Question

How many lone pairs of electrons are represented in the Lewis structure of a phosphate ion (PO43-)?

Answer 1

Answer :  The number of lone pairs of electrons represented in the Lewis-dot structure of phosphate ion are, 22

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

The valance electrons are shown by 'dot'.

The given molecule is, phosphate ion $(PO_4^{3-})$

As we know that phosphorous has '5' valence electrons and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in phosphate ion $(PO_4^{3-})$ is,

$(PO_4^{3-})$ = 5 + 4(6) + 3 = 32

According to Lewis-dot structure, there are 10 number of bonding electrons and 22 number of non-bonding electrons or lone pair electrons.

The Lewis-dot structure of phosphate ion is shown below.

Answer 2

Hey

P has 5 valence electrons
O has 6 valence electrons (each for 4 oxygen)
And finally, for every negative charge, there is an additional valence electron

We should add these all up = 5 + 24 + 3 = 32 valence electrons

With this, we can be guided to illustrate the lewis structure as P as central atom and the 3 oxygen each with a single bond with P and 1 oxygen with a double bond with P. We place the valence electrons until octet rule is satisfied,
we will be left with 12 lone pairs for phosphate ion.

Hoped I Helped