Draw the structure of bromous acid, HBrO2

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Draw the structure of bromous acid, HBrO2. Optimize formal charges.

The Correct Answer and Explanation is:

The structure of bromous acid (HBrO2) consists of the following:

Steps to Draw the Structure:

  1. Determine the number of valence electrons:
    • H: 1 valence electron
    • Br: 7 valence electrons
    • O: 6 valence electrons per oxygen atom (2 oxygen atoms)
    Total electrons = 1 + 7 + 12 = 20 valence electrons.
  2. Initial bonding:
    • Place H attached to one of the oxygen atoms (hydroxyl group -OH).
    • Attach the Br to the other oxygen atoms with single bonds.
    • Connect the remaining oxygen atoms via single bonds to Br.
  3. Distribute remaining electrons:
    • Place lone pairs on oxygen atoms to complete their octets.
    • If necessary, form a double bond between Br and one of the oxygen atoms to minimize formal charges and ensure the most stable structure.

Final Structure:

  • Bromous acid (HBrO2) has a central Br atom with single bonds to two oxygen atoms and one single bond to the hydroxyl group (-OH).
  • The Br atom forms a double bond with one of the oxygen atoms.
  • The structure will be:
    HO-Br=O
    with the second oxygen atom (double-bonded to Br) having two lone pairs of electrons.

Formal Charges:

  • Br has a formal charge of 0 in the optimal structure because it follows the octet rule with a double bond and two lone pairs on oxygens.
  • Each oxygen atom should have a formal charge of 0, as they satisfy their octet by bonding with the bromine and having lone pairs.
  • Hydrogen has no formal charge.

explanation:

Bromous acid (HBrO2) is an oxoacid where the central bromine atom is bonded to two oxygen atoms and one hydroxyl group (-OH). The structure must account for the number of valence electrons and bond connectivity while minimizing formal charges. The octet rule is key in determining the number of bonds between atoms.

Starting with the bonding skeleton, the hydrogen atom attaches to one oxygen atom to form the hydroxyl group (-OH). The central bromine atom then bonds with two oxygen atoms. To optimize the structure and formal charges, a double bond is formed between the bromine and one oxygen atom. This arrangement ensures that the formal charges are balanced.

In this structure, the bromine atom does not exceed its valence shell of electrons, and it maintains neutrality with a formal charge of 0. The two oxygen atoms involved in the bonding also retain zero formal charge. The hydroxyl group (-OH) ensures that the hydrogen atom carries no formal charge either.

This optimal structure minimizes electron repulsion, maintains formal charges at zero where possible, and follows the typical bonding patterns of bromine and oxygen. The overall stability of the molecule is determined by having no formal charges and satisfying the valence shell electron rules for each atom. The final structure is both energetically favorable and chemically consistent with the properties of bromous acid.

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