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To determine the Lewis structure for the covalent compound nitrosyl fluoride (NOF), follow these detailed steps:
1. Determine the total number of valence electrons:
- Nitrogen (N) has 5 valence electrons.
- Oxygen (O) has 6 valence electrons.
- Fluorine (F) has 7 valence electrons.
- The total number of valence electrons in NOF is:
[tex]\[ 5 (\text{from N}) + 6 (\text{from O}) + 7 (\text{from F}) = 18 \text{ valence electrons} \][/tex]
2. Create the skeletal structure:
- Place the least electronegative atom (Nitrogen) in the center.
- Connect it to Oxygen and Fluorine with single bonds initially.
3. Assign electrons to complete the octets:
- Start by satisfying the outer atoms (Fluorine and Oxygen), then move to the central atom (Nitrogen).
Here is a step-by-step construction of the Lewis structure:
Step 1: Draw the initial bonds.
[tex]\[ \text{N - O - F} \][/tex]
Step 2: Distribute remaining electrons to complete Fluorine's octet.
Fluorine needs 6 additional electrons (3 lone pairs):
[tex]\[ \text{N - O - } F \bullet \bullet \bullet \bullet \bullet \bullet \][/tex]
Step 3: Complete the octet for Oxygen.
Oxygen needs 4 additional electrons (2 lone pairs):
[tex]\[ \text{N - O} \bullet \bullet \bullet \bullet \text{- } F \bullet \bullet \bullet \bullet \bullet \bullet \][/tex]
Step 4: Determine the best Lewis structure:
Nitrogen currently has 2 electrons from its bonds and needs more to complete its octet. Oxygen would form one double bond with Nitrogen, sharing 2 more electrons:
[tex]\[ \text{N} = \text{O} \text{- } F \bullet \bullet \bullet \bullet \bullet \bullet \][/tex]
With lone pairs:
[tex]\[ \text{N} \bullet\bullet = \text{O} \bullet\bullet \text{- } F \bullet \bullet \bullet \bullet \bullet \bullet \][/tex]
Resulting Lewis structure:
[tex]\[ \begin{array}{c c c c c c} & F & : & .\ .\ .\ .\ & & \\ & | & & & & \\ N & = & O & : & .\:. \end{array} \][/tex]
- The Nitrogen (N) has 1 lone pair.
- The Oxygen (O) has 2 lone pairs.
- The Fluorine (F) has 3 lone pairs.
Therefore, the final Lewis structure is:
[tex]\[ \text{N} (=\text{O}) - \text{F}: \text{N with 1 lone pair, O with 2 lone pairs, F with 3 lone pairs.} \][/tex]
There are 18 valence electrons in total and each atom has an octet.
1. Determine the total number of valence electrons:
- Nitrogen (N) has 5 valence electrons.
- Oxygen (O) has 6 valence electrons.
- Fluorine (F) has 7 valence electrons.
- The total number of valence electrons in NOF is:
[tex]\[ 5 (\text{from N}) + 6 (\text{from O}) + 7 (\text{from F}) = 18 \text{ valence electrons} \][/tex]
2. Create the skeletal structure:
- Place the least electronegative atom (Nitrogen) in the center.
- Connect it to Oxygen and Fluorine with single bonds initially.
3. Assign electrons to complete the octets:
- Start by satisfying the outer atoms (Fluorine and Oxygen), then move to the central atom (Nitrogen).
Here is a step-by-step construction of the Lewis structure:
Step 1: Draw the initial bonds.
[tex]\[ \text{N - O - F} \][/tex]
Step 2: Distribute remaining electrons to complete Fluorine's octet.
Fluorine needs 6 additional electrons (3 lone pairs):
[tex]\[ \text{N - O - } F \bullet \bullet \bullet \bullet \bullet \bullet \][/tex]
Step 3: Complete the octet for Oxygen.
Oxygen needs 4 additional electrons (2 lone pairs):
[tex]\[ \text{N - O} \bullet \bullet \bullet \bullet \text{- } F \bullet \bullet \bullet \bullet \bullet \bullet \][/tex]
Step 4: Determine the best Lewis structure:
Nitrogen currently has 2 electrons from its bonds and needs more to complete its octet. Oxygen would form one double bond with Nitrogen, sharing 2 more electrons:
[tex]\[ \text{N} = \text{O} \text{- } F \bullet \bullet \bullet \bullet \bullet \bullet \][/tex]
With lone pairs:
[tex]\[ \text{N} \bullet\bullet = \text{O} \bullet\bullet \text{- } F \bullet \bullet \bullet \bullet \bullet \bullet \][/tex]
Resulting Lewis structure:
[tex]\[ \begin{array}{c c c c c c} & F & : & .\ .\ .\ .\ & & \\ & | & & & & \\ N & = & O & : & .\:. \end{array} \][/tex]
- The Nitrogen (N) has 1 lone pair.
- The Oxygen (O) has 2 lone pairs.
- The Fluorine (F) has 3 lone pairs.
Therefore, the final Lewis structure is:
[tex]\[ \text{N} (=\text{O}) - \text{F}: \text{N with 1 lone pair, O with 2 lone pairs, F with 3 lone pairs.} \][/tex]
There are 18 valence electrons in total and each atom has an octet.
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