Obtain the oxidation number for the element noted in each of the following.
Nb in NbO2
Br in KBrO4
Mn in KMnO4
Also I want to know how to do it.
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To determine the oxidation number of an element in a compound, you should follow these steps:
1. Identify the oxidation state of any known elements. In this case, potassium (K) in KBrO4 has a known oxidation state of +1.
2. Assign variables to the oxidation states of the remaining elements, if they are unknown. Let's use "x" for the oxidation state of the element in question.
3. Utilize the known oxidation state and assign it to the element. Remember, the total oxidation numbers of all elements in a compound must sum up to zero.
Now, let's apply these steps to each compound you mentioned:
1. Nb in NbO2:
Since the compound is neutral, the sum of the oxidation numbers must be zero. Let's assign "x" to the oxidation state of Nb.
2x + (2 * -2) = 0 (since there are two oxygen atoms)
2x - 4 = 0
2x = 4
x = 2
Therefore, the oxidation number of Nb in NbO2 is +2.
2. Br in KBrO4:
As mentioned earlier, the oxidation state of potassium (K) in KBrO4 is +1, and the sum of the oxidation numbers in the compound must equal zero.
So, let's assign "x" to the oxidation state of Br.
+1 + x + 4(-2) = 0 (4 oxygen atoms, each with an oxidation state of -2)
x - 6 = 0
x = +6
Therefore, the oxidation number of Br in KBrO4 is +6.
3. Mn in KMnO4:
Following the same logic as above, let's assign "x" to the oxidation state of Mn.
+1 + x + 4(-2) = 0
x - 6 = 0
x = +6
Therefore, the oxidation number of Mn in KMnO4 is +6.
So, to summarize:
The oxidation number of Nb in NbO2 is +2.
The oxidation number of Br in KBrO4 is +6.
The oxidation number of Mn in KMnO4 is +6.