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

Question

Experimentally it was found that a metal oxide has formula \(\mathrm{M}_{0.98} \mathrm{O} .\) 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

(d) 4.08%

Step 1: Understand Ratios

The chemical formula can be expressed as \(\text{M}_{0.98}\text{O}\), indicating a non-stoichiometric compound with 0.98 moles of \(\text{M}\) for every mole of \(\text{O}\). The deviation suggests the presence of different oxidation states of \(\text{M}\).

Step 2: Equations Comparing Charges

To balance the charge, let \(x\) be the fraction of \(\text{M}^{3+}\) ions and \(0.98-x\) be the fraction of \(\text{M}^{2+}\) ions. The total charge contributed by \(x\) moles of \(\text{M}^{3+}\) ions is \(3x\) and by \(0.98-x\) moles of \(\text{M}^{2+}\) ions is \(2(0.98-x)\). The oxide ion contributes a charge of -2.

Step 3: Charge Balance Equation

Set up the charge balance equation: \[3x + 2(0.98 - x) = 2\]

Step 4: Solve Charge Balance Equation

Solve for \(x\):\[3x + 1.96 - 2x = 2\] \[x + 1.96 = 2\]\[x = 2 - 1.96 = 0.04\]

Step 5: Calculate Fraction of \(\text{M}^{3+}\)

The fraction of metal ions existing as \(\text{M}^{3+}\) is \(0.04\) out of \(0.98\) moles:Fraction = \(\frac{0.04}{0.98}\)

Step 6: Convert Fraction to Percentage

Convert the fraction to a percentage:\(\frac{0.04}{0.98} \times 100\% = 4.08\%\)

Key Concepts

Non-stoichiometry

In chemical formulas, non-stoichiometry refers to the condition where the proportions between elements in a compound deviate from simple whole numbers. This occurs when compounds are not strictly formed according to a precise fixed ratio of atoms. In the case of the metal oxide given in the exercise, the formula is represented as \( \mathrm{M}_{0.98} \mathrm{O} \). Here, "0.98" signifies that for every oxygen atom, there are only approximately 0.98 atoms of the metal \( \text{M} \). This deviation implies that some form of imperfection or variation takes place within the crystal lattice, causing the metal to exist in two different oxidation states.

Non-stoichiometric compounds are often the result of defects in the solid's crystal structure or the partial filling of lattice sites, and they can exhibit interesting electrical, magnetic, or optical properties due to these irregularities. It's fascinating to observe how real-world substances don't always follow the neat, tidy rules of stoichiometry seen in basic chemical equations. Instead, they sometimes adopt a more complex or unconventional structure.

Oxidation States

Oxidation states in chemistry signify the degree of oxidation or loss of electrons an atom has undergone. For metal \( \text{M} \) in the given oxide, it exists in two states: \( \text{M}^{2+} \) and \( \text{M}^{3+} \). Simply put, \( \text{M}^{2+} \) means that each metal atom has lost two electrons, whereas \( \text{M}^{3+} \) denotes the loss of three electrons.

The presence of these different oxidation states within the same compound indicates that a variety of electron environments or energy levels are accessible, allowing the metal to attain stability via differing charge states. This is not uncommon in transition metals, as they can exhibit variable oxidation due to their partially filled d-orbitals, which provide flexible electron configurations.

• \( \text{M}^{2+} \): Loses two electrons.
• \( \text{M}^{3+} \): Loses three electrons.

Understanding oxidation states is crucial for predicting the behavior of a substance in chemical reactions, particularly the redox (reduction-oxidation) processes.

Charge Balance

In chemical systems, maintaining charge balance is essential. For every positive charge, there should be an equivalent negative charge to achieve electrical neutrality. In the context of our metal oxide problem, the goal involves ensuring that all the positive charges from the metal ions (\( \text{M}^{2+} \) and \( \text{M}^{3+} \)) balance out the negative charges from the oxygen ions (\( \text{O}^{2-} \)).

Charging balancing for the formula \( \text{M}_{0.98}\text{O} \) requires setting up an equation that summarizes the contribution of each component's charge. Based on the stepwise solution:

• The charge contributed by \( x \) moles of \( \text{M}^{3+} \) ions is \( 3x \).
• The charge contributed by \( 0.98-x \) moles of \( \text{M}^{2+} \) ions is \( 2(0.98-x) \).
• The charge contributed by one mole of \( \text{O}^{2-} \) ions is \( -2 \).

Setting up the equation and solving it allows us to determine the fraction of metal that exists in each oxidation state, ultimately leading to solving step 5 of the exercise.

Fraction Calculation

Fraction calculation converts the number from part-to-whole relationships into comparable units such as percentages. When we find the fraction of metal existing as \( \text{M}^{3+} \) in \( \text{M}_{0.98}\text{O} \), we identify what portion of the total 0.98 moles of metal atoms are actually \( \text{M}^{3+} \).

Based on the exercise solution, we determined the fraction \( x \) as 0.04. Calculating this into a fraction, we perform:\[\text{Fraction of } \text{M}^{3+} = \frac{0.04}{0.98}\]To express this fraction as a percentage, it involves simple multiplication by 100 to convert the fraction into a percentage:
\[\frac{0.04}{0.98} \times 100\% = 4.08\%\]This percentage ultimately helps communicate the result more intuitively, suggesting that 4.08% of the metal \( \text{M} \) in the compound exists in the \( \text{M}^{3+} \) oxidation state. Converting fractions to percentages is a powerful tool to ensure clear numerical understanding in chemical calculations.