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Question: Analysis shows that nickel oxide has formula \(N{{i}_{0.98}}O\) . The percentage of Ni present as \(...

Analysis shows that nickel oxide has formula Ni0.98ON{{i}_{0.98}}O . The percentage of Ni present as Ni2+N{{i}^{2+}} and Ni3+N{{i}^{3+}} respectively will be:
A. 99%,1%99\%,1\%
B. 98%,2%98\%,2\%
C. 97%,3%97\%,3\%
D. 96%,4%96\%,4\%

Explanation

Solution

Hint: Electrical neutrality of a compound is maintained in any compound. Therefore, total positive charge on cations would be equal to total negative charge on anions.

Complete step-by-step answer:
This type of defect is known as metal deficiency defect which comes under non stoichiometric defect. In this type of defect elements which show multiple oxidation states get oxidized to higher oxidation states at some sites. In these types of crystals, it is difficult to maintain stoichiometry.
The formula of nickel oxide given in formula is Ni0.98ON{{i}_{0.98}}O

According to the formula, there are 98 Ni ions and 100 oxide ions.

Out 98 Ni ions,
Let us assume that the number of x ions is in +2 oxidation state.

Therefore, 98−x ions will be the number of ions present in +3 oxidation state.

Oxide ion has −2 charges.

To maintain the electrical neutrality, total positive charge on cations is equal to total negative charge on anions.

& 2x+3(98-x)+100\left( -2 \right)=0 \\\ & 2x+294-3x-200=0 \\\ & -x+94=0 \\\ & -x=-94 \\\ & x=94 \\\ & \\\ \end{aligned}$$ Fraction of $N{{i}^{2+}}$ ions =$\dfrac{94}{98}=0.96$ Fraction of $N{{i}^{3+}}$ ions =$\dfrac{4}{98}=0.04$ Hence, the fractions of nickel that exist as $N{{i}^{2+}}$ and $N{{i}^{3+}}$ are 0.96 and 0.04 respectively. In percentage $N{{i}^{2+}}$ are present in 96% $N{{i}^{3+}}$ ions are present in 4%. The correct option is D. Note: This is an example of a non stoichiometric defect. Nickel shows two oxidation states that is +2 and +3 that is why at some sites nickel is oxidized to +3 oxidation state. Due to which nickel and oxygen are not present in 1:1 ratio. stoichiometry is not maintained but the amount of positive charge on cation is equal to the amount of negative charge on anions. Therefore, applying this it came out that $N{{i}^{2+}}$ is present 96% and $N{{i}^{3+}}$ is present 4%.