Draw a Lewis dot diagram for NF2H and use the oxidation state method of electron bookkeeping to determine how many electrons each atom should be assigned.
We can't draw Lewis structures on this board.
:::F-N:-F:::
-H
To draw a Lewis dot diagram for NF2H, we need to determine the number of valence electrons for each atom.
First, let's find the valence electrons for each atom:
N: Nitrogen is in group 5 of the periodic table, so it has 5 valence electrons.
F: Fluorine is in group 7, so it has 7 valence electrons.
H: Hydrogen is in group 1, so it has 1 valence electron.
Now, let's determine how many electrons each atom should be assigned using the oxidation state method of electron bookkeeping. In this method, we assign electrons based on the following rules:
1. Each bond between two atoms is counted as two electrons.
2. Each lone pair of electrons on an atom is counted as two electrons.
For NF2H, nitrogen (N) will be the central atom. It will form single bonds with two fluorine atoms (F) and a single bond with a hydrogen atom (H).
N:
- Nitrogen forms three single bonds, so it will be assigned 6 electrons for these bonds (3 x 2 = 6).
- Nitrogen has 5 valence electrons, so it will be assigned 5 - 6 = -1 electrons.
F:
- Fluorine forms a single bond with nitrogen, so it will be assigned 2 electrons for this bond.
- Fluorine has 7 valence electrons, so it will be assigned 7 - 2 = 5 electrons.
H:
- Hydrogen forms a single bond with nitrogen, so it will be assigned 2 electrons for this bond.
- Hydrogen has 1 valence electron, so it will be assigned 1 - 2 = -1 electron.
The Lewis dot diagram for NF2H with the assigned electrons is as follows:
H
|
F - N - F
The nitrogen atom has a charge of -1, while the fluorine and hydrogen atoms have a charge of -1.
To draw a Lewis dot diagram for NF2H, we need to determine the total number of valence electrons in the molecule.
The symbol N represents nitrogen, which is in Group 5A of the periodic table and has 5 valence electrons. The symbol F represents fluorine, which is in Group 7A and has 7 valence electrons. The symbol H represents hydrogen, which is in Group 1A and has 1 valence electron.
N: 5 valence electrons
F: 7 valence electrons (2 fluorine atoms)
H: 1 valence electron (1 hydrogen atom)
Now, let's determine the total number of valence electrons:
Total valence electrons = (5 from N) + (2 × 7 from 2 F atoms) + (1 from H)
Total valence electrons = 5 + 14 + 1 = 20
Next, we'll draw the Lewis dot diagram for NF2H:
N: ·
F: ··
F: ··
H: ·
Now, let's assign oxidation states to each atom using the oxidation state method of electron bookkeeping:
The oxidation state of hydrogen (H) is usually +1 in compounds.
The oxidation state of fluorine (F) is usually -1 in compounds.
The oxidation state of nitrogen (N) can vary, but we can determine it using the following equation:
Total charge on molecule = Sum of oxidation states
Since the molecule is neutral, the total charge is 0. Therefore:
Oxidation state of N + (2 × Oxidation state of F) + Oxidation state of H = 0
Let's replace the oxidation states:
Oxidation state of N + (2 × -1) + (1 × +1) = 0
Simplifying the equation:
Oxidation state of N - 2 + 1 = 0
Oxidation state of N - 1 = 0
Oxidation state of N = +1
So, using the oxidation state method, we assign the following oxidation states:
N: +1
F: -1
H: +1
I hope this explanation helps you understand how to draw a Lewis dot diagram and determine oxidation states using the oxidation state method of electron bookkeeping!