Chemistry 102: Acids: Draw Lewis Structure for phosphoric acid that is consistent with the following behavior

Question

13.

The value of K_a1 of hypophosphorous acid, H₃PO₂, is nearly the same as the K_a1 of phosphoric acid, H₃PO₄. Draw a Lewis structure for phosphoric acid that is consistent with this behavior.

Draw the Lewis structure with the formal charges minimized. Do not include formal charges in your structure. Include hydrogen atoms and nonbonding electrons in the structure.

Answer 1

The structure you want has one P=O double bond and three P-O-H single bonds, with phosphorus at the centre. Here it is in text, since I cannot draw here:

          O
          ||
  H - O -  P  - O - H
          |
          O
          |
          H

Every oxygen carries two lone pairs. Phosphorus has none.

Why the clue about H₃PO₂ forces this structure. Both formulas start with three hydrogens, so at a glance you might expect both to be triprotic and to behave differently. They are not. Hypophosphorous acid is monoprotic: only one of its hydrogens sits on an oxygen. The other two are bonded straight to phosphorus, and a P-H bond is not acidic in water.

          O
          ||
  H  -    P  - O - H
          |
          H

Now look at what the two molecules share. Each has exactly one P=O double bond and at least one O-H group attached to that same phosphorus. K_a1 only describes the first proton to leave, and in both cases that proton departs from an identical local environment: an O-H on a phosphorus bearing a P=O. Identical environment, nearly identical K_a1. That agreement is the evidence that phosphoric acid must contain a P=O double bond.

What the alternative would look like, and why it is rejected. You could instead draw phosphorus with four single bonds, three to O-H groups and one to a lone oxygen carrying a negative charge, giving phosphorus a +1. That obeys the octet rule strictly, but it introduces separated formal charges, and the problem tells you to minimise them. Converting that P-O^- single bond into a P=O double bond drops every formal charge to zero:

Phosphorus: 5 valence electrons, 5 bonds, no lone pairs, so 5 - 0 - 5 = 0. Double-bonded oxygen: 6 - 4 - 2 = 0. Each hydroxyl oxygen: 6 - 4 - 2 = 0. Every atom neutral.

That does put ten electrons around phosphorus, which is fine. Phosphorus is in period 3 and can hold an expanded octet, unlike nitrogen directly above it, which cannot. This is exactly the situation the "minimise formal charges" instruction is pointing you toward.

One idea worth carrying forward: count acidic hydrogens by where they sit, not by the formula. H₃PO₄ is triprotic and H₃PO₂ is monoprotic despite both being written H₃P-something. Only hydrogens attached to an electronegative atom, oxygen here, come off in water. The same rule explains why acetic acid, HC₂H₃O₂, releases one proton and not four.