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Grade: 12

                        

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.]

11 years ago

Answers : (4)

AskiitianExpert Shine
10 Points
							

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.

 

11 years ago
Shivam Rajput
13 Points
							
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
one year ago
Yash Chourasiya
askIITians Faculty
246 Points
							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.

I hope this answer will help you.
5 months ago
Yash Chourasiya
askIITians Faculty
246 Points
							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.

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya
4 months ago
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