The oxidation number of each element in the compound  is:

The oxidation number of chlorine is .  There are 3 chlorine atoms present in the compound, so 

Because this is a neutral atom, the overall charge is 0.  Therefore we can set 

Therefore, 

We can then solve for the oxidation number for 

Recall that a reducing agent is being oxidized and is losing electrons. Thus, when looking at the equation, look for which oxidation state is becoming more positive.

Start by assigning oxidation states to the elements.

For the reactants:

 has an oxidation state of .

 has an oxidation state of .

 has an oxidation state of 

 has an oxidation state of .

Now, assign the oxidation state of the products.

 has an oxidation state of .

 has an oxidation state of .

 has an oxidation state of .

 has an oxidation state of .

Only  undergoes a loss of electrons. It must be the reducing agent.

Let's start by looking at the equation and assigning oxidation numbers based off of the general rules we are given:

Let's start by looking at the reactants, namely  . Knowing that an atom in its elemental form has an oxidation number of ,   has an oxidation number equal to .

Next let's look at . This is a neutral compound so know the oxidation number of the whole compound must equal . Going off of our general rules, the oxidation number of fluorine is equal . Since we have 3 fluorines we multiply the oxidation number by  to get the cumulative oxidation number of all the fluorines together to get . So now we must solve for the oxidation of Cl, which is unknown. This is done through a simple algebraic equation: 

so . Therefore the oxidation number of Chlorine is .

Now we have the oxidation numbers of all the elements present in this equation that are on the reactant side. I recommend writing down their oxidation numbers next to each other, so the elements that are oxidized and reduced can quickly be determined.

Oxidation numbers of elements in reactants:

Now let's look at the product side of the equation and determine the oxidation numbers of the elements there. First let's look at . Knowing the general rules of oxidation numbers we know that fluorine has an oxidation number of . Since there are  fluorines we multiply  which equals . We can now solve for the oxidation number of Uranium.

, therefore   has an oxidation number of .

Finally, let's look . We can once again use the general rule of oxidation numbers that fluorine has a  oxidation number to simplify this. Therefore 

, therefore Cl has an oxidation number of  in the product.

Let's now write down the oxidation numbers of all the elements in the products.

Oxidation numbers of elements in products:

Now let's compare the oxidation numbers of the elements in the reactants with those in the products.  goes from a  oxidation number to a , therefore it is the element that's oxidized (oxidation represents a loss in electrons).  goes from  to , therefore it is reduced (reduction means gaining electrons). Finally 's oxidation number is  in both the products and reactants therefore it is neither oxidized nor reduced. This means that  is the only element that's oxidized.

Let's start by going through each answer case-by-case applying the elementary rules of oxidation states we are given. The most applicable of these rules in this problem are the oxidation state of hydrogen in a compound is generally equal to , and the oxidation state of oxygen in a compound is generally equal to .

Let's start by looking at . We know this compound as a whole is neutrally charged (equal to  overall charge), and applying our knowledge of the general oxidation state of hydrogen in a compound and knowing there are  hydrogens we know that the   hydrogens have a total charge of  together, and nitrogen has an unknown oxidation state , so for our equation we have:

solving for  gives , giving us the answer we need. Therefore the nitrogen atom in  has an oxidation state of , which is the correct answer.

Let's still look at the other cases, however:

As this compound is uncharged we know it's net charge is equal to 

Using our knowledge of oxidation states of hydrogen and oxygen and counting the number of hydrogens and oxygen in this compound, we can determine that the total charge of all the hydrogens together is equal to , and that the total charge of the one oxygen is .

Just as we did above we can create an equation and solve for our unknown.

therefore , so the oxidation state of nitrogen in  is . This means this isn't the right compound.

Next let's look at :

We can quickly apply our knowledge of oxidation states of oxygen knowing that the oxidation state of oxygen in compounds is generally . Since there are  oxygens we must multiply   to find out the total charge contributed by all the oxygens, which is .

So our equation for this becomes:

therefore  and the oxidation state of nitrogen is , which isn't the answer we are looking for.

Finally the last compound is :

With this compound we can apply the rule we know about hydrogen's oxidation state in a compound being equal to . Since there are  hydrogens, we must multiply  which equals .

Since there are two nitrogens, our equation is:

and . Therefore the oxidation state of nitrogen is equal to .

