# Chemistry - How do I figure out the hybridization of a particular atom in a molecule?

## Solution 1:

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.

$\ce{NH3}$

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.

## Solution 2:

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: $$H= \frac{1}{2}(5+3-0+0)=4 \to \ce{sp^3}$$

## Solution 3:

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.