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QuestionChemistry

Draw one Lewis structure (including all bonds and all non-bonding electron pairs/lone pairs) of BrO 3 − ​ (bromate ion) that follows the octet rule. Determine the number of valence electrons in the molecule. Predict the ideal bond angles. Determine the electron group and molecular geometry according to the VSEPR method for the bromine atom.
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Answer

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Step 1:
I'll solve this problem step by step, following the specified LaTeX formatting guidelines:

Step 2:
: Determine the total number of valence electrons

Total valence electrons: $$7 + 18 + 1 = 26$$ electrons
- Bromine (Br): 7 valence electrons - Oxygen (O): 6 valence electrons × 3 = 18 valence electrons - Negative charge: + 1 electron

Step 3:
: Arrange the atoms

- Place bromine in the center - Attach three oxygen atoms to bromine - Distribute remaining electrons as lone pairs to satisfy octets

Step 4:
: Draw the Lewis structure

- Create single bonds between Br and each O - Distribute remaining electrons as lone pairs - Ensure each atom (except Br) has a complete octet - One oxygen will have a formal negative charge Detailed Lewis structure characteristics: - Two oxygen atoms have single bonds and complete octets - One oxygen atom has a single bond and a negative charge - Bromine has four electron groups (three single bonds, one lone pair)

Step 5:
: Determine ideal bond angles

- Ideal bond angle: $$109.5°
- Electron group geometry around bromine: Tetrahedral

Step 6:
: VSEPR Analysis

- Electron group geometry: Tetrahedral - Molecular geometry: Trigonal pyramidal - Electron groups around central Br: 4 (3 bonds, 1 lone pair)

Final Answer

The Lewis structure of BrO_{3}^{-} follows the octet rule with: - 26 total valence electrons - Tetrahedral electron geometry - Trigonal pyramidal molecular geometry - Ideal bond angle of 109.5°