Let's start by examining each of the compounds we are given case-by-case and applying the general rules of oxidation numbers we know.

Let's start with :

We know that based off of the general rules of oxidation states, Oxygen generally has an oxidation number of , since we have  oxygens we must multiply  to get the total cumulative charge of all the oxygens. This equals .

Now we can solve for the oxidation state of phosphorus:

Since we have  Phosphoruses we multiply our unknown  by .

.

This is equal to  because the compound has an overall  net charge.

Solving for  gives you  therefore the oxidation state of Phosphorus in this compound is equal to .

Next let's looking at :

Knowing that an atom in its elemental form has an oxidation state of . This compound has an oxidation number of .

Now let's look at :

Knowing that the oxidation number of hydrogen is  and that there are  hydrogens, the total oxidation of all the hydrogens together is , therefore our equation is:

so , therefore the oxidation state of Phosphorus in this compound is .

Now let's consider :

Using our elementary rules for the oxidation numbers, we know oxygen has a  oxidation number. We can also determine the oxidation number of chlorine, using the rule stating that the oxidation number of halogens is usually . Since there are three chlorines we must multiply , which gives us .

Therefore our equation is:

therefore , so the oxidation number of phosphorus in this case equals .

Finally let's look at :

Using the rules stated above for , we obtain the equation:

Therefore  meaning the oxidation number of .

This means that the compound with the lowest oxidation state is , therefore it is the right answer.

Let's first consider the overall compound . Since it is negatively charged overall it is equal to . Applying our elementary rules of oxidation states we know that halogens generally have an oxidation state of , so this means chlorine has an oxidation state of . Since there are  chlorines we must multiply  to get the charge of all the chlorines together. This is equal to .

We can now solve for oxidation state of gold through creating an equation as shown below:

Simplifying this we get that , therefore the oxidation state of gold is equal to 

Reduction is defined as the gain of electrons so we want to find the element that gains electrons/becomes negatively charged.

Once again our equation is 

Let's start on the reactant side with :

Following our general rules of oxidation states, we know that an atom in its elemental form has an oxidation state equal to zero, therefore  has an oxidation state equal to .

Next, consider : Based off of our general oxidation state rules we known the oxidation number of oxygen is , since we have two oxygens we must multiply . This means that the overall combined charge of the oxygens in this compound is equal to .

We can now solve for the oxidation state of  using this equation, which set equal to zero because the net charge of the compound is .

, therefore 

Now let's consider :

Applying the rule for the oxidation number of oxygen being  and the rule of hydrogen being equal to +1, we obtain the equation:

where  is the oxidation number of sulfur.

This simplifies so that , therefore the oxidation number of .

Moving on to the products, let's consider :

We know that the oxidation state of Oxygen is  based off the rules mentioned above. We can also determine that the oxidation number of  is equal to  because lead is most stable when it loses two electrons.

Now we must solve for sulfur's oxidation state:

therefore , therefore the oxidation state of sulfur .

As for , we can just use the general oxidation rules mentioned above to determine    

Looking at the changes of oxidation numbers between elements in the products and reactants we see that there is no change in the oxidation state of sulfur, hydrogen, and oxygen between the products and reactants, but lead goes from  to , so therefore it is reduced.

In order to determine the oxidation state of copper in this compound we first must note the fact that compound has an overall charge of .

We can also apply the general rules of oxidation numbers we know to determine the oxidation numbers of hydrogen and oxygen. Since hydrogen is a group  element, it has an oxidation number of . Since there are two hydrogens we must multiple  which equals   to get the  total charge of both hydrogens. 

As for oxygen, based off of our general rules of oxidation numbers, we know that oxygen normally has an oxidation number of , so knowing the oxidation numbers of oxygen and hydrogen, and net the charge of the compound, we can setup an equation to solve for the oxidation number of copper as shown below:

therefore , therefore copper has an oxidation number equal to .

In order to determine the oxidation number we must apply the general rules of oxidation numbers we know. One of these rules states the oxidation number of fluorine is always . Since there are  fluorines that means the fluorines together contribute a  charge to the compound. Since the overall charge of the compound is , we can create an equation with an unknown (charge of sulfur) and solve for it to find the oxidation number of sulfur.

This equation is  The reason why it is  instead of just  is because there are  sulfurs.

therefore the oxidation number of sulfur is .