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Relative Strengths of Bonds

  1. Dec 11, 2013 #1


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    Consider the bond dissociation enthalpy data:

    C-C, C≡C; N-N, N≡N. Bond dissociation enthalpies are, respectively: 348 kJ/mol, 812 kJ/mol, 163 kJ/mol, and 945 kJ/mol.

    Explain the data with regard to simple electronegativity considerations, which would lead one to wrongly suppose that the singly bond nitrogen atoms should have higher bond dissociation enthalpies than the singly bond carbon atoms.

    2. Relevant equations

    Bond dissociation enthalpy is affected by atomic radii, electronegativity, and orbital stability.

    3. The attempt at a solution

    1) Singly bond carbon has a higher bond dissociation enthalpy than nitrogen-nitrogen because of orbital stability. Carbon by itself has an electron configuration of 2p^2. The one bond formed between one carbon atom and another carbon atom causes each carbon atom to have an electron configuration of 2p^3. This imparts great stability - more than what a pair of electrons would to singly bonded nitrogen. Nitrogen has an electron configuration of 2p^3. The addition of a shared electron pair would bump nitrogen up to 2p^4. This is a half-filled p-orbital plus 1 extra electron in the first subshell. Electron-electron repulsion causes the nitrogen-nitrogen atom to be less stable.

    2) Orbital considerations also apply to the triple-bonded species. Nitrogen has an electron configuration of 2p^3. Three bonds, or the sharing of six electrons, would bump each nitrogen up to a noble gas configuration; nitrogen will become isoelectronic with neon. On the other hand, three electrons would bump carbon up to fluorine. Neon is more stable than fluorine. Therefore, it makes sense that triple-bonded nitrogen is more stable than triple-bonded carbon.


    1) How sound is my logic? I understand that it is nature to reach the most stable state possible - or the state with the lowest potential energy. Is this the law behind the stability of the molecules considered above?
  2. jcsd

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