Solution:
We need to find out where nitrogen has the lowest (most negative) oxidation state among the options. To do this, we will calculate the oxidation state of nitrogen in each compound.
(a) In NH2OH (Hydroxylamine):
Let nitrogen's oxidation state be x. Hydrogen (H) is +1 and oxygen (O) is -2 .
There are 2 H atoms, so 2×(+1)=+2. There is 10 atom, so (−2). There is also 1 extra H (in OH ), so another +1 .
x+2(+1)+(−2)+(+1)=0
x+1=0
x=−1
So, in hydroxylamine, nitrogen is at -1 .
(b) In NH4Cl (Ammonium chloride):
This compound has an ammonium ion (NH4+)and a chloride ion (Cl−).
Let nitrogen's oxidation state be x. Hydrogen is +1 , chlorine is -1 .
There are 4 H atoms, so 4×(+1)=+4.
x+4×1−1‌=0
x+4(+1)‌=+1
x+4‌=1
x‌=−3
So, in ammonium chloride, nitrogen is at -3 .
(c) In N2H4 (Hydrazine):
Let nitrogen's oxidation state be x. Hydrogen is +1 .
There are 2 nitrogen and 4 hydrogen atoms.
2x+4(+1)‌=0
2x‌=−4
x‌=−2
So, in hydrazine, nitrogen is at -2 .
(d) In HNO2 (Nitrous acid):
Let nitrogen's oxidation state be x. Hydrogen is +1 , oxygen is -2 .
There is 1 H and 2 O .
(+1)+x+2(−2)=0
1+x−4=0
x−3=0
x=+3
So, in nitrous acid, nitrogen is at +3 .
Now, let's compare the results: −1,−3,−2,+3. The smallest value is -3 . This happens in NH4Cl (ammonium chloride).
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