VSEPR and Polarity

Chapter 08

Valence shell electron-pair repulsion (VSEPR) theory is a model used to predict the geometry (shape) of molecules by considering the number of electron pairs around a central atom. A core principle to this theory states that valence electron repel each other leading to a geometry that minimizes the repulsion. Electron pairs are lone electrons, lone electron pairs, or bonding electrons of any type around a central atom, the total of which is referred to as the steric number. Lone electron pairs “take up more space” than bonding electrons. Lone elctrons “take up less space” than electron pairs or bonding electrons. Finally, there are two different geometries that can be assigned.

• Electron domain geometry (EDG) - the geometry when considering all lone electron pairs and bonding domains around a central atom
• Molecular geometry (MG) - the geometry when considering only bonding domains around a central atom

Molecular geometries for simple compounds. Electron domain geometries are given in the first column. Molecular geometries are identical to the electron domain geometries for structures with zero lone electron pairs. X represents an atom and E represnts a non-bonding electron domain (lone electron or lone electron pair).

Let us visualize electron pairs of various steric numbers and rationalize the geometry that minimizes the repulsions.

Steric number = 2

Consider two electron domains (steric number = 2), represented as balloon-shaped three-dimensional space that the electron cloud occupies, connected at a central point. The energy is lowest when these two clouds are furthest apart from each other at 180 ° giving a linear geometry.

We can apply this concept to a molecule such as carbon dioxide. For carbon dioxide, the steric number for carbon is 2. The geometry that minimizes the repulsion of the electron pairs is a linear electron domain geometry and a linear molecular geometry with a bond angle of 180°.

Carbon dioxide is a non-polar molecule meaning it does not have a dipole. A dipole (or dipole moment) is a separation of positive and negative charge within a molecule or ion and can be estimated by considering the polarity of a bond as well as its direction. Particles with a dipole are said to have a polarity (i.e. are polar) which affect the interaction behavior that the particle can have with other particles.

Carbon dioxide has two polar bonds, each equal in magnitude and pointing in opposite directions. These vectors cancel each other out resulting in a molecule with no net dipole moment.

Visualize linear CO₂

Linear

Key Characteristics

• 0 lone electron pairs
• 180° bond angle
• EDG: linear
• MG: linear

Non-polar

Linear structures whose terminal atoms are the same are non-polar (i.e. do not have a dipole).

Polar

Linear structures whose terminal atoms are different are polar (i.e. have a dipole).

Bond polarity shown in black and dipole moment in blue.

The multi-centered acetylene structure can be broken down into two central carbon atoms, each with two electron domains around them.

Steric number = 3

Now consider how three electron domains (steric number = 3). The maximum separation between these pairs is 120 °. The geometry that minimizes the repulsion of the electron pairs is a trigonal planar geometry.

Visualize trigonal planar borane, BH₃

Trigonal planar

Key Characteristics

• 0 lone electron pairs
• 120° bond angle
• EDG: trigonal planar
• MG: trigonal planar

Non-polar

Trigonal planar structures whose terminal atoms are the same are non-polar (i.e. do not have a dipole).

Polar

Trigonal planar structures whose terminal atoms are different are polar (i.e. have a dipole).

Bent

Key Characteristics

• 0 lone electron pairs
• < 120° bond angle if not a radical
• > 120° bond angle if a radical
• EDG: trigonal planar
• MG: bent

All bent molecules are polar (i.e. have a dipole).

Polar

Steric number = 4

Consider four electron domains (steric number = 4). The geometry that minimizes the repulsions between these pairs is a tetrahedral geometry.

The bond angles are not equivalent if a lone electron pair is present.

Visualize tetrahedral methane, CH₄

Tetrahedral

Key Characteristics

• 0 lone electron pairs
• 109.5° bond angle if all terminal atoms are the same
• EDG: tetrahedral
• MG: tetrahedral

Non-polar

Tetrahedral molecules whose terminal atoms are the same are non-polar (i.e. do not have a dipole).

Polar

Tetrahedral molecules whose terminal atoms are not the same are polar (i.e. have a dipole).

Trigonal pyramidal

Key Characteristics

• 1 lone electron pair
• < 109.5° bond angle
• EDG: tetrahedral
• MG: trigonal pyramidal

All trigonal pyramidal structures have a dipole if all atoms are not the same.

Polar

Visualize trigonal pyramidal ammonia, NH₃

Bent

Key Characteristics

• 2 lone electron pairs
• << 109.5° bond angle
• EDG: tetrahedral
• MG: bent

All bent molecules are polar (i.e. have a dipole).

Visualize bent water, H₂O

Steric number = 5

A steric number of 5 results in a trigonal bipyramidal geometry. Terminal atom positions are labeled as axial (ax) or equatorial (eq). Note the bond angles between the atom positions.

Visualize trigonal bipyramidal phosphorus pentafluoride, PF₅

Trigonal bipyramidal

Key Characteristics

• 0 lone electron pairs
• 90° axial bond angle
• 120° equatorial bond angle
• EDG: trigonal bipyramidal
• MG: trigonal bipyramidal

Non-polar

Trigonal bipyramidal structures are non-polar if

• all the terminal atoms are the same, or
• the atoms at the axial positions and the atoms at the equatorial positions are the same.

