Exercise 2
Question 16
Analysis shows that nickel oxide has the formula Ni0.98 O1.00. What fractions of nickel exist as Ni2+and Ni3+ions?
The formula of nickel oxide is Ni0.98 O1.00.
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) = +2x
And, total charge on Ni3+ions = (98 - x)(+3) = 294 - 3x
Since, the compound is neutral, we can write: 2x+ (294 - 3x) + ( - 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 Ni2+ = 94/98 = 0.959
And, fraction of nickel that exists as Ni3+ = 4/98 = 0.041
Alternatively, fraction of nickel that exists as Ni3+= 1 - 0.959 = 0.041
