Problem 147 - A Complete Resource Book in Chemistry for JEE Main

Question

Experimentally it was found that a metal oxide has. formula \(\mathrm{M}_{0.98}\) O. Metal \(\mathrm{M}\), is present as \(\mathrm{M}^{2+}\) and \(\mathrm{M}^{3+}\) in its oxide. Fraction of the metal which exists as \(\mathrm{M}^{3+}\) would be (a) \(6.05 \%\) (b) \(5.08 \%\) (c) \(7.01 \%\) (d) \(4.08 \%\)

Step-by-Step Solution

Answer

The fraction of metal as \( \mathrm{M}^{3+} \) is approximately 4%, so option (d) 4.08% is correct.

Step 1: Understand the Problem

We have a metal oxide with the formula \( \mathrm{M}_{0.98} \mathrm{O} \). The metal \( \mathrm{M} \) is present in two oxidation states: \( \mathrm{M}^{2+} \) and \( \mathrm{M}^{3+} \). We need to find the fraction of metal \( M \) that exists as \( \mathrm{M}^{3+} \), expressed as a percentage.

Step 2: Set Up Equations

For each oxygen atom, we need a total charge of \(+2\) to maintain neutrality. Let \( x \) be the fraction of \( M \) as \( \mathrm{M}^{3+} \), thus the fraction as \( \mathrm{M}^{2+} \) is \( 0.98 - x \) since there are 0.98 moles of \( M \). The charge balance equation is:\[ 2(0.98 - x) + 3x = 2 \times 1 \]

Step 3: Solve the Equation

Substitute the values and solve for \( x \):\( 2(0.98) - 2x + 3x = 2 \) gives:\[ 1.96 + x = 2 \]Solve for \( x \):\[ x = 2 - 1.96 = 0.04 \].

Step 4: Calculate Fraction as Percentage

The fraction of \( M \) as \( \mathrm{M}^{3+} \) is \( x = 0.04 \). To convert this to a percentage, multiply by 100:\[0.04 \times 100 = 4\% \].

Step 5: Choose the Correct Option

From the given options, 4.08% is closest to our result of 4%. Hence, option (d) is the correct answer.

Key Concepts

Oxidation States

The concept of oxidation states is vital for understanding how elements form compounds by losing or gaining electrons. In a compound, the oxidation state of an element signifies the degree of oxidation; essentially, it's the charge the atom would have if all bonds were purely ionic. For our example problem involving a metal oxide, the metal \( \mathrm{M} \) exists in two oxidation states, \( \mathrm{M}^{2+} \) and \( \mathrm{M}^{3+} \).

Here's how these oxidation states play a role:

• \( \mathrm{M}^{2+} \) means the metal has lost two electrons.
• \( \mathrm{M}^{3+} \) means the metal has lost three electrons.

By understanding the distribution of these oxidation states in the metal oxide, chemists can determine the exact combination of ions present, which affects properties like reactivity and solubility.
Thus, finding out how many atoms exist in each oxidation state helps ensure the metal oxide has the correct chemical formulation. This ties directly into solving the given problem, where the aim was to determine the distribution of these oxidation states as a fraction.

Charge Balance

Charge balance is a critical principle in chemistry, especially for compounds and reactions.
The total charge in a stable compound must be zero, meaning the positive and negative charges have to balance. For ionic compounds this is particularly evident. Oxides, like the one in the problem, rely on charge balance to figure out how many ions of each type are present.

In the example with metal oxide \( \mathrm{M}_{0.98}\mathrm{O} \):

• Oxygen's oxidation state is \( -2 \), so each oxygen contributes a negative two charge.
• The metal ions must together supply a positive two charge per constituent oxygen to maintain electrical neutrality.

The charge balance equation ensures that the sum of the products of ion charges and their respective amounts are equal to zero. We solved this balance to find out that part of the metal \( M \) is present as \( \mathrm{M}^{3+} \), which contributes differently to the charge than \( \mathrm{M}^{2+} \).
The goal was to set up and solve the equation \( 2(0.98 - x) + 3x = 2 \times 1 \), ensuring the cumulative charge matched the charge provided by the oxygen, showing the substance's neutrality.

Percent Composition

Percent composition is the percentage by mass of each element in a compound. While usually discussed in terms of mass, in certain problems like ours, it can also refer to the percentage by mole or fraction of particular states or species, like different oxidation states.

To find the percent composition of \( \mathrm{M}^{3+} \) in our compound:

• First, we calculated the mole fraction of the \( \mathrm{M}^{3+} \) ion. In our discussion, this was found as \( 0.04 \) from the equation \( 2(0.98 - x) + 3x = 2 \).
• We then converted this mole fraction into a percentage by multiplying by 100, obtaining \( 4\% \).

This calculation helped determine how much of the metal \( M \) exactly exists as \( \mathrm{M}^{3+} \). Understanding the percent composition of different ions helps in predicting the chemical behavior of the substance, influencing properties like conductivity, reactivity, and the types of reactions it can undergo.
It also connects to practical applications, such as in materials science where altering compositions impacts the final properties of metallic compounds.