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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.

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

It has a hybridization of $sp^3$.

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 $\ce{H2CO}$?

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3 Answers 3

up vote 23 down vote accepted

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 - $sp$ - the hybridization of one $s$ and one $p$ orbital produce two hybrid orbitals oriented $180^\circ$ apart.
  • Trigonal planar - $sp^2$ - the hybridization of one $s$ and two $p$ orbitals produce three hybrid orbitals oriented $120^\circ$ from each other all in the same plane.
  • Tetrahedral - $sp^3$ - the hybridization of one $s$ and three $p$ orbitals produce four hybrid orbitals oriented toward the points of a regular tetrahedron, $109.5^\circ$ apart.
  • Trigonal bipyramidal - $dsp^3$ or $sp^3d$ - the hybridization of one $s$, three $p$, and one $d$ 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 - $d^2sp^3$ or $sp^3d^2$ - the hybridization of one $s$, three $p$, and two $d$ 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 $sp^3$, 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.
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You can find the hybridization of the atom by finding its steric number:

steric number = no of atoms bonded (to the atom you are finding the hyb. of) + lone pairs with that atom.

if Steric no comes to be 4, there is sp3 hybridization in the atom.

if steric no comes to be 3, there is sp2 hybridization in the atom.

if steric no comes to be 2, there is sp hybridization in the atom.

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

  • $V$: no of valence electrons in central atom
  • $X$: no of monovalent atoms around the central atom
  • $C$: +ve charge on cation
  • $A$: -ve charge on anion

$$H=4 \implies sp^3,\;2\implies sp,\;3\implies sp^2,\;5\implies sp^3d...$$

e.g.: in $\ce{PH3}$ $$H= \frac{1}{2}(5+3-0+0)=4 \implies sp^3$$ in $\ce{H2S}$ $$H= \frac{1}{2}(6+2-0+0)=4 \implies sp^3$$

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Welcome to chemistry.SE! The help center contains the FAQ you may need to know about the workings of our community. Also, this can be helpful in making your future posts better. – Ϻ.Λ.Ʀ. Jan 14 at 19:34
@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° – ron Jan 15 at 17:34

protected by ManishEarth May 21 at 16:21

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