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The following table lists the bond dissociation energy ($E_{diss}$) for single covalent bonds formed between C and atoms A, B, D, E. Which of the atoms has smallest size?


(a) D
(b) E
(c) A
(d) B

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  • Smaller the atom, more effective will be the orbital overlap and larger will be bond dissociation energy.
  • The bond dissociation enthalpy is the change in enthalpy when one mole of covalent bonds of a gaseous covalent compound is broken to form products in the gas phase.
Answer: E
When a bond is strong, there is a higher bond energy because it takes more energy to break a strong bond. This correlates with Bond Order and Bond Length. When the Bond Order is higher, Bond Length is shorter, and the shorter the Bond Length means a greater the Bond Energy because of increased electric attraction.
The Bonds has been ordered in descending order of $E_{diss}$ (Bond dissociation enthalpy). Thus, E has the smallest size.
answered Feb 25, 2014 by mosymeow_1

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