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Step-by-Step Solution
Step 1: Write the chemical formula and identify the oxidation state of oxygen
The given metal oxide is $M_{0.98}O$. Since the oxidation state of oxygen (O) is $-2$, we proceed by letting the average oxidation state of the metal $M$ be $x$.
Step 2: Set up the charge balance equation
In a neutral compound, the total positive charge must balance the total negative charge. Hence, we write:
$0.98 \times x + 1 \times (-2) = 0$
Here, $0.98 \times x$ represents the total positive contribution (from metal $M$) and $1 \times (-2)$ is the negative contribution from one oxygen atom.
Step 3: Solve for the average oxidation state of $M$
Rearranging the equation:
$0.98x = 2 \quad\Rightarrow\quad x = \frac{2}{0.98} = \frac{200}{98} \approx 2.04$
This value (about 2.04) lies between 2 and 3, implying that in the sample, some metal ions are in the $+2$ state ($M^{2+}$), and others are in the $+3$ state ($M^{3+}$).
Step 4: Assume 100 atoms of the metal $M$
Let $y$ be the number of $M^{3+}$ ions out of the 100 metal atoms. Therefore, the number of $M^{2+}$ ions will be $(100 - y)$.
Step 5: Write the total positive charge contributed by the metal
• Each $M^{3+}$ ion contributes $+3$ units of charge, so $y$ such ions contribute $3y$.
• Each $M^{2+}$ ion contributes $+2$ units of charge, so $(100 - y)$ such ions contribute $2 \times (100 - y) = 200 - 2y$.
Total charge from 100 metal atoms = $3y + 200 - 2y = 200 + y$.
Step 6: Calculate the average charge per metal atom
The average charge per metal atom among these 100 atoms is:
$\frac{200 + y}{100} = 2 + \frac{y}{100}.$
But we already know from Step 3 that the average oxidation state (charge) per metal atom must be $\frac{200}{98}.$
Step 7: Equate the two expressions and solve for $y$
Set the average charge expressions equal to each other:
$\frac{200 + y}{100} = \frac{200}{98}.$
Cross-multiplying gives:
$200 + y = \frac{200}{98} \times 100 = \frac{20000}{98} \approx 204.08.$
Therefore,
$y \approx 204.08 - 200 = 4.08.$
Step 8: Interpret the result for the fraction of $M^{3+}$
Out of 100 metal atoms, $4.08$ atoms (on average) are in the $+3$ oxidation state. Hence, the fraction of the metal present as $M^{3+}$ is:
$\frac{4.08}{100} \times 100\% = 4.08\%.$
Final Answer
The fraction of the metal that exists as $M^{3+}$ is $4.08\%.$
