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. Fraction of the metal which exists as M 3+ would be: (1) 4.08% (2) 6.05% (3) 5.08% (4) 7.01%

To determine the fraction of the metal that exists as M³⁺ in the metal oxide with the formula M_0.98O, we can start by establishing the relationship between the oxidation states of the metal ions and the overall charge balance in the compound. The metal M can exist in two oxidation states: M²⁺ and M³⁺.

Understanding the Composition of the Metal Oxide

The formula M_0.98O indicates that there are 0.98 moles of metal M for every mole of oxygen (O). Since oxygen typically has a charge of -2, the total negative charge contributed by oxygen in this compound is:

• Charge from oxygen = 1 mole O × -2 = -2

Setting Up the Charge Balance Equation

Let’s denote the fraction of M that is in the +3 oxidation state as x. Consequently, the fraction of M in the +2 oxidation state would be (0.98 - x). The total positive charge contributed by the metal can be expressed as:

• Charge from M³⁺ = 3x
• Charge from M²⁺ = 2(0.98 - x)

Now, we can write the equation for the total positive charge:

3x + 2(0.98 - x) = 2

Simplifying the Equation

Expanding the equation gives us:

• 3x + 1.96 - 2x = 2

Combining like terms results in:

• x + 1.96 = 2

Now, isolating x:

• x = 2 - 1.96
• x = 0.04

Calculating the Fraction of M³⁺

Now that we have determined that x = 0.04, we can find the fraction of M that exists as M³⁺ relative to the total amount of metal M:

• Fraction of M³⁺ = x / 0.98
• Fraction of M³⁺ = 0.04 / 0.98

Calculating this gives:

• Fraction of M³⁺ ≈ 0.0408

Converting to Percentage

To express this fraction as a percentage, we multiply by 100:

• Percentage of M³⁺ ≈ 0.0408 × 100 = 4.08%

Final Answer

Thus, the fraction of the metal that exists as M³⁺ in the oxide M_0.98O is approximately 4.08%. Therefore, the correct answer is (1) 4.08%.