I'm learning how to apply the VSEPR theory to Lewis structures and in my homework, I'm being asked to provide the hybridization of the central atom in each Lewis structure I've drawn.

I've drawn out the Lewis structure for all the required compounds and figured out the arrangements of the electron regions, and figured out the shape of each molecule. I'm being asked to figure out the hybridization of the central atom of various molecules.

I found a sample question with all the answers filled out: $\ce{NH3}$

It is $sp^3$ hybridized.

Where does this come from? I understand how to figure out the standard orbitals for an atom, but I'm lost with hybridization.

My textbook uses $\ce{CH4}$ as an example. Carbon has $2s^22p^2$, but in this molecule, it has four $sp^3$. I understand the purpose of four (there are four hydrogens), but where did the "3" in $sp^3$ come from?

How would I figure out something more complicated like $\ce{H2CO}$?


If you can assign the total electron geometry (geometry of all electron domains, not just bonding domains) on the central atom using VSEPR, then you can always automatically assign hybridization. Hybridization was invented to make quantum mechanical bonding theories work better with known empirical geometries. If you know one, then you always know the other.

  • Linear - $\ce{sp}$ - the hybridization of one $\ce{s}$ and one $\ce{p}$ orbital produce two hybrid orbitals oriented $180^\circ$ apart.
  • Trigonal planar - $\ce{sp^2}$ - the hybridization of one $\ce{s}$ and two $\ce{p}$ orbitals produce three hybrid orbitals oriented $120^\circ$ from each other all in the same plane.
  • Tetrahedral - $\ce{sp^3}$ - the hybridization of one $\ce{s}$ and three $\ce{p}$ orbitals produce four hybrid orbitals oriented toward the points of a regular tetrahedron, $109.5^\circ$ apart.
  • Trigonal bipyramidal - $\ce{dsp^3}$ or $\ce{sp^3d}$ - the hybridization of one $\ce{s}$, three $\ce{p}$, and one $\ce{d}$ orbitals produce five hybrid orbitals oriented in this weird shape: three equatorial hybrid orbitals oriented $120^\circ$ from each other all in the same plane and two axial orbitals oriented $180^\circ$ apart, orthogonal to the equatorial orbitals.
  • Octahedral - $\ce{d^2sp^3}$ or $\ce{sp^3d^2}$ - the hybridization of one $\ce{s}$, three $\ce{p}$, and two $\ce{d}$ orbitals produce six hybrid orbitals oriented toward the points of a regular octahedron $90^\circ$ apart.

I assume you haven't learned any of the geometries above steric number 6 (since they are rare), but they each correspond to a specific hybridization also.


For $\ce{NH3}$, which category does it fit in above? Remember to count the lone pair as an electron domain for determining total electron geometry. Since the sample question says $\ce{NH3}$ is $\ce{sp^3}$, then $\ce{NH3}$ must be tetrahedral. Make sure you can figure out how $\ce{NH3}$ has tetrahedral electron geometry.

For $\ce{H2CO}$

  1. Start by drawing the Lewis structure. The least electronegative atom that is not a hydrogen goes in the center (unless you have been given structural arrangement).
  2. Determine the number of electron domains on the central atom.
  3. Determine the electron geometry using VSEPR. Correlate the geometry with the hybridization.
  4. Practice until you can do this quickly.

You can find the hybridization of an atom by finding its steric number:

The steric number = the number of atoms bonded to the atom + the number of lone pairs the atom has.

If the steric number is 4, the atom is $\mathrm{sp^3}$ hybridized.

If the steric number is 3, the atom is $\mathrm{sp^2}$ hybridized.

If the steric number is 2, the atom is $\mathrm{sp}$ hybridized.

  • $\begingroup$ What about methyl free radical ? 3 + 1 = 4, so sp3 ? No that's doesn't hold true $\endgroup$ – Tilak Maddy Oct 18 '17 at 9:22
  • $\begingroup$ @TilakMaddy A single electron isn't counted as a lone "pair", AFAIK; though there might be other examples to counter this formula. Shortcut formulae don't always hold true. $\endgroup$ – dryairship Mar 19 '18 at 1:23

Hybridization is given by the following formula: $$H= \frac{1}{2} (V + X - C + A)$$ Where:

  • $V$ = number of valence electrons in central atom
  • $X$ = number of monovalent atoms around the central atom
  • $C$ = positive charge on cation
  • $A$ = negative charge on anion

$$H=4 \to \ce{sp^3},\;2\to \ce{sp,\;3}\to \ce{sp^2}...$$

e.g.: in $\ce{NH3}$, the hybridization of $\ce{N}$ atom is: $$H= \frac{1}{2}(5+3-0+0)=4 \to \ce{sp^3}$$

  • 11
    $\begingroup$ @user12757 The problem with this method is that is does not always predict the correct hybridization. For example, $\ce{PH3}$ is not $\ce{sp^3}$ hybridized; it is basically unhybridized with an H-P-H angle around 90° $\endgroup$ – ron Jan 15 '15 at 17:34
  • 1
    $\begingroup$ @ron Wouldn't the currently accepted answer also suffer from the same fault? Unless anyone actually knows that the phosphorus in phosphine is unhybridized, they will always end up counting one lone pair and three sigma bonds for $sp3$ hybridisation... $\endgroup$ – Gaurang Tandon Jan 17 '18 at 10:01
  • $\begingroup$ @GaurangTandon Yes, it would. $\endgroup$ – ron Jan 25 '18 at 16:05

protected by ManishEarth May 21 '15 at 16:21

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