What is the oxidation number of sulfur (s) in NaHSO3?
I know that Na is 1 but I do not know what else to do. Please explain, thank you for your help!!
Well, sulfur is a bit of a trickster when it comes to oxidation numbers. In compounds like NaHSO3, we can assign oxidation numbers based on some general rules. Let's break it down.
First, we know that the oxidation number of Na is +1, given that it's in Group 1 of the periodic table.
Second, let's assume the oxidation number of oxygen (O) is -2. Since we have three oxygen atoms in NaHSO3, the total oxidation number contributed by oxygen is -6.
Now comes the tricky part. We need to figure out the oxidation number of hydrogen (H) and sulfur (S) by using the rule that the sum of the oxidation numbers in a compound should be equal to the charge of the compound, which is zero in this case.
Since there is only one hydrogen atom, we can assume its oxidation number is +1.
So, to balance out the oxidation numbers, we have +1 for Na, -2 for three oxygens (-6 in total), and +1 for hydrogen. This means the oxidation number of sulfur must be +4 to give a total of 0.
Therefore, the oxidation number of sulfur (S) in NaHSO3 is +4. Now you know how to sulfurize your chemistry knowledge!
To determine the oxidation number of sulfur (S) in NaHSO3, we need to look at the charges of the other elements in the compound and apply some rules.
1. The oxidation number of sodium (Na) is +1 in almost all compounds, as it readily loses one electron to achieve a stable electron configuration.
2. Oxygen (O) usually has an oxidation number of -2, unless it is combined with fluorine (F) or in certain peroxides where its oxidation number is -1.
3. Hydrogen (H) typically has an oxidation number of +1 when bonded to a nonmetal like sulfur.
4. Since the overall charge of the compound NaHSO3 is neutral, the sum of all the oxidation numbers must be zero.
Now let's put these rules into practice:
We know the oxidation number of Na is +1. Let's assign this value.
The oxidation number of oxygen is typically -2. In NaHSO3, there are three oxygen atoms. So -2 x 3 = -6.
The oxidation number of hydrogen is +1. In NaHSO3, there is one hydrogen atom. So +1 x 1 = +1.
Let's assign the oxidation number of sulfur as "x."
Now, we can add up all the oxidation numbers:
(+1) + (-6) + (+1) + (x) = 0
Simplifying the equation:
-4 + x = 0
We can solve for "x" by bringing -4 to the other side of the equation:
x = +4
Therefore, the oxidation number of sulfur (S) in NaHSO3 is +4.
'cause the sum of every atoms's N.O of this molecule is equal to 0
NaHSO3.
Na is +1.
H is +1.
O is -2 so 3*-2 = -6
Therefore, S must be +4?