# VSEPR Theory Part III

**by Janet Gray Coonce, MS**

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Transcription notes for Part III begins at Video time 21:18.

If instead of 3 equal regions of electron density (single bonds), we have only 2 bonds and an unshared pair of electrons, we could represent that by the notation, **AB _{2}u**.

The electronic geometry would still be **trigonal planar**, but as predicted by the **VSEPR theory**, the **lone pair** of electrons will** repel** or **bend** the less electronegative atoms sharing electrons. The **molecular geometry is bent.** The bond angle will therefore be close to but less than 120 degrees.

Consider a compound represented by the formula **AB _{4}**. An example is methane, CH

_{4}.

Here we have **4 valence shell electron pair regions repelling each other** to get **as far apart from the other regions as possible.** The orbitals are **sp ^{3} hybridized.**. This is a

**tetrahedral electronic and molecular geometry**. There are

**no lone pairs**of electrons so the

**electronic and molecular geometry are the same**.

The 3 dimensional model illustrates tetrahedral geometry. All atoms bonded to the central atom are equal and symmetric so there is no net dipole. All the bond angles are the same and equal to 109.5 degrees.

If one of the atoms were different, then there would be unequal charges surrounding the central atom resulting in a** net dipole.** An example would be **methanol, CH _{3}0H**. The oxygen atom represented by the silver ball is more electronegative than the hydrogens (green) at the other end of the central carbon atom (blue). Methanol is therefore a

**polar**compound.

In summary, if all atoms around the central atom are the same, we would predict no net dipole. If one of them are different, we would predict a **dipole** with **the electrons being pulled toward the more electronegative region**. A dipole would also be predicted if 2 of the atoms were different.

If one of the regions in the four sp_{3 }hybridized orbitals were occupied by a** lone pair** of electrons, the formula would be represented by **AB _{3}u.**

There still would be 4 regions of electron density, **3 bonds** and one pair of **unshared** electrons. **Ammonia gas, NH _{3} is an example.**

The **electronic geometry is tetrahedral** as represented by the balloon model. There are still **4 regions of electron density.** There are** 3 pairs of shared electrons** (purple balloons) which participate in bonds and **one unshared pair of lone electrons** represented by the **yellow balloon**. The lone pair of electrons are more electronegative and will **repel** the less negative shared pairs of electrons. The electronegative forces of the lone pair will repel the electron regions involved in bonding with the atoms and result in** trigonal pyramid molecular geometry**.

This is the molecular model of a **AB _{3}u** molecule. Note that the 4 atoms are not in the same plane. The atoms in the molecule are geometrically arranged in a

**trigonal pyramid**configuration. We would expect there to be

**a dipole**with the lone pair of electrons above the carbon atom to be much more electronegative than the hydrogen atoms at the other end of the molecule. We would therefore predict that ammonia gas would be a polar molecule.

If there are 4 regions of electronic density (**sp3 hybridized orbitals**) surrounding a central atom with 2 pairs of electrons shared in bonds and 2 lone pair of unshared electrons, we could represent this theoretical molecule by the formula **AB _{2}u_{2}**. A familiar example is

**H**

_{2}0.

The **electronic geometry** is** tetrahedral**. The **unshared electrons will repel** the less electronegative atoms (B) which will result in a** bent molecular geometry**. The bond angle (B-A-B) will therefore be **less than** 109.5 degrees which is the angle which would exist if the molecular geometry were not bent but tetrahedral.

Here force is applied to the green atoms to illustrate how the forces from the lone pairs of electrons repel the green atoms. This illustrates why the **VSEPR theory** predicts** a bond angle of less than** 109.5 degrees.

