Find the bond enthalpy (in kJ/mol) of 'three centre two electron bond' in B 2 H 6 from the following data: Δ f H° [BH 3 (g)] = 98 kJ/mol; Δ f H° [B 2 H 6 (g)] = 44 kJ/mol; ΔH Atomization [B(

Question

Find the bond enthalpy (in kJ/mol) of 'three centre two electron bond' in B 2 H 6 from the following data: Δ f H° [BH 3 (g)] = 98 kJ/mol; Δ f H° [B 2 H 6 (g)] = 44 kJ/mol; ΔH Atomization [B(s)] = 560 kJ/mol; ΔH Atomization [H 2 (g)] = 436 kJ/mol.

TestHub · Accepted Answer

BH 3 (g) → B(g) + 3H(g): ΔH = 560 + 3(218) - 98 = 1116 kJ, so B-H = 372 kJ/mol. B 2 H 6 (g) → 2B(g) + 6H(g): ΔH = 2(560) + 3(436) - 44 = 2384 kJ. Four terminal B-H bonds: 4 × 372 = 1488 kJ. Remaining 2384 - 1488 = 896 kJ is shared by two three-centre two-electron bonds, so each is 896/2 = 448 kJ/mol. The answer is 0448.

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