For a weak electrolyte AxBy that partially dissociates into x cations and y anions, the degree of dissociation (α) is related to the van't Hoff factor (i) by which correct expression?

Introduction / Context:
In solutions of electrolytes, the van't Hoff factor i accounts for the effect of dissociation or association of solute particles on colligative properties such as boiling point elevation and freezing point depression. For weak electrolytes that partially dissociate into ions, there is a useful relationship between the observed van't Hoff factor and the degree of dissociation α. This question focuses on a general electrolyte of the form AxBy that dissociates into x cations and y anions and asks which expression correctly links α and i.

Given Data / Assumptions:

• The electrolyte has the formula AxBy and dissociates as AxBy ⇌ x A^y+ + y B^x−.
• The total number of ions formed from one formula unit on complete dissociation is v = x + y.
• The degree of dissociation α is the fraction of formula units that have dissociated.
• The van't Hoff factor i is the ratio of the actual number of solute particles in solution to the number of formula units originally dissolved.
• The electrolyte is weak, so α is between 0 and 1.

Concept / Approach:
If one mole of AxBy is dissolved, before dissociation there is 1 mole of solute particles. After partial dissociation, a fraction α of the formula units dissociate into ions and a fraction (1 − α) remains undissociated. The total number of particles in solution becomes (1 − α) moles of undissociated AxBy plus α v moles of ions, where v = x + y is the total number of ions from complete dissociation of one formula unit. Therefore, total particles = 1 + α (v − 1). The van't Hoff factor i is defined as this total divided by the initial 1 mole, so i = 1 + α (v − 1). Solving this expression for α gives α = (i − 1) / (v − 1) = (i − 1) / (x + y − 1).

Step-by-Step Solution:
Step 1: Write the dissociation of AxBy in solution: AxBy ⇌ x A^y+ + y B^x−. Step 2: Define v as the total number of ions produced per formula unit on complete dissociation, v = x + y. Step 3: For 1 mole of AxBy, the number of undissociated formula units is (1 − α), and the number of dissociated formula units is α. Step 4: Each dissociated unit produces v particles, so the number of ions is α v. Total solute particles = (1 − α) from undissociated units plus α v from ions = 1 + α (v − 1). Step 5: By definition, van't Hoff factor i = (actual number of particles) / (initial number of formula units) = 1 + α (v − 1). Rearranging for α gives α = (i − 1) / (v − 1) = (i − 1) / (x + y − 1).

Verification / Alternative check:
Consider a simple electrolyte like AB that dissociates into A+ and B−, so x = 1, y = 1, and v = 2. Using the formula, i = 1 + α (2 − 1) = 1 + α, and α = (i − 1) / (2 − 1) = i − 1. This matches standard textbook relationships for simple 1:1 electrolytes. For an electrolyte like CaCl2 with x = 1 and y = 2, v = 3, and the same general relationship applies: i = 1 + α (3 − 1) and α = (i − 1) / 2. These examples confirm that the denominator should be (x + y − 1) and that α increases linearly with i in this way.

Why Other Options Are Wrong:
Expressions that place (x + y + 1) or (x + y − 1) in the numerator and (i − 1) in the denominator invert the correct relationship and would give unrealistic values of α, sometimes greater than 1 or negative. Options with (x + y + 1) in the denominator do not match the derived formula for total particles, which clearly involves (v − 1) = (x + y − 1), not (v + 1). The option α = i / (x + y) ignores the constant term 1 in the expression for i and does not reduce correctly for simple electrolytes like AB. Therefore, these alternative formulas contradict the proper derivation and are incorrect.

Common Pitfalls:
Students often memorise the formula i = 1 + α (v − 1) but then rearrange it incorrectly due to algebra errors, leading to wrong expressions for α. Another frequent mistake is to confuse v with x or y alone, or to forget to subtract 1 in the denominator when solving for α. To avoid these errors, rederive the formula step by step when in doubt, starting from the definition of van't Hoff factor as actual particles divided by initial formula units. Always keep track of the fact that some fraction of solute remains undissociated.

Final Answer:
For a weak electrolyte AxBy, the correct relationship is α = (i − 1) / (x + y − 1).