To calculate the number of bond pairs in the hydronium ion (\(H_3O^+\)), we first need to determine its Lewis structure. The hydronium ion consists of a central oxygen atom bonded to three hydrogen atoms and carries a positive charge.
Here's how to analyze it:
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Count Valence Electrons:
- Oxygen has 6 valence electrons.
- Each hydrogen atom has 1 valence electron, totaling 3 from three hydrogen atoms.
- The positive charge indicates that we remove one electron from the total count.
So, the total number of valence electrons is: \[ 6 \text{ (from O)} + 3 \times 1 \text{ (from H)} - 1 \text{ (for the positive charge)} = 8 \text{ valence electrons.} \]
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Draw the Lewis Structure:
- Place the oxygen atom in the center and bond it to three hydrogen atoms.
- Each O-H bond uses 2 valence electrons.
- Three O-H bonds will use \(3 \times 2 = 6\) electrons, leaving 2 electrons.
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Consider the Remaining Electrons:
- The remaining 2 electrons can be used to complete the octet of oxygen.
- Since only three bonds are formed with hydrogen atoms, there are no lone pairs on the oxygen atom.
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Identify Bond Pairs:
- Each O-H bond is a bond pair.
- In \(H_3O^+\), there are 3 O-H bonds.
Therefore, the hydronium ion \(H_3O^+\) has 3 bond pairs.