CramX Logo

Q
QuestionChemistry

Which is the Lewis structure for H^3PO^4? B) $H \square \ddot{0} \square H$ C) $\ddot{0}$ : D) D) $\quad \ddot{H}$
12 months agoReport content

Answer

Full Solution Locked

Sign in to view the complete step-by-step solution and unlock all study resources.

Step 1:
: Determine the total number of valence electrons in the Lewis structure for H^3PO^4.

5 (P) + 3 (H) = 8 \text{ valence electrons}
Phosphorus (P) is in the 5th period and group VA of the periodic table, so it has 5 valence electrons. Each hydrogen (H) atom has 1 valence electron. Therefore, the total number of valence electrons in H^3PO^4 is:

Step 2:
: Draw the skeletal structure of H^3PO^4.

Phosphorus is the central atom, and four sites are occupied by hydrogen atoms or single bonds. The skeletal structure is: ``` H - P - O | H | H ```

Step 3:
: Place the remaining valence electrons on the outer atoms as lone pairs.

Since there are 8 valence electrons, and 8 electrons are already in the skeletal structure, there are no more electrons to place.

Step 4:
: Check the formal charges of the atoms in the Lewis structure.

FC(O) = 6 - 4 - \frac{1}{2}(0) = 2 - 1 = 1
The formal charge of an atom in a Lewis structure is calculated as follows: For the central phosphorus atom: For each hydrogen atom: For each oxygen atom: Since the formal charges are reasonable (nearby atoms have opposite charges), the Lewis structure is acceptable.

Step 5:
: Compare the given options to the correct Lewis structure.

The correct Lewis structure for H^3PO^4 is: ``` H - P : | H | H ``` where the double-bonded oxygen has two lone pairs of electrons. Comparing the given options, the closest match is option D): ![img- 3.jpeg](img- 3.jpeg) However, the double-bonded oxygen atom should have two lone pairs of electrons. Therefore, the correct option should be: ![img- 5.jpeg](img- 5.jpeg)

Final Answer

The closest correct option is D'), which includes the double-bonded oxygen atom with two lone pairs of electrons.