Question

All the metal halides have negative enthalpies of formation. Arrange the following halides from most to least negative enthalpy of formation:
(A) Fluoride,
(B) Chloride,
(C) Bromide,
(D) Iodide

• A. (A), (C), (B), (D)
• B. (A), (B), (C), (D)
• C. (D), (B), (C), (A)
• D. (A), (B), (D), (C)

Hint:

Enthalpy of formation is the energy released when one mole of a compound is formed from its elements in their standard states. The smaller ions have stronger electrostatic forces of attraction, which leads to greater lattice energy and higher negative enthalpy of formation.

Answer 1

The Correct Option is C

The enthalpy of formation of a metal halide is mainly related to its lattice energy, which is the energy released when the ions form the solid lattice.
• Fluoride (F-): Fluoride ions are the smallest halide ions, which results in a high lattice energy and therefore, most negative enthalpy of formation when they combine with metals.
• Chloride (Cl-): Chloride ions are larger than fluoride, which leads to a smaller lattice energy than fluorides but still greater than the other halides, with correspondingly less negative enthalpy of formation.
• Bromide (Br-): Bromide ions are larger than chloride ions, resulting in even smaller lattice energy and correspondingly lower negative enthalpy of formation.
• Iodide (I-): Iodide ions are the largest among halides, leading to the smallest lattice energy and thus the least negative enthalpy of formation.
Thus, the order of enthalpy of formation from most negative to least negative is Fluoride > Chloride > Bromide > Iodide.