahedral, trigonal pyramidal, trigonal bipyramidal.

Explanation:

The VSPER theory states that the bonds of sharing electrons and the lone pairs of electrons will repulse as much as possible. So, by the repulsion, the molecule will have some shape.

In the ion PO₄³⁻, the central atom P has 5 electrons in its valence shell, so it needs 3 electrons to be stable. Oxygen has 6 electrons at the valence shell and needs 2 to be stable. 3 oxygens share 1 pair of electrons with P, and the two lone pair remaining in P is shared with the other O, then the central atom makes 4 bonds and has no lone pairs, the shape is tetrahedral.

In the ion H₃O⁺, the central atom O has 6 electrons in its valence shell and needs 2 electrons to be stable. The hydrogen has 1 electron, and need 1 more to be stable. The hydrogens share 1 pair of electrons with the oxygen, then it remains 3 electrons at the central atom, and the VSPER theory states that the shape will be a trigonal pyramidal.

In the AsF₅, the central atom As has 5 valence electrons, and F has 1 electron in its valence shell, so each F shares one pair of electrons with As, and there are no lone pairs in the central atom. For 5 bonds without lone pairs, the shape is trigonal bipyramidal.