12
$\begingroup$

When we are talking about hybridization, in $\mathrm{sp/sp^2/sp^3}$ hybridization, does s stand for sigma bond and p for pi bond ?

$\endgroup$
16
+50
$\begingroup$

You got it backwards.

The $\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$, $\mathrm{f}$ orbitals stand for sharp, principal, diffuse, and fundamental. Wikipedia: Electron Configuration § Notation:

The choice of letters originates from a now-obsolete system of categorizing spectral lines as "sharp", "principal", "diffuse" and "fundamental" (or "fine"), based on their observed fine structure: their modern usage indicates orbitals with an azimuthal quantum number, l, of $0$, $1$, $2$ or $3$ respectively.


Then, $\sigma$, $\pi$, $\delta$, $\phi$, being the Greek equivalents of $\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$, $\mathrm{f}$ respectively, came to designate bonds or molecular orbital symmetry, because $\mathrm{s}$ is the first orbital to form $\sigma$ bonds, $\mathrm{p}$ the first orbital to form $\pi$ bonds, $\mathrm{d}$ the first orbital to form $\delta$ bonds, and $\mathrm{f}$ the first orbital to form $\phi$ bonds.


The $\mathrm{s}$ and $\mathrm{p}$ in $\mathrm{sp^3}$ are precisely the $\mathrm{s}$ orbital and the $\mathrm{p}$ orbital. $\mathrm{sp^3}$ means that $1$ $\mathrm{s}$ orbital mixes with $3$ $\mathrm{p}$ orbitals to create $4$ hybrid orbitals known as $\mathrm{sp^3}$ orbitals.

$\endgroup$
  • $\begingroup$ Thanks, so what is s and p in sp3 or sp2? $\endgroup$ – Vedant Oct 16 '16 at 6:09
  • $\begingroup$ s orbital and p orbital. $\endgroup$ – DHMO Oct 16 '16 at 6:09
  • $\begingroup$ Sorry, I forgot to address this. $\endgroup$ – DHMO Oct 16 '16 at 6:09
  • 3
    $\begingroup$ The last part is a bit incorrect and only applies to ideal tetrahedral molecules like methane strictly. The notation for hybrid orbitals is more accurately $$\ce{sp^{n} {=} s^{\frac{1}{n+1}}p^{\frac{n}{n+1}}}.$$ Therefore one part $\ce{s}$ mixes with one part $\ce{p}$ to form one hybrid orbital $\ce{sp^3}$. There can be different types of hybrid orbitals at the same atom. $\endgroup$ – Martin - マーチン Oct 16 '16 at 7:49
4
$\begingroup$

It is important to note that while s orbitals cannot form π bonds, p orbitals are very capable at forming σ bonds. Similarly, neither of those two can form δ bonds but d-orbitals can and will form σ and π bonds.

An s orbital does not have a nodal plane. A p orbital has exactly one and a d orbital exactly two. (Note: I am not counting the ‘inner nodes’, e.g. inside a 2s orbital.) This definition was interpreted for bonds to determine whether a bond has zero, one or two nodal planes that fully contain the bond axis. A σ bond is rotation symmetric and does not have a nodal plane. A π bond has exactly one and a δ bond exactly two.

The $\ce{Cl-Cl}$ bond in chlorine, for example, is a σ bond formed by two p orbitals. And similarly, in transition metal complexes $\mathrm{d}^\unicode[Times]{x3c0}\mathrm{p}^\unicode[Times]{x3c0}$ bonds are common.

The $\mathrm{sp}^n$ nomenclature stands for hybrid orbitals; others have already described them.

$\endgroup$
2
$\begingroup$

No,the s,p,d,f notations are initial letters of words Sharp,Principal,Diffused and Fundamental

$\endgroup$

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.