Using the balloons to show the tetrahedral electronic geometry of H_{2}0 notice that the **2 lone pair** (yellow balloons) of unshared electrons take up more space on one end of the oxygen atom than the space taken up by the bonds sharing electrons with hydrogen (purple balloons). These will create even more electronegative force than the single pair of unshared electrons we saw in NH_{3}. We would therefore predict an even smaller bond angle than we saw in NH_{3}. That is the bond angle is quite a bit less than 109 degrees which would exist if all 4 regions of electronegativity surrounding the central atom were equal. So the H-O-H bond is bent even more than the H-N-H bonds we saw in NH_{3}. We would also predict **water to be polar** with the more electronegative forces in the region of the unpaired electrons and the less electronegative forces bonding to the hydrogen atoms.

Think about this for a minute. What can you predict? Since we predict NH_{3} to be polar, what do you predict would happen if you bubbled NH_{3} gas through a polar liquid such as distilled water (pH = 7)? Would the resultant solution be an acid or a base?

If there are 5 equal electron regions** (dsp3 hybridized)** participating in bonds surrounding a central atom the formula can be represented by **AB _{5}**.

**PCl**is an example of a molecule with

_{5}gas**5 equal bonds**between a central atom of phosphorus and 5 atoms of chlorine. The

**electronic and molecular geometry is trigonal bipyramid**. TheB-A-B

**bond angles in the axial plane**are

**identical and equal to 90 degrees**. The B-A-B

**bond angles in the equatorial plane**are also

**identical and equal to 120 degrees.**

If one of the 5 electron regions of a molecule with **bipyramid electronic geometry** has a non-bonded lone pair of electrons it could be represented by the formula **AB _{4}u**.

**SFl**is an example.

_{4}gasWhile the **electronic geometry** is still** trigonal bipyramid**, the **molecular geometry** is a** “see-saw.”**

Here the AB_{4}u **molecular geometry, “see-saw”** is demonstrated.

If two of the 5 electron regions of a molecule with **bipyramid** electronic geometry has a non-bonded lone pair of electrons it could be represented by the formula **AB _{3}u_{2}**. The

**electronic geometry is trigonal bipyramid**. The

**orbitals are dsp**. The

^{3}hybridized**molecular geometry is “T-shaped.”**

**ClF _{3}** is an example of a molecule with the

**AB**

_{3}u_{2 }configuration. The bond angles are 90 degrees and the atoms are in a T-shaped configuration.

If 3 of the regions in a molecule with 5 electron regions have a lone pair of electrons, the configuration is expressed by the notation **AB _{2}U_{3}**. The orbitals are

**dsp**. The

^{3}hybridized**electronic geometry**is still

**trigonal bipyramid**.

The **molecular geometry is linear**, the **bond angles are 180 degrees**. **XeF _{2}** is an example of a molecule with this configuration. It is a strong fluorinating agent.

If there are **6 equal electronic bonding regions around a central atom**, the configuration can be expressed by the notation** AB _{6}**.

A central atom with 6 sigma bonds is **d ^{2}sp^{3} hybridized** and is

**octahedral**in shape. An example is

**SF**The central atom has 2d + 1s + 3p = 6 hybridized orbitals. The

_{6}.**bond angles are all equal at 90 degrees.**The electronic geometry is

**octahedral**. The

**molecular geometry**is

**octahedral**.

If one of the sigma bonds is replaced by a pair of unshared electrons, the configuration is expressed by the notation **AB _{5}u**. There are still

**6 regions**of electron density. The orbitals are

**d**.

^{2}sp^{3}hybridizedThe molecule would have a **square pyramid geometry**. **BrF _{5}** is an example. Like ClF

_{3}it is also an interhalogen compound and very reactive.

If a second of the sigma bonds is replaced by a pair of unshared electrons, the configuration is expressed by the notation **AB _{4}u_{2}**. There are still 6 regions of electron density. The orbitals are still

**d**.

^{2}sp^{3}hybridizedThe molecule is still has** d ^{2}sp^{3} hybridized electronic orbitals** in an

**octahedral geometric configuration**but the molecular geometry is

**square planar**. All 5 atoms are in the same plane with the bonded atoms forming a square around the central atom.

**XeF**is an example.

_{4}

Transcription by James C. Gray MD FACOG