Q16 Analysis shows that nickel oxide has the formula Ni0.98O1.00.
What fractions of nickel exist as Ni2+ and Ni3+ ions?
Formula is Ni0.98O1.00
So the ration of Ni : O = 98:100
So if there are 100 atoms of oxygen then 98 atoms of Ni
Let number of atoms of Ni+2 = x
Then number of atoms of Ni+3 = 98–x
Charge on Ni = charge on O
So that oxygen has charge –2
3(98–x) + 2x = 2 (100)
294 –3x +2x = 200
–x = – 94
x = 94
Percentage of Ni+2 =
(atom of Ni+2/total number of atoms of Ni)100
=100(94/98)*100
=
96%
Percentage of Ni+3
=100 – Ni+2
=100 – 96
= 4 %
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Short Cut method for this
ReplyDeleteAvg. Charge of Ni(oxidation number)= 0.98x-2=0
-> x=2/0.98=2.04
%age of Ni2+ = Avg.Charge -mass of 2nd isotope /difference btw 2 isotopes
i.e 2.04-3/1×100=96%
Your choice now either go by my way or his ...
JEE ENTRANCE STUDENT
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