A) 99%, 1%
B) 98%, 2%
C) 97%, 3%
D) 96%, 4%
Correct Answer: D
Solution :
If amount of O-atoms=100 then, amount of Ni-atoms = 98 Let \[N{{i}^{2+}}=x\], then \[N{{i}^{3+}}=98-x\] Total charge on 100 \[{{O}^{2-}}ions=200\] Total charge on \[N{{i}^{2+}}\]and \[N{{i}^{3+}}=2x+3(98-x)\] For neutrality, \[2x+3(98-x)=200\] Or, \[x=94\] % of \[N{{i}^{2+}}=\frac{94}{98}\times 100=96%\]and % of \[N{{i}^{3+}}=100-96=4%\]
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