Examine this Lewis structure for the phosphate ion, PO4 How many valence electrons are around the P atom in this structure?
a. Nine Eight
b. Four
c. Twelve
d. None of the above
The correct answer and explanation is :
Let’s carefully analyze the phosphate ion, $\text{PO}_4^{3-}$, focusing on the valence electrons around the phosphorus (P) atom in its Lewis structure.
Step 1: Understand the phosphate ion and its Lewis structure
- Phosphorus (P) is the central atom.
- It is bonded to four oxygen atoms (O).
- The overall charge of the ion is -3.
- Phosphorus is in group 15, so it has 5 valence electrons.
- Each oxygen atom has 6 valence electrons (group 16).
Step 2: Common Lewis structure for $\text{PO}_4^{3-}$
- Phosphorus typically forms five bonds in phosphate, but in the Lewis structure for $\text{PO}_4^{3-}$, phosphorus forms four single bonds to oxygen atoms.
- Each P–O bond counts as 2 electrons shared.
- Oxygen atoms carry the negative charges to account for the -3 charge overall (usually 3 oxygens have a single bond with P and one has a double bond in resonance forms).
- In the most simplified basic structure without resonance, P has 4 single bonds to O atoms.
Step 3: Counting valence electrons around phosphorus
- Valence electrons around P means the electrons that belong or are shared by phosphorus in bonds or lone pairs.
- Phosphorus has no lone pairs in the phosphate ion.
- Each P–O bond has 2 electrons, and P shares those electrons in the bond.
- Since there are 4 P–O bonds, phosphorus is involved in $4 \times 2 = 8$ electrons.
Step 4: Does phosphorus violate the octet rule?
- Phosphorus can expand its octet because it is in period 3, so it can have more than 8 electrons.
- In some resonance structures of phosphate, phosphorus can have 10 electrons (with a double bond).
- But in the basic tetrahedral Lewis structure of $\text{PO}_4^{3-}$, phosphorus shares 8 electrons.
Answer: Eight (8) valence electrons around phosphorus in the standard Lewis structure for phosphate ion.
Explanation in ~300 words
The phosphate ion $\text{PO}_4^{3-}$ consists of a central phosphorus atom bonded to four oxygen atoms. When drawing the Lewis structure, phosphorus, which has 5 valence electrons, forms bonds with four oxygen atoms, each having 6 valence electrons. The overall charge of -3 is typically distributed on the oxygen atoms, making the ion stable.
In the Lewis structure, phosphorus forms four single covalent bonds to oxygen atoms. Each bond consists of a pair of shared electrons (2 electrons per bond). Since phosphorus has no lone pairs in this structure, the only electrons directly around phosphorus are the bonding electrons shared with oxygen atoms.
Counting these electrons, we find:
- Four P–O bonds × 2 electrons per bond = 8 electrons
This means phosphorus has 8 valence electrons around it in the Lewis structure of the phosphate ion, satisfying the octet rule. Although phosphorus can expand its octet (since it is in period 3 of the periodic table), the typical Lewis structure shows 8 electrons around it.
This rules out other answer choices:
- 9 or 12 electrons would suggest hypervalency or lone pairs, which aren’t present in the basic structure.
- 4 electrons would correspond to phosphorus only having single electrons or incomplete bonding, which is not the case here.
- None of the above is incorrect because 8 electrons is the correct count.
Thus, the correct choice is:
a. Eight