Analysis shows that nickel oxide has the formula $N i_{0.98} O$ . The percentage of Ni present as $N i^{2 +}$ and $N i^{3 +}$ respectively will be:

99 %,

1 %

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98 %,

2 %

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97 %,

3 %

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96 %,

4 %

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Solution

The correct option is C $96 %,$ $4 %$
If the amount of O-atoms $= 100$

then, the 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 $N i^{3 +} = 2 x + 3 (98 - x)$

For neutrality, $2 x + 3 (98 - x) = 200$
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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