The intermolecular potential energy for the molecules A, B, C and D given below suggests that :
Show Hint
Read the graph: the depth of each well gives the bond enthalpy, the position of the minimum gives the bond length, and the steepness (curvature) of the well gives the stiffness of the bond.
Step 1: From the potential energy curves, the minima are approximately: A–A about \(-450\ \mathrm{kJ\ mol^{-1}}\) at about 70 pm; A–B about \(-550\ \mathrm{kJ\ mol^{-1}}\) at about 100 pm; A–C about \(-420\ \mathrm{kJ\ mol^{-1}}\) at about 130 pm; A–D about \(-340\ \mathrm{kJ\ mol^{-1}}\) at about 150 pm. Step 2: Bond enthalpy equals the depth of the well, so the largest bond enthalpy belongs to A–B (deepest), not A–A. Option (1) is wrong. Step 3: The shortest bond length is where the minimum occurs at the smallest distance, which is A–A, not A–D (A–D has the longest). Option (3) is wrong. Step 4: Electronegativity of D cannot be deduced from these curves. Option (4) is wrong. Step 5: The A–B curve is the deepest and has the steepest sides around its minimum, i.e. the largest curvature \(\left(\frac{d^2V}{dr^2}\right)\), which means the stiffest bond. Answer: (2) A–B has the stiffest bond
