Put the following compounds in order of INCREASING oxidation number of N:
N2H4, NO3-1, N2, NH3 ?
Do you know how to determine the oxidation number for N in these compounds?
I'll do the first two for you.
#1 rule. Compounds or elements have a charge of zero.
#2 rule.The oxidation state of all of the elements in the material must add to zero (if it's a compound) or to the charge on the ion if it's an ion.
For N2H4. H is +1 so 4 of those = +4. That makes 2N = -4 or each N must be -2.
How do you know H is +1. It's ALWAYS +1 except in hydrides such as LiH, NaH, KH, AlH3, etc.
For NO3^-. O is -2 each and 3*-2 = -6 total. You want to leave a -1 charge on the ion; therefore, N must be +5. You can check that to make sure.
N = +5
O = 3*-2 = -6
+5 + )-6) = -1 which is the charge on the ion.
How do I know O is -2. It is ALWAYS -2 except in peroxides or superoxides such as H2O2 (it's -1 in peroxides), K2O2 (another peroxide) or KO2 (it's -1/2 in superoxides).
Now you do the other two. Clearly explain why you're stuck if you need more help.
To determine the increasing order of oxidation numbers of nitrogen (N) in the given compounds, we need to understand the concept of oxidation states.
Oxidation states are assigned to individual atoms in a compound to indicate the apparent charge that atom would have if the compound was composed of ions. Oxidation numbers can be determined by considering electronegativity differences, electron sharing, and the distribution of electrons in a molecule.
Let's analyze each compound to find the oxidation number of nitrogen:
1. N2H4 (hydrazine):
In hydrazine, the sum of the oxidation numbers of all atoms in a neutral compound is zero. Let's assume the oxidation number of nitrogen is x. Since hydrogen (H) is less electronegative than nitrogen, it has an oxidation number of +1. We have two hydrogens, so their total oxidation number contribution is +2. Therefore, the sum of the oxidation numbers in N2H4 is 2x + 2 = 0. Solving this equation, we find that x = -1. Hence, the oxidation number of nitrogen in N2H4 is -1.
2. NO3^- (nitrate ion):
In the nitrate ion, the overall charge is -1. Oxygen (O) generally has an oxidation number of -2, so three oxygen atoms contribute a total charge of -6. We know the overall charge of the nitrate ion is -1, so the combined oxidation numbers of the three oxygen atoms is -6. Hence, the oxidation number of nitrogen in NO3^- is +5, which balances out the negative charge from the oxygens.
3. N2 (nitrogen gas):
In nitrogen gas, the oxidation number of each nitrogen atom is 0. In uncombined elemental form, the oxidation state of an element is always zero.
4. NH3 (ammonia):
In ammonia, we again consider that the sum of oxidation numbers in a neutral compound is zero. Hydrogen has an oxidation number of +1 as before, and we have three hydrogens, contributing +3 to the total sum. Assuming the oxidation number of nitrogen in NH3 is x, we have x + 3 = 0. Therefore, x = -3. Thus, the oxidation number of nitrogen in NH3 is -3.
Now, let's arrange the compounds in increasing order of oxidation number of nitrogen:
N2 (0) < N2H4 (-1) < NH3 (-3) < NO3^- (+5)
Therefore, the increasing order of oxidation numbers of nitrogen is: N2 < N2H4 < NH3 < NO3^-