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

Draw the structure of bromous acid, HBrO2. Optimize formal charges. Draw the structure of bromous acid, HBrO2. Optimize formal charges. Draw the Lewis structure of NCl3. Include lone pairs.

The Correct Answer and Explanation is:

1. Bromous Acid (HBrO2)

Step 1: Lewis Structure of HBrO2

Valence Electrons:
- Bromine (Br) is in group 17, so it has 7 valence electrons.
- Oxygen (O) is in group 16, so each oxygen has 6 valence electrons.
- Hydrogen (H) has 1 valence electron.

So, the total valence electrons are:
7 (Br)+6 (O)×2+1 (H)=20 electrons7 \, (\text{Br}) + 6 \, (\text{O}) \times 2 + 1 \, (\text{H}) = 20 \, \text{electrons}7(Br)+6(O)×2+1(H)=20electrons

Step 2: Skeleton Structure

Place the bromine atom (Br) in the center as it’s less electronegative than oxygen.
Connect the oxygen atoms to bromine with single bonds and add hydrogen to one of the oxygen atoms, since it’s an acid.

The structure looks like this:

mathematicaCopyEdit   H - O - Br - O

Step 3: Completing Octets and Adding Lone Pairs

Add lone pairs to oxygen atoms to satisfy their octet.
Bromine can exceed the octet rule because it’s in period 4, so it can have more than 8 electrons.

The full structure with lone pairs:

mathematica    H        O:      Br:      O:
      |         |       ||      ||
      O - Br - O:  :

Step 4: Formal Charges

Now, let’s calculate the formal charges:

Hydrogen (H): 1 bond, 0 lone pairs = 1 valence electron → formal charge = 0.
Oxygens: Oxygen has 6 valence electrons; each bonded to hydrogen or bromine. We distribute the lone pairs to minimize formal charge.
Bromine: It has 7 valence electrons, 3 bonds, and a lone pair.

When formal charges are optimized, the structure remains close to neutral, with the oxygen atoms having negative formal charges and bromine remaining neutral.

2. Lewis Structure of Nitrogen Trichloride (NCl3)

Step 1: Valence Electrons

Nitrogen (N): 5 valence electrons.
Chlorine (Cl): 7 valence electrons (for 3 chlorine atoms).

Total valence electrons:
5 (N)+7 (Cl)×3=26 electrons5 \, (\text{N}) + 7 \, (\text{Cl}) \times 3 = 26 \, \text{electrons}5(N)+7(Cl)×3=26electrons

Step 2: Skeleton Structure

Place nitrogen (N) in the center as it is less electronegative than chlorine.
Attach each chlorine (Cl) to nitrogen (N) with single bonds.

The initial skeleton looks like this:

  Cl - N - Cl
        |
       Cl

Step 3: Completing Octets

Chlorine atoms will have 3 lone pairs each to satisfy the octet rule.
Nitrogen will have one lone pair and be involved in three single bonds with chlorine.

The full structure with lone pairs:

ruby     :Cl:       :Cl:
       |         |
   :Cl: - N - :Cl:
       |  
      :Cl:

Step 4: Formal Charges

Chlorines (Cl): Each chlorine atom has 3 lone pairs and 1 bond. Their formal charge is 0.
Nitrogen (N): Nitrogen has 5 valence electrons, 3 bonds, and 1 lone pair. Its formal charge is also 0.

Thus, the final structure is NCl3, where all atoms have formal charges of 0.

Conclusion:

HBrO2: The structure is based on octet rules, with bromine able to accommodate more electrons, while formal charges are minimized by distributing lone pairs appropriately on oxygen.
NCl3: The structure places chlorine atoms around nitrogen, satisfying the octet for chlorine and ensuring formal charge neutrality for all atoms.

1. Bromous Acid (HBrO2)

2. Lewis Structure of Nitrogen Trichloride (NCl3)

Conclusion:

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