 # 1 mole of N2H4 loses10 moles of electrons to form a new compond Y. assuming that all nitrogen appears in the new compound , what is the oxidation state of nitrogen in Y  ? [there is no change in the oxidation state of hydrogen.]

13 years ago

Hi

Oxidation state of NItrogen in n2h4 is -2.  2(-2) + 4(+1)= 0

therefore two N makes it -4 , since they lose 10 e-  , hence total charge is +6, and since H is still +1 ,  Two N divides the +6 .

then ox state of N is +3 in the new compound Y.

3 years ago
1 mole N2H4 loses 10mole e-
So, 1 N2H4 loses 10e-
So,  new compound Y=(N2H4)10+
Charge on y is 10+
Let charge on N be x.
2x+4=10
x=+3

1 Yash Chourasiya
2 years ago
Hello Student

The oxidation state of N in hydrazine is -2.
1 mole of hydrazine contains 2 moles of N and loses 10 moles of electrons. which means, 1 N atom will lose 5 electrons. So, its oxidation number will increase by 5.
Hence, the oxidation number of N in compound Y will be−2+5=+3. Yash Chourasiya
2 years ago
Dear Student

Oxidation state of NItrogen in N2H4 is -2.
2(-2) + 4(+1)= 0

Therefore two N makes it -4 , since they lose 10 e- , hence total charge is +6, and since H is still +1 , Two N divides the +6 .

then oxidation state of N is +3 in the new compound Y.