What is the oxidation number for each atom in NH4F?

N=+3, H=-4, F=+1
N=-3, H=+1, F=-1***
N=+3, H+1, F=-1
N=-3, H=-1, F=+1

To determine the oxidation number for each atom in NH4F (ammonium fluoride), we need to apply certain rules.

1. The sum of the oxidation numbers in a molecule or ion must equal the overall charge of the molecule or ion. In the case of NH4F, it is a neutral molecule, so the sum of the oxidation numbers is zero.

2. The oxidation number of hydrogen (H) is usually +1, except when it is bonded to a metal, in which case it can be -1.

3. The oxidation number of fluorine (F) is always -1 in compounds.

Using these rules, we can determine the oxidation number for each atom in NH4F.

In NH4F, we have one nitrogen atom (N) and four hydrogen atoms (H) bonded to it, along with one fluorine atom (F).

Since we know the oxidation number of fluorine is -1, and there is only one fluorine atom in the compound, the contribution to the sum of oxidation numbers from fluorine is -1.

Let's assume the oxidation number of nitrogen is "x" and the oxidation number of hydrogen is "y."

Since there are four hydrogen atoms each with an oxidation number of +1, the total contribution from hydrogen would be +4.

Now, applying the rule that the sum of oxidation numbers in a neutral molecule is zero, we can set up an equation:

x + 4y - 1 = 0 (the sum must equal zero)

To solve for "x" and "y," we can rearrange the equation:

x = 1 - 4y

Now, we substitute this expression for "x" into the equation:

(1 - 4y) + 4y - 1 = 0

When we simplify this equation, we find that no matter the value of "y," the solution becomes:

x = -3

So, the oxidation number of nitrogen (N) in NH4F is -3.

Thus, the correct answer is:

N = -3, H = +1, F = -1

The correct oxidation numbers for each atom in NH4F are: N = -3, H = +1, F = -1.

correct