**Question:**

Analysis shows that nickel oxide has the formula Ni0.98O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions?

**Solution:**

The formula of nickel oxide is Ni0.98O1.00.

Therefore, the ratio of the number of Ni atoms to the number of O atoms,

Ni : O = 0.98 : 1.00 = 98 : 100

Now, total charge on 100 O2− ions = 100 × (−2)

= −200

Let the number of Ni2+ ions be *x*.

So, the number of Ni3+ ions is 98 − *x*.

Now, total charge on Ni2+ ions = *x*(+2)

= +2*x*

And, total charge on Ni3+ ions = (98 − *x*)(+3)

= 294 − 3*x*

Since, the compound is neutral, we can write:

2*x* + (294 − 3*x*) + (−200) = 0

⇒ −*x* + 94 = 0

⇒ *x* = 94

Therefore, number of Ni2+ ions = 94

And, number of Ni3+ ions = 98 − 94 = 4

Hence, fraction of nickel that exists as $\mathrm{Ni}^{2+}=\frac{94}{98}$

= 0.959

And, fraction of nickel that exists as $\mathrm{Ni}^{3+}=\frac{4}{98}$

= 0.041

Alternatively, fraction of nickel that exists as Ni3+ = 1 − 0.959

= 0.041

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