To draw the Lewis structure for the phosphate ion (PO43-), we first need to determine the total number of valence electrons. Phosphorus (P) has 5 valence electrons, and each oxygen (O) has 6, giving us:
- 1 x 5 (for P) = 5
- 4 x 6 (for O) = 24
- Total = 5 + 24 + 3 (accounting for the 3- charge) = 32 electrons
Now, we begin drawing the structure. Place phosphorus in the center, bonded to four oxygen atoms. Connect each oxygen atom to phosphorus with single bonds initially. Each single bond contributes 2 electrons, accounting for 8 electrons used. Next, we will allocate the remaining electrons.
After forming the single bonds, we have:
- Total used: 8 electrons
- Remaining: 32 – 8 = 24 electrons
Now, distribute the remaining 24 electrons as lone pairs on the oxygen atoms. Each oxygen will need 6 electrons to fill its octet (2 in the bond and 6 as lone pairs), and since there are four oxygen atoms to account for, we have enough electrons to do so:
- Each O gets 6 lone pair electrons (4 O x 6 = 24).
The final structure should show P with single bonds to four oxygens, and each oxygen surrounding it with three lone pairs.
Regarding the geometry, the central atom phosphorus is surrounded by four oxygen atoms. The arrangement around the central atom, based on VSEPR theory, is tetrahedral due to the four bonding pairs of electrons, leading to a bond angle of approximately 109.5 degrees.
To determine if the molecule is polar or nonpolar, we look at the electronegativity of the atoms. Oxygen is more electronegative than phosphorus, creating polar bonds. However, because of the symmetrical tetrahedral geometry, the dipoles from each P-O bond will cancel each other out. Thus, despite the polar bonds, the overall structure is nonpolar.
In summary, the Lewis structure for PO43- shows a central phosphorus atom with four single bonds to oxygen atoms in a tetrahedral shape, and since the molecular geometry is symmetrical, the phosphate ion is nonpolar.