0
$\begingroup$

I have learned hybridisation as a concept created to explain the same $\ce{C-H}$ bond lengths in $\ce{CH4}$.

I found it easy to find the type of hybridisation in different molecules, but my problem is how to find whether an atom in a molecule under goes hybridisation or not.

For example, $\ce{N}$ goes under hybridisation in $\ce{NH3}$, but there are some examples where $\ce{N}$ won't hybridise (not sure abt $\ce{N2}$, if you may suggest). In $\ce{ClF3}$, $\ce{Cl}$ is hybridised but in $\ce{PCl5}$ $\ce{Cl}$ is not.

I want the reason for undergoing hybridisation.

$\endgroup$
  • 6
    $\begingroup$ Hybridization occurs in our heads, rather than in atoms. $\endgroup$ – Ivan Neretin Mar 25 '17 at 7:27
  • $\begingroup$ Then when should we think about it. $\endgroup$ – user428838 Mar 25 '17 at 7:43
  • $\begingroup$ The thing is, orbitals and everything concerned with that is an oversimplified model to explain a lot of chemistry rather okayish. If you want to get a better understanding of how it might actually work, you need to start with quantum chemistry. $\endgroup$ – Fl.pf. Mar 25 '17 at 8:15
  • $\begingroup$ As it has already been said hybridisation is a theory we use so as to explain things (e.g. why the bonds of methane are the same) that otherwise do not make sense. In general you need to use each time the theory that best describes the result. Other times you'll use the valence bond theory or other times you'll use the molecular orbitals theory. $\endgroup$ – Αντώνιος Κελεσίδης Mar 25 '17 at 10:34
  • 1
    $\begingroup$ If you really want to, you could consider Cl in PCl5 to be sp3 hybridized with 3 of those sp3 orbitals occupied by lone pairs $\endgroup$ – xasthor Mar 25 '17 at 10:50
1
$\begingroup$

In general, you should never assume a need for hybridisation unless you have a bonding situation (such as a tetrahedron or related tetravalent structures) which cannot be explained without hybridisation.

As soon as you have a lone pair on any atom, it is more favourable to have this lone pair in an unhybridised s orbital rather than in a hybridised $\mathrm{sp}^n$ orbital. This can be seen for example in phosphane, hydrogen sulphide, etc. in which practically unhybridised central atoms are possible and thus the bond angles show pure p orbitals taking part in the $\ce{A-H}$ bonds.

If you consider very small central atoms (e.g. nitrogen in ammonia) or very large ligands (e.g. chlorine in $\ce{PCl3}$), then having a perfectly unhybridised central atom — meaning a bonding angle of $90^\circ$ — would introduce too much steric strain between the ligands. Thus, the overall system is more favourable if the central atom’s bonding orbitals are partially hybridised, i.e. if the bond angles are enlargened slightly to reduce steric stress. The final bond angle and thus the final central atom’s configuration is a carefully balanced compromise between a hybridisation with a p contribution as high as possible and the ligands separated maximally.


By the way, chlorine is practically never hybridised and it is most certainly unhybridised in $\ce{ClF3}$. Rather, this molecule should be understood as having:

  1. a single $\ce{Cl-F}$ σ bond using a p orbital both of chlorine and of fluorine

  2. a four-electron-three-centre bond using a p orbital of each of the three atoms that form the linear $\ce{F\bond{...}Cl\bond{...}F}$ fragment (i.e. a p orbital of chlorine). This bond can be understood with the following resonance structures:

    $$\ce{F-\overset{+}{Cl}\bond{...}\overset{-}{F} <-> \overset{-}{F}\bond{...}\overset{+}{Cl}-F}$$

  3. one chlorine lone pair in a p orbital perpendicular to all $\ce{C-F}$ bonds

  4. one chlorine lone pair in an s-type orbital.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.