SELECT * FROM question_mgmt as q WHERE id=1748 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=9 AND chapterId=56 and ex_no='2' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 12th chemistry The Solid State analysis shows that nickel oxide has the formula n

Question:

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

Answer:

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


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