QQuestionChemistry
QuestionChemistry
Which is the Lewis structure for H^3PO^4?
B) $H \square \ddot{0} \square H$
C) $\ddot{0}$ :
D)
D) $\quad \ddot{H}$
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Answer
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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):  However, the double-bonded oxygen atom should have two lone pairs of electrons. Therefore, the correct option should be: 
Final Answer
The closest correct option is D'), which includes the double-bonded oxygen atom with two lone pairs of electrons.
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