Question

A sample of a metal oxide has formula $M _{0.83} O _{1.00}$ The metal $M$ can exist in two oxidation states $+2$ and $+3$ In the sample of $M _{0.83} O _{1.00}$, the percentage of metal ions existing in $+2$ oxidation state is___$ \%$ (nearest integer)

Hint:

In stoichiometry problems, carefully balance the charges and mole ratios to determine the proportions of each oxidation state.

Answer 1

Correct Answer: 59

 

Let the amount of metal in the \( +2 \) oxidation state be \( x \), and the amount in the \( +3 \) oxidation state be \( 0.83 - x \).

From the charge balance equation:

\[ 2x + 3(0.83 - x) = 0.83 \]

Simplifying the equation:

\[ 2x + 2.49 - 3x = 0.83 \quad \Rightarrow \quad -x + 2.49 = 0.83 \quad \Rightarrow \quad -x = 0.83 - 2.49 = -1.66 \]

\[ x = 0.49 \]

Thus, the percentage of metal ions in the \( +2 \) oxidation state is:

\[ \frac{0.49}{0.83} \times 100 = 59\% \]

Answer 2

The accurate response is 59%.
By solving the equation \(2x+3(0.83−x)=2\),
we obtain the value of x as 0.49.
Therefore, the percentage of metal ions present in the +2 oxidation state
\((\% M^{2+})\) can be calculated as \(\frac{(0.83-0.49)}{0.83} \times 100\),
which simplifies to \(59\%\).
So, the correct answer is 59.