oxidation state of nitrogen in n2h4
In this compound N2H4 the oxidation number on each atom of nitrogen is -2 and oxidation number on each hydrogen atom is +1 so the over all on nitrogen atoms has -4 and on hydrogen atoms is +4 so overall compound is neutral .
To determine the oxidation state of nitrogen (N) in N2H4 (hydrazine), we need to assign oxidation states to each individual atom.
Let's assume the oxidation state of hydrogen (H) is +1 and the oxidation state of oxygen (O) is -2.
Since hydrazine (N2H4) is a neutral molecule, the sum of the oxidation states must equal zero.
Let x be the oxidation state of nitrogen (N).
The compound N2H4 contains two nitrogen atoms (N), so we can write the equation:
2(x) + 4(+1) = 0
2x + 4 = 0
2x = -4
Dividing both sides by 2, we get:
x = -2
Therefore, the oxidation state of nitrogen (N) in N2H4 is -2.
To determine the oxidation state of nitrogen in a compound like N2H4 (hydrazine), we need to follow a few steps.
1. Assign a variable (x) to the oxidation state of nitrogen.
2. Since hydrazine is a neutral compound, the sum of the oxidation states of all the atoms must be zero.
3. The total number of hydrogen atoms in N2H4 is 4 (two per nitrogen atom), and hydrogen always has an oxidation state of +1.
4. The sum of the oxidation states of the nitrogen atoms must equal zero, so we can write the equation: 2x + 4(+1) = 0.
Solving this equation, we get:
2x + 4 = 0
2x = -4
x = -2
Therefore, the oxidation state of nitrogen in N2H4 is -2.
It depends on the rules for assigning oxidation numbers.
My rule for Hydrogen is :
# Hydrogen has an oxidation state of +1 except when bonded to more electropositive elements such as sodium, aluminium, and boron, as in NaH, NaBH4, LiAlH4, where each H has an oxidation state of -1.\
So here, N2H4 means nitrogen is the more electropositive, so H is -1, leaving N to be +2
N is +2