Problem: Draw Lewis structures of all the important resonance forms of (b) HBrO4 (HOBrO3)

FREE Expert Solution
  • For the Lewis structure of HBrO4, Br appears to have 7 valence electron, H as 1 and O has 6. Total valence electrons is 32. ((1(1) + 6(4) + 7(1)  = 32)
  • Br will appear as central atom since its less electronegative than O (H can't be central atom)
  • Draw the initial structure of HBrOwhere all O atoms is attached to Br and H on one of the O atom. This structure will try follow the octet rule to complete 8 electrons around each atom except for Br which has an expanded octet
  • Os around Br (except the OH) will try to have 1 electron (3 lone pairs in total) to complete the octet while Br will also take another electron forming 2 lone pairs
  • Calculating for the formal charges:

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Problem Details

Draw Lewis structures of all the important resonance forms of 

(b) HBrO4 (HOBrO3)


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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Resonance Structures concept. You can view video lessons to learn Resonance Structures. Or if you need more Resonance Structures practice, you can also practice Resonance Structures practice problems.

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Based on our data, we think this problem is relevant for Professor Dixon's class at UCF.

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Our data indicates that this problem or a close variation was asked in Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition. You can also practice Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition practice problems.