What would be the the oxidation number of (N):
* NaNO3
* NH4Cl
Isnt N +V in both?
It is +5 in NaNo3 but it is -3 in NH4Cl
To determine the oxidation number of an element in a compound, you need to consider a few rules:
1. The oxidation number of an atom in its free state (in an elemental form) is always zero.
2. The sum of the oxidation numbers of all atoms in a neutral compound is zero.
3. The sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge of the ion.
Now let's apply these rules to find the oxidation number of nitrogen (N) in the given compounds:
1. NaNO3:
Since the overall compound is neutral, the sum of the oxidation numbers of all atoms should be zero. The oxidation number of sodium (Na) is always +1, and oxygen (O) is usually -2.
Let's denote the oxidation number of nitrogen as x.
Therefore, (+1) + x + (3 × -2) = 0
Simplifying the equation: +1 + x - 6 = 0
x - 5 = 0
x = 5
So, the oxidation number of nitrogen (N) in NaNO3 is +5.
2. NH4Cl:
Again, we need to consider that the compound is neutral. The oxidation number of hydrogen (H) is always +1, and chlorine (Cl) is usually -1.
Let's denote the oxidation number of nitrogen as y.
Therefore, (+1) + 4 × (+1) + y + (-1) = 0
Simplifying the equation: +1 + 4 + y - 1 = 0
y + 4 = 0
y = -4
So, the oxidation number of nitrogen (N) in NH4Cl is -4.