JEE Advanced levelSingle correct optionDifficulty 3 of 3pt214
For the three hypothetical bonds below, bond energies in kJ/mol are given as E(A−A),E(B−B),E(A−B):
(i) P-Q: 160, 200, 320
(ii) R-S: 160, 260, 280
(iii) T-U: 200, 260, 250.
Using Pauling's relation with Δ converted to eV (1eV=96.5 kJ/mol), the bonds in increasing order of polarity are
A(i) < (ii) < (iii)
B(iii) < (ii) < (i)
C(ii) < (iii) < (i)
D(iii) < (i) < (ii)
Answer: B
Compute Δ=E(A−B)−E(A−A)E(B−B). For (i): 320−178.9=141.1 kJ/mol, so Δχ=1.21. For (ii): 280−204.0=76.0 kJ/mol, so Δχ=0.89. For (iii): 250−229.0=21.0 kJ/mol, so Δχ=0.47. The order of polarity is (iii) < (ii) < (i).