Draw Lewis structures for each of the following compounds.

Draw Lewis structures for each of the following compounds. (Assign lone pairs and radical electrons where appropriate.) (a) bromoamine (NH2Br) (b) fluoroethyne (C2HF) (c) ethanol (C2H5OH)

The Correct Answer and Explanation is:

Here are the correct Lewis structures and explanations for the given compounds:

(a) Bromoamine (NH₂Br)

Lewis Structure:

mathematicaCopyEdit  H
  |
H–N–Br
  ..

Explanation:
Bromoamine consists of nitrogen (N), hydrogen (H), and bromine (Br). Nitrogen has 5 valence electrons, each hydrogen has 1, and bromine has 7. Nitrogen forms three bonds: two with hydrogen atoms and one with bromine. Nitrogen then has one lone pair. Bromine forms a single bond with nitrogen and has three lone pairs to complete its octet. Each hydrogen is fully satisfied with one bond. There are no radical electrons.

(b) Fluoroethyne (C₂HF)

Lewis Structure:

mathematicaCopyEditH–C≡C–F

Explanation:
Fluoroethyne contains two carbon atoms, one hydrogen, and one fluorine. Each carbon has 4 valence electrons, hydrogen has 1, and fluorine has 7. The two carbon atoms are connected by a triple bond (one sigma and two pi bonds). One carbon is bonded to hydrogen, and the other to fluorine. Fluorine, having formed one bond, retains three lone pairs. Each atom achieves a full octet except hydrogen, which only needs two electrons. There are no unpaired electrons in this molecule.

(c) Ethanol (C₂H₅OH)

Lewis Structure:

mathematicaCopyEdit      H   H   H
      |   |   |
H–C–C–O–H
    |   
    H

Explanation:
Ethanol consists of two carbon atoms, six hydrogen atoms, and one oxygen atom. The first carbon (left) forms four bonds (three with hydrogen, one with the second carbon). The second carbon forms bonds with two hydrogens, the first carbon, and the hydroxyl group (–OH). The oxygen in the hydroxyl group forms two bonds (one with carbon and one with hydrogen) and has two lone pairs. Each atom follows the octet rule (except hydrogen), and there are no radical electrons in this molecule.

All structures are drawn with appropriate lone pairs and bond types, and no radical electrons are present in any of the three compounds.