Physical Chemistry> oxidation state...

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.]

satyapreet singh , 15 Years ago

Grade 12

4 Answers

AskiitianExpert Shine

Last Activity: 15 Years ago

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.

Shivam Rajput

Last Activity: 5 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

Yash Chourasiya

Last Activity: 4 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.

I hope this answer will help you.

Yash Chourasiya

Last Activity: 4 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.

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