give the oxidation number of nitrogen in the following:
1. NH2OH
2. N2F4
3. NH4+
4. HNO2
Here is a good site to help you in determining oxidation states(numbers).
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I'll be happy to check your work.
Sure! Here are the oxidation numbers of nitrogen in the given compounds:
1. NH2OH: The sum of the oxidation numbers in a neutral molecule is always zero. Since hydrogen is typically assigned an oxidation number of +1 and oxygen is usually assigned -2, we can set up the equation:
x + 2 * (+1) + (-2) + (+1) = 0
Simplifying the equation, we get:
x + 2 + (-2) + 1 = 0
x = -1
Therefore, the oxidation number of nitrogen in NH2OH is -1.
2. N2F4: In fluorine compounds, fluorine usually has an oxidation number of -1. Since the compound is neutral, the sum of the oxidation numbers must be zero. Setting up the equation:
2x + 4 * (-1) = 0
Simplifying the equation, we get:
2x - 4 = 0
2x = 4
x = +2
Therefore, the oxidation number of nitrogen in N2F4 is +2.
3. NH4+: Ammonium ion (NH4+) has an overall charge of +1. Since the sum of the oxidation numbers of all the atoms in an ion equals the charge of the ion, we can set up the equation:
x + 4 * (+1) = +1
Simplifying the equation, we get:
x + 4 = 1
x = -3
Therefore, the oxidation number of nitrogen in NH4+ is -3.
4. HNO2: Nitrous acid (HNO2) is a neutral molecule. Again, the sum of the oxidation numbers of all the atoms in a neutral molecule is zero. We can set up the equation:
(+1) + x + (-2) = 0
Simplifying the equation, we get:
x - 1 - 2 = 0
x = +1
Therefore, the oxidation number of nitrogen in HNO2 is +1.
To determine the oxidation number of nitrogen in a compound, you can follow some general rules:
1. For neutral compounds, the sum of the oxidation numbers must be zero.
2. For polyatomic ions, the sum of the oxidation numbers must be equal to the charge of the ion.
3. The oxidation number of hydrogen is always +1, and oxygen is typically -2.
Now let's determine the oxidation number of nitrogen in each compound you mentioned:
1. NH2OH:
Let's assume that the oxidation number of nitrogen is x.
The sum of the oxidation numbers in NH2OH can be set up as follows:
(+1) + (x) + (-2) + (+1) + (-2) = 0
Simplifying this equation, we have:
x - 2 = 0
x = +2
Therefore, the oxidation number of nitrogen in NH2OH is +2.
2. N2F4:
Let's assume that the oxidation number of nitrogen in N2F4 is x.
The sum of the oxidation numbers in N2F4 can be set up as follows:
(x) + (x) + 4(-1) = 0
Simplifying this equation, we have:
2x - 4 = 0
2x = 4
x = +2
Therefore, the oxidation number of nitrogen in N2F4 is +2.
3. NH4+:
The sum of the oxidation numbers in NH4+ must be equal to the charge of the ion, which is +1.
Let's assume that the oxidation number of nitrogen is x.
The sum of the oxidation numbers in NH4+ can be set up as follows:
(+1) + (x) + 4(+1) = +1
Simplifying this equation, we have:
x + 4 = 0
x = -3
Therefore, the oxidation number of nitrogen in NH4+ is -3.
4. HNO2:
Let's assume that the oxidation number of nitrogen in HNO2 is x.
The sum of the oxidation numbers in HNO2 can be set up as follows:
(+1) + (x) + 2(-2) = 0
Simplifying this equation, we have:
x - 4 = 0
x = +4
Therefore, the oxidation number of nitrogen in HNO2 is +4.
Remember to always consider the general rules for assigning oxidation numbers and set up the relevant equation to solve for the unknown oxidation number.