I"m confused because when ns draw every one of the Os attached with single bonds to p the officially charges don"t do sense. However when I include a dual bond to one of the Os the officially charges room -1 to 3 the the oxygens. I"m puzzled if there have to be a double bond or no for the PO4 3- structure.


If we begin at the beginning by illustration the Lewis dot framework we start with valence electrons.

You are watching: How many lone pairs of electrons are represented in the lewis structure of a phosphate ion (po43-)?

P has 5 ve

each O has 6 ve therefore that"s a total of 4x6 = 24 ve

We include 3 more for the 3- fee for a full of 32 valence electrons

So, this can be written with P at the center and 4 O atom attached with single bonds and all O atom would have actually 3 lone pairs. This setup gives a formal fee of +1 on P and -1 on all O atoms. No good.

Adding a twin bond to among the O atoms results in a formal fee of zero top top P, zero top top the O v the twin bond, and also -1 ~ above the other 3 O atoms. This is fine since the charge on the phosphate anion is 3- and also so the -1 on every of 3 oxygens accounts because that this. You shouldn"t be confused because you have done that correctly. Also recall the P deserve to have an increased octet (which the does in the dot framework we just drew).

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