Lone electron pairs are omitted for clarity.

Polar

Trigonal bipyramidal structures are polar if

• at least one of the terminal atoms are different, or
• the atoms in the axial and/or terminal positions are different

Seesaw

Key Characteristics

• 1 lone electron pair
• < 90° axial bond angle
• < 120° equatorial bond angle
• EDG: trigonal bipyramidal
• MG: seesaw

All seesaw structures are polar (i.e. have a dipole).

Polar

Visualize seesaw sulfur tetrafluoride, SF₄

T-shaped

Key Characteristics

• 2 lone electron pairs
• < 90° axial bond angle
• EDG: trigonal bipyramidal
• MG: T-shaped

All T-shaped structures are polar (i.e. have a dipole).

Polar

Linear

Key Characteristics

• 3 lone electron pairs
• 180° axial bond angle
• EDG: trigonal bipyramidal
• MG: Linear

Non-polar

Linear structures whose terminal atoms are the same are non-polar (i.e. do not have a dipole).

Polar

Linear structures whose terminal atoms are different are polar (i.e. have a dipole).

Steric number = 6

A steric number of 6 results in an octahedral geometry. Terminal atom positions are labeled as axial (ax) or equatorial (eq). Note the bond angles between the atom positions.

Octahedral

Key Characteristics

• 0 lone electron pairs
• 90° bond angles
• EDG: octahedral
• MG: octahedral

Non-polar

Octahedral structures are non-polar if

• all the terminal atoms are the same, or
• the atoms at the axial positions and the atoms at the equatorial positions are the same.

Lone electron pairs are omitted for clarity.

Visualize octahedral sulfur hexafluoride, SF₆

Polar

Octahedral structures are polar if

• one of the terminal atoms are different, or
• one of the equatorial atoms are different from one another, or
• one of the axial atoms are different from one another, or
• one of the equatorial atoms are different from one another and one of the axial atoms are different from one another

Lone electron pairs are omitted for clarity.

Square pyramidal

Key Characteristics

• 1 lone electron pair
• < 90° bond angles
• EDG: octahedral
• MG: square pyramidal

All square pyramidal structures are polar (i.e. have a dipole).

Polar

Lone electron pairs on terminal atoms omitted for clarity.

Visualize square pyramidal bromine pentafluoride, BrF₅

Square planar

Key Characteristics

• 2 lone electron pairs
• 90° bond angles
• EDG: octahedral
• MG: square planar

Non-polar

Square pyramidal structures are non-polar (i.e. do not have a dipole) if

• all terminal atoms are the same, or
• opposing terminal atoms are the same

Lone electron pairs on terminal atoms omitted for clarity.

Polar

Square pyramidal structures are polar (i.e. do not have a dipole) if

• one of the terminal atoms are the different, or
• opposing terminal atoms are different

Lone electron pairs on terminal atoms omitted for clarity.

Visualize square pyramidal xenon tetrafluoride, XeF₄

T-shaped

Key Characteristics

• 3 lone electron pairs
• < 90° bond angles
• EDG: octahedral
• MG: T-shaped

All T-shaped structures are polar (i.e. have a dipole).

Polar

Visualize square pyramidal iodine trifluoride, IF₃

Linear

Key Characteristics

• 4 lone electron pairs
• 180° bond angles
• EDG: octahedral
• MG: linear

Non-polar

Linear structures whose terminal atoms are the same are non-polar (i.e. do not have a dipole).

Polar

Linear structures whose terminal atoms are different are polar (i.e. have a dipole).

Multi-centered molecules

VSEPR can be applied to molecules with more than one central atom. Simply apply the rules of VSEPR to each central atom to determine the geometry about that atom.

Propane, CH₃CH₂CH₃

Visualize propane

Acetic acid, CH₃COOH

Visualize acetic acid

Benzene, C₆H₆

Visualize benzene

Glycine, NH₂CH₂COOH

Visualize glycine

Bonds

Bond lengths

Bond lengths are commonly reported in Ångstroms or, in SI units, picometers. The conversion is given below.

\[1~\mathring{\text{A}} = 10^2~\mathrm{pm}\]

Average covalent bond lengths (in Å) are provided below.

Single-bond lengths (in Å)

	H	C	N	O	F	Si	P	Si	Cl	Br	I
H	0.74	1.10	0.98	0.94	0.92	1.45	1.38	1.32	1.27	1.42	1.61
C		1.54	1.47	1.43	1.41	1.94	1.87	1.81	1.76	1.91	2.10
N			1.40	1.36	1.34	1.87	1.80	1.74	1.69	1.84	2.03
O				1.32	1.30	1.83	1.76	1.70	1.65	1.80	1.99
F					1.28	1.81	1.74	1.68	1.63	1.78	1.97
Si						2.34	2.27	2.21	2.16	2.31	2.50
P							2.20	2.14	2.09	2.24	2.43
Si								2.08	2.03	2.18	2.37
Cl									2.00	2.13	2.32
Br										2.28	2.47
I											2.66

Multi-bond lengths (in Å)

Bond	Length
C=C	1.34
C=N	1.27
C=O	1.22
N=O	1.15
C≡C	1.21
C≡N	1.15
C≡O	1.13
N≡O	1.08

Bond order

Bonds can be described by their bond order which describes the type of bond.

