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The energy required to break 76gm gaseous fluorine in to free gaseous atom is 180kcal at 25*C the bond energy of F-F bond will be

The energy required to break 76gm gaseous fluorine in to free gaseous atom is 180kcal at 25*C the bond energy of F-F bond will be

Grade:12

1 Answers

SAGAR SINGH - IIT DELHI
878 Points
13 years ago

Dear student,

The energies of bonds between atoms of substantially different electronegativities tend to be high, e.g., the 110 kcal of the H-O bond. Another important example is the bonds between oxygen and carbon atoms in carbon dioxide, CO2. The carbon atom shares two pairs of electrons with each of the oxygen atoms, and each of these double bonds has a bond energy of 187 kcal (or about 93 kcal for each shared pair of electrons). The larger the bond energy, the more energy is needed to break the bond. Thus bonds between atoms of differing electronegativities are apt to be very strong and stable.

On the other hand, the energies of bonds between atoms of similar electronegativity tend to be smaller. Of course, where the atoms are the same (e.g., O=O, C-C) there is no electronegativity difference and we are not surprised to find lower bond energy values. Each of the shared electron pairs in the oxygen molecule is worth 58 kcal for a total of 116 in each molecule, as we saw above. The bond energy of the C-C bond is 80 kcal. There is a slight difference in electronegativity between carbon and hydrogen, and a bond energy of 98 kcal. The bonds with smaller bond energies are, by definition, easier to break. Thus these bonds are weaker and less stable.

 

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Sagar Singh

B.Tech, IIT Delhi

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