Bond energy for N-N is 163Kj/mol whereas bond energy for N=N is 418kj/mol even though it should be lesser than twice bond energy for N-N. Why is that? For carbon it is: Bond energy for C-C is 348KJ/mol Bond energy for C=C is 614Kj/mol

My book explains the reason for carbon but it contradicts with the case of nitrogen.

=======================The reason can be described both qualitatively and quantitatively. Qualitatively, you can think of it in the following fashion: There are four sigma levels when one looks at the MOs formed from the 2s2s and 2p2p orbitals of N2N2 : σs,σ∗s,σp,σ∗pσs,σs∗,σp,σp∗. These levels interact as they have same symmetry and the σpσp level, being third highest in energy, is pushed up more (think level repulsion from second order perturbation theory), making that interaction weaker. ππ levels do not have others of the same symmetry to mix with, making them stabler (especially because overlap is massive as the system is small).

Quantitatively, within the Atomic Orbital basis of Nitrogen, the Hamiltonian matrix is pretty big (considering only the 2s2s and 2p2p levels, we have 8 orbitals and thus an 8 by 8 matrix of elements ⟨m|H|n⟩⟨m|H|n⟩ where m,nm,n are two of the 8 orbitals mixing (and they can be the same-like 2s2s on same atom). I am not writing the 64 element matrix out for obvious reasons, but symmetry makes a host of elements zero, simplifying our work. We are left with block diagonalized bits, two 2×22×2s for the 2 pi levels (which work out simply) and one 4×44×4 for the sigma level.

The σσ Hamiltonian matrix is still too big for me to fully write out, but it contains terms like interaction between 2pz2pz on one atom and 2s2s on the other. Even 2p2p on one atom and 2s2s on the same are not perfectly non-interacting now (they were previously orthogonal when there was no second atom, but the perturbation to the Hamiltonian from the presence of the second atom throws it off)-although this is truly speaking the smallest interaction to consider.

Now for some numbers. Please note that I making lots of approximations and so this is not really a reflection of real N2N2 but is simply a crude estimate. Ionization of 2p2p for Nitrogen is 14.53eV. I looked at some NIST data (Energy Levels of Neutral Nitrogen ( N I )) and seems like ss about 8eV below in energy. Bond-strength of N2N2 is about 941 kJ/mol and so we can guess the coupling to be ~3eV3eV ish (roughly 1/3rd of that). As an approximation, let us claim that individual orbital energies themselves are not greatly perturbed by the presence of the other atom (coupling is though-which is why we have bonding). Lets claim that the p−pp−p coupling is V,s−pV,s−pcoupling is 0.75V0.75V and s−ss−s coupling is 0.50V0.50V (p has greater coupling due to directionality). Note that I made the numbers up (although they 'seem' reasonable) and thus they need not reflect reality.

The dashed curve represent how energies will look if the σsσs and σpσp do not communicate with each other. As you can observe, the 2p bonding (cyan ) line is distinctly less bonding than before and the 2s antibonding orbital is quite a bit antibonding. Although my numbers are all guesses, overall qualitative behavior ought to be the same.

Found this but coud not understand.


closed as unclear what you're asking by Todd Minehardt, airhuff, Pritt says Reinstate Monica, paracetamol, Buttonwood Jul 8 '17 at 12:44

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    $\begingroup$ What molecules are you asking about when you use n-n and n=n? I assume its not $\ce{N2}$ as this has a dissociation energy of $945\, \pu{kJmol^{-1}}$. But the nature of sigma and pi bonding is different so why should they be equal, other than by accident? $\endgroup$ – porphyrin Jul 7 '17 at 21:32
  • $\begingroup$ Welcome to ChemSE! To avoid errors in asking and answering questions on ChemSE, please adhere to proper use of dimensions, which includes "k" (lowercase) as prefix for $\pu{1E3}$, "m" (lowercase) for $\pu{1E-3}$, but "M" (capital) for $\pu{1E6}$. Use element symbols as IUPAC does without alteration; e.g. $\ce{PU}$ refers to phosphor and uranium (if there is such a compound?), $\ce{Pu}$ to plutonium. Learn mathjax works, allowing you to type single, double, triple bonds, too: chemistry.meta.stackexchange.com/questions/86/… $\endgroup$ – Buttonwood Jul 8 '17 at 12:43
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    $\begingroup$ Guys the question is very clear, (though I was befuddled myself and close voted it) and Oscar Lanzi has answered it satisfactorily. $\endgroup$ – Pritt says Reinstate Monica Jul 8 '17 at 15:03

The $\pi$ bond in $N=N$ is actually not stronger than the $\sigma$ bond. Rather the $\sigma$ bond is stronger in $N=N$ than in $N-N$, and this plus the $\pi$ bond gives the double-bonded linkage more than twice the binding energy.

So what makes the $N=N$ stronger? Fewer valence electrons are there when you have a double bond, that's why you form double bonds to begin with. You have $12$ valence electrons around a pair of double-bonded nitrogens versus $14$ with a single bond. Having fewer valence electrons means less repulsion between them to intefere with forming a stronger, shorter $\sigma$ bond.


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