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Q. Analysis shows that an oxide ore of nickel has formula $Ni_{0.98} O_{1.00}$ The percentage of nickel as $Ni^{3+}$ ions is nearly

The Solid State

Solution:

Let no. of nickel ions $= 98$
$\therefore $ No. of oxide ions $= 100 $
Total negative charge on $O^{2-}$ ions
$= 2 \times 100 = 200$
Let no. of $Ni^{3+}$ ions $= x$
$\therefore $ No. of $Ni^{2+}$ ions $= 98 - x$
$x \times 3 + (98 - x) \times 2 = 200$
$3x - 2x = 200 - 196$
$x = 4$
Percentage of nickel as $Ni^{3+} = \frac{4}{98} \times 100$
$ = 4.08$