How is H3PO4 Lewis structured

1) For the H2PO4-

The central atom is the P, with 5 electrons on the last shell. It shares 1 electron with one of the O(1) through a single covalent bond and another electron with O(2) again through a single covalent bond. The P now has 7 electrons and Oxygens 1 and 2 also have 7.

Oxygens 1 and 2 form single covalent bonds with the Hydrogens 1 and 2. Everybody becomes stable now, the Os with 8 electrons and the Hs with 2 ("octet" and "duet")

The P atom also forms a single covalent bond with O(3). P then have now 8 electrons and O(3) has 7. The negative charge (an extra electron) now goes to O(3) which now have 8 electrons in the valence shell.

The only atom left is O(4), which has 6 electrons in the last shell. The P atom shares (without loosing) 2 of its 8 electrons through what is known as a "coordinating' or "dative" bond with this O(4). Everybody is now stable, according to the so called "octet" and "duet" rules.

2) For the XeOF2

The Xe forms 2 single covalent bonds with the 2 F atoms. These had 7 with one more electron from the Xe now now have 8 in the last shell. The Xe now have 10 electrons, 8 before and 2 extra from the Fs. It does not obey the "octet" rule in such "strange" structure.

The last bond is between the Xe and O. The Xe atom form a "coordinated" or "dative" bond with the O, "lending" two of its 10 electrons to the O atom, which had 6 and now have 8.

This is the "Lewis Electronic Structure" version. If you want a more "Quantum" explanation, please check the ref., and see the hybridation of the Xe electronic orbitals (dsp3).

Source(s): For the XeOF2