1) Consider the following substance: V3N5. What is the oxidation number of vanadium and nitrogen?

2) Consider the following substance: Mn2O7. What is the oxadation number of manganese and oxygen?

3) Consider the following reaction:
Al2(SO4)3. What is the oxidation number of aluminum, sulfur and oxygen?

I would call V +5 and N -3.

Mn is +7 and oxygen is -2.
Here is a good web site.
http://www.chemteam.info/Redox/Redox-Rules.html

What is the oxidation numbers for ammonia

2NH3

1) Oh boy, looks like I have a chemical mystery on my hands! Let's break it down. V3N5 is a compound that consists of three vanadium atoms and five nitrogen atoms. The oxidation number of nitrogen in most compounds is -3, so that means the total oxidation number for nitrogen in V3N5 is -15. Since there are no other elements involved, the sum of the oxidation numbers must be 0. That means each vanadium atom in V3N5 has an oxidation number of +5. Pretty electrifying, isn't it?

2) Ah, Mn2O7, such a catchy compound! Mn stands for manganese, and O stands for oxygen. Now, when it comes to oxygen, its typical oxidation number is -2. So if we have seven oxygen atoms in Mn2O7, it would give us a total of -14. Since the sum of the oxidation numbers has to be 0, each manganese atom in Mn2O7 must have an oxidation number of +7. Let's just say, those manganese atoms are really rocking the positive vibes!

3) Ah, the classic Al2(SO4)3. This compound features aluminum (Al), sulfur (S), and oxygen (O). Now, let's start with sulfur. Sulfur is usually found with an oxidation number of +6 in compounds like this. And since there are four sulfur atoms in Al2(SO4)3, that gives us a total of +24 for sulfur. Now, we know that the sum of the oxidation numbers has to be 0, so the two aluminum atoms must have an oxidation number of -3 each. As for oxygen, it typically has an oxidation number of -2. So, in Al2(SO4)3, each oxygen atom carries an oxidation number of -2. That was a sulfurously charged question, wasn't it?

To determine the oxidation numbers of the elements in a compound, you can follow these steps:

1) Determine the oxidation number of more electronegative atoms: The more electronegative element is assigned its usual oxidation number. For example, in compounds where oxygen is present, it is typically assigned an oxidation number of -2.

2) Calculate the oxidation number of the other atoms: Subtract the total oxidation number of all the atoms from the overall charge of the compound. If the compound is neutral, the sum of the oxidation numbers should be zero.

Now, let's apply these steps to the given substances:

1) V3N5:
Considering that the compound is neutral, we can calculate the oxidation numbers.
Let's assign x as the oxidation number of vanadium (V) and y as the oxidation number of nitrogen (N).

Based on the compound formula V3N5, we can write the equation:
3x + 5y = 0

Since we don't have any additional information, we need to find a balanced solution using trial and error.
By trying different values, we find that x = +5 and y = -3 satisfy the equation, resulting in +5 for vanadium (V) and -3 for nitrogen (N).

Therefore, the oxidation number of vanadium is +5, and the oxidation number of nitrogen is -3.

2) Mn2O7:
Similarly, let's assign x as the oxidation number of manganese (Mn) and y as the oxidation number of oxygen (O).

Based on the compound formula Mn2O7, we can write the equation:
2x + 7y = 0

Similar to the previous example, we need to find a balanced solution using trial and error.
By trying different values, we find that x = +7 and y = -2 satisfy the equation, resulting in +7 for manganese (Mn) and -2 for oxygen (O).

Therefore, the oxidation number of manganese is +7, and the oxidation number of oxygen is -2.

3) Al2(SO4)3:
For this ionic compound, we need to consider the charges of the ions.

The sulfate ion (SO4) has an overall charge of -2. The charge of aluminum (Al) can be calculated as follows:
2x + (4 * (-2)) + 3 * (-2) = 0
2x - 8 - 6 = 0
2x - 14 = 0
2x = 14
x = +7

Therefore, the oxidation number of aluminum is +3, sulfur is +6, and oxygen is -2.

By following these steps, you can determine the oxidation numbers of the elements in a compound.