Question #76fa4

2 Answers
May 3, 2016

Answer:

Nitrogen will be in a #+3# oxidation state.

Explanation:

The idea here is that you can use the number of moles of electrons lost by one mole of hydrazine, #"N"_2"H"_4#, to determine how many moles of electrons were lost by one mole of nitrogen.

Since hydrogen's oxidation state is said to remain unchanged, you can say for a fact that all the moles of electrons lost by hydrazine were actually lost by nitrogen.

Now, look at hydrazine's molecular formula, which contains

  • two atoms of nitrogen, #2 xx "N"#
  • four atoms of hydrogen, #4 xx "H"#

Notice that one mole of hydrazine contains two moles of nitrogen.

This of course means that when #1# mole of hydrazine loses #10# electrons, these electrons are actually coming from #2# moles of nitrogen.

As a result, you can say that every mole of nitrogen present in one mole of hydrazine will lose #5# moles of electrons.

If you take this down to the level of a single atom, you can say that one atom of nitrogen will lose #5# electrons.

In hydrazine, hydrogen has a #color(blue)(+1)# oxidation state, which means that nitrogen has an oxidation state equal to

#2 xx ON_"N" + 4 xx ON_"H" = 0#

#2 xx ON_"N" = 0 - 4 * (color(blue)(+1))#

#"ON"_N = (-4)/2 = -2#

Now, when each nitrogen atom loses #5# electrons, its oxidation state increases by #5#, i.e. it's being oxidized. You will have

#color(red)(|bar(ul(color(white)(a/a)color(black)(stackrel(color(blue)(-2))("N")_2"H"_4 -> 2stackrel(color(blue)(+3))("N") + 10"e"^(-))color(white)(a/a)|)))#

Here one atom of nitrogen loses #5# electrons, so two atoms of nitrogen will lose #10# electrons.

The oxidation state of nitrogen in this new compound #"Y"# will thus be equal to #color(blue)(+3)#.

May 4, 2016

Answer:

+3

Explanation:

Let ON of N in #N_2H_4# be x. H ,being more electropositive than N ,it will have +1ON and species being a molecule total ON wil be zero.
So #2*x+4*(+1)=0=>x=-2#
By the problem
the oxidation process may be written as
#N_2H_4->Y("*")#
#"*molecule contains 2 N-atoms"#

One mole of #N_2H_4# loses 10 moles of electrons to form a new compound Y
So One molecule of #N_2H_4# loses 10 electrons to form one molecule of new compound Y which contains 2 N-atoms As there is no change of ON of H-atom , 2N-atoms of #N_2H_4# loses 10 electrons causing total 10 unit increase in ON of 2 N-atoms .

Now if the ON of 2 N-atoms of Y be y then total increase in ON of N-atoms#=(2y-2*(-2)=2y+4)#
By the given condition of the problem , we have
#2y+4=10=>y=+3#

Hence the oxidation state of N in the compound Y is +3