Ncert Exercises & Solutions

Match the species in Column I with the type of hybrid orbitals in Column II.

Column I	Column II
A. SF₄	1. sp³d²
B. IF₅	2. d²sp³
C. NO₂⁺	3. sp³d
D. NH₄⁺	4. sp³
	5. sp

All the similar bonds in a molecule do not have the same bond enthalpies. e.g., in

H_{2} O (H - O - H)

molecule after the breaking of first O-H bond, the second O-H bond undergoes some change because of changed chemical environment.

Therefore, in polyatomic molecules the term mean or average bond enthalpy is used. It is obtained by dividinng total bond dissociation enthalpy by the number of bonds broken.

e . g ., H_{2} O (g) \to H (g) + OH (g);

∆_{a} H_{1}^{0} = 502 kJ {mol}^{- 1} OH (g) \to H + O (g);

∆_{a} H_{2}^{0} = 427 kJ {mol}^{- 1}

Average O-H bond enthalpy = \frac{502 + 427}{2} = 464.5 kJ {mol}^{- 1}

The bond enthalpies of O-H bond in C₂H₅OH and H₂O are different because of the different chemical (electronic) environment around oxygen atom.

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