Question

Analysis shows that nickel oxide has the formula $N{{i}_{0.98}}{{O}_{1.0}}$. What fractions of nickel exist as $N{{i}^{2+}}$ and $N{{i}^{3+}}$ions?

Answer 1

Hint: The solid compounds have not the exact amount of the metal as in their formula; this is due to the defect in solids called as metal deficiency defect. Any chemical compound gains electrical neutrality by having the same number of positive cations and the negative anions. The defects occur due to the missing ions.

Complete answer:
We have been given an analysis of nickel oxide that has the formula $N{{i}_{0.98}}{{O}_{1.0}}$, this shoes that nickel ions are deficient and is 0.98 while oxygen is 1.0. This is due to the deficiency of nickel ions. We have to take out the fraction of $N{{i}^{2+}}$ and$N{{i}^{3+}}$ions.
So assuming $N{{i}^{2+}}$as x, $N{{i}^{3+}}$as (98 – x) and O as 100. We know that the molecules are electrically neutral, means
Number of cation = number of anion
This implies that, $N{{i}^{2+}}$ + $N{{i}^{3+}}$= ${{O}^{2-}}$ (charge on oxygen is – 2)
So, multiplying the charges with their assumed values we have:
$2x+3(98-x)=2\times 100$
$2x+294-3x=200$, solving for x, we have,
x = 94.
So, fraction can be calculated as:
For $N{{i}^{2+}}$(x)=$\dfrac{94}{98}=0.96$
For $N{{i}^{3+}}$(98 – x) =$\dfrac{98-94}{98}=0.4$
Hence, the fractions of nickel that exist a s$N{{i}^{2+}}$ and $N{{i}^{3+}}$ ions are 0.96 and 0.4 respectively.

Note:
By taking out $N{{i}^{2+}}$ fraction to be 0.96 we can take out the fraction of$N{{i}^{3+}}$to be 1.0 – 0.96 = 0.4. These types of proportions of metals decide the chemical formula of the compounds. They arise due to the less nickel ions in its crystal lattice and give rise to imperfection in solids. Metal deficiency defect is a type of non – stoichiometric defect.