Question

Experimentally it was found that a metal oxide has formula ${M_{0.98}}O$. Metal M, is present as ${M^{2 + }}$ and ${M^{3 + }}$ in its oxide. The fraction of the metal which exists as ${M^{3 + }}$ would be:
A.$5.08\%$
B.$7.01\%$
C.$4.08\%$
D.$6.05\%$

Answer 1

Metals react with oxygen to form metal oxide, since oxygen has a $- 2$ oxidation state to form a stable oxide the metal should be present in the $+ 2$. According to the given question for every one oxide ion there $0.98$ metal ions.

Complete step by step solution:
We know that since the oxidation state of oxygen is $- 2$, so for Metal oxide $= {M_{0.98}}O$ to be neutral, the total oxidation state has to be $+ 2$for ${M_{0.98}}O$. For this, let us assume that the molecules of ${M_{0.98}}O$ are $100$.
Let us assume the fraction of${M^{3 + }}$be $x$. Then we can say that the fraction of ${M^{2 + }}$ will be $(0.98 - x)$
Now, for the compound to be neutral, we will balance the negative and the positive charge and for that we will multiply $x$ with $3$ and $(0.98 - x)$ with $2$:
$3x + (0.98 - x) \times 2 = 2 \\\ \Rightarrow 3x + 1.96 - 2x = 2 \\\ \Rightarrow x = 2 - 1.96 \\\ \Rightarrow x = 0.04 \\\$
So, we have $0.04$ fraction of ${M^{3 + }}$ and the percentage of metal in $+ 3$ state would be:
$\Rightarrow {M^{3 + }} = \dfrac{{0.04}}{{0.98}} \times 100 \\\ \Rightarrow {M^{3 + }} = 4.08\% \\\$
Therefore, the correct answer is (C).

Note: With increasing oxidation state the acidity of an element also increases. The transition group elements possess seven different oxidation states and hence are more considered least acidic to more acidic with increasing oxidation state and more basic to less basic with the same condition.