Answered

Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

Drag the tiles to the correct locations. Each tile can be used more than once, but not all tiles will be used. One or more locations will remain empty.

Nitrosyl fluoride has the chemical formula NOF. Nitrogen has five valence electrons, oxygen has six, and fluorine has seven. Complete the Lewis structure for this covalent compound.

N
[tex]$\square$[/tex]
[tex]$\square$[/tex]
[tex]$\square$[/tex]
[tex]$\square$[/tex]
[tex]$\square$[/tex]

Sagot :

GinnyAnswer
In order to provide the required Lewis structure for Nitrosyl fluoride (NOF), we need to understand the distribution of the valence electrons for the given atoms. The chemical formula of the compound is NOF, which means it contains one nitrogen (N), one oxygen (O), and one fluorine (F) atom.

Let's count the total number of valence electrons in the molecule:
- Nitrogen (N) has 5 valence electrons.
- Oxygen (O) has 6 valence electrons.
- Fluorine (F) has 7 valence electrons.

Adding them together gives us:
5 (from N) + 6 (from O) + 7 (from F) = 18 valence electrons in total.

Now, let's arrange these valence electrons according to the rules of the Lewis structure:

1. Write the symbols for the atoms, and connect them with single bonds. Usually, the least electronegative atom is placed at the center, except for hydrogen. Here, nitrogen (N) is the least electronegative atom.
N - O - F

2. Distribute the remaining valence electrons to fulfill the octet rule (or duet rule for hydrogen). Each dash (bond) represents 2 electrons.

3. Start by connecting the atoms with single bonds and then distribute the remaining electrons to complete the octets.

Drawing the Lewis structure for NOF:
1. Place single bonds between N, O, and F:
N - O - F

2. Each N-O-F bond initially counts as 2 electrons each. Thus, we initially have:
3 bonds * 2 electrons/bond = 6 valence electrons used.
Remaining electrons: 18 - 6 = 12

3. Complete the octets starting with the outer atoms (F and O), and then the central atom (N).

- Fluorine (F) already has 1 bond (2 electrons). To complete its octet, it needs 6 more electrons:
F: bonds (2) + lone pairs (6) = 8 electrons

- Update the electron count:
12 - 6 = 6 electrons remaining

- Oxygen (O) also needs to complete its octet. It currently has 2 electrons from the bond with N and F. Add 4 more electrons:
O: bonds (2) + lone pairs (4) = 8 electrons

- Update the electron count:
6 - 4 = 2 electrons remaining

- Now, place the remaining 2 electrons on nitrogen (N). Originally, nitrogen had 2 bonds (N-O and N=O). Adding the 2 remaining electrons:
N: bonds (4) + lone pairs (2) = 6 electrons - violating the octet rule.

Considering this, we can adjust and create a double bond between N and O to complete its octet for both atoms.

Final Lewis structure for NOF:

..
:N=O:
..
..
|
F
..

Submit this final Lewis structure for your reference.
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.