• single bond → bond order = 1
• double bond → bond order = 2
• triple bond → bond order = 3

Bond orders can be an average value. For example, ozone (O₃) has two resonance structures, each containing a single and double bond. The average bond order for the bond between each oxygen atom would be 1.5.

Bond energies

Bond dissociation energy (BDE, D₀, or DH°) is a measure of the strength or binding energy of a chemical bond between two atoms or fragments. These energies can be expressed as the enthalpy change for a homolytic bond cleavage between two atoms given as

\[\mathrm{R}\!\!-\!\!\mathrm{X} ~~\longrightarrow~~ \mathrm{R} + \mathrm{X}\]

where a homolytic cleavage involves the equal distribution of the bonding electrons between both atoms or fragments resulting in two radicals.

The bond dissociation energy can be estimated from standard heats of formation such that

\[DH^{\circ}(\mathrm{R}\!\!-\!\!\mathrm{X}) = \Delta_{\mathrm{f}}H^{\circ}(\mathrm{R}) + \Delta_{\mathrm{f}}H^{\circ}(\mathrm{X}) - \Delta_{\mathrm{f}}H^{\circ}(\mathrm{RX})\]

Average bond dissociation energies (in kJ mol^–1) are given below.

Average single bond dissociation energies (in kJ mol^–1)

	H	C	N	O	F	Si	P	Si	Cl	Br	I
H	436	413	391	463	565	328	322	347	432	366	299
C		346	305	358	485	–	–	272	339	285	213
N			163	201	283	–	–	–	192	–	–
O				146	–	452	335	–	218	201	201
F					155	565	490	284	253	249	278
Si						222	–	293	381	310	234
P							201	–	326	–	184
Si								226	255	–	–
Cl									242	216	208
Br										193	175
I											151

Average multi-bond dissociation energies (in kJ mol^–1)

Bond	Bond Energy
N=N	418
N≡N	945
C=N	615
C≡N	887
O=O (in O2)	498
C=C	610
C≡C	835
C=O	745
C=O (in CO2)	83
C≡O	1046

The enthalpy of a reaction can be estimated by considering the bonds broken and the bonds formed in the reaction. In this exercise, we consider all covalent bonds for all the reactants to be broken followed by a rearrangement of the atoms and formation of all the bonds formed for all the products.

\[\Delta_{\mathrm{r}} H^{\circ} = \sum DH^{\circ}(\mathrm{bonds~broken}) - \sum DH^{\circ}(\mathrm{bonds~formed})\]

This requires being able to draw Lewis structures for all species involved.

Consider the following combustion reaction under standard conditions.

\[2~\mathrm{H_2(g)} + \mathrm{O_2(g)} \longrightarrow \mathrm{2~H_2O(g)}\]

Determine the number of bonds broken by the reactants and their bond energies.

• 2(H–H) → 2(436 kJ mol^–1) = 872 kJ mol^–1
• 1(O=O) → 1(498 kJ mol^–1) = 498 kJ mol^–1

Next, determine the bonds formed (and bond energies) in the products.

• 4(O–H) → 4(463 kJ mol^–1) = 1 852 kJ mol^–1

Finally, estimate the enthalpy of reaction.

\[\begin{align*} \Delta_{\mathrm{c}} H^{\circ} &= \sum DH^{\circ}(\mathrm{bonds~broken}) - \sum DH^{\circ}(\mathrm{bonds~formed}) \\[1.5ex] &= \left ( 872 + 498 \right )~\mathrm{kJ~mol^{-1}} - \left ( 1~852 \right )~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -482~\mathrm{kJ~mol^{-1}} \end{align*}\]

The formation of water from hydrogen and oxygen gas is very exothermic.

Let us compare the enthalpy of this reaction using standard thermodynamic values.

\[\begin{align*} \Delta_{\mathrm{c}} H^{\circ}(\mathrm{H_2O, g}) &= \!\!\!\!\sum_{products,~p} \!\!\! \nu_p~\Delta_{\mathrm{f}} H_{p}^{\circ} ~~ - \!\!\!\!\sum_{reactants,~r} \!\!\! \nu_r~\Delta_{\mathrm{f}} H_{r}^{\circ} \\[1.5ex] &= \bigg [ 2~\Delta_{\mathrm{f}}H^{\circ}(\mathrm{H_2, g}) + \Delta_{\mathrm{f}}H^{\circ}(\mathrm{O_2, g}) \bigg ] - \bigg [ 2~ \Delta_{\mathrm{f}}H^{\circ}(\mathrm{H_2O, g})\bigg ] \\[1.5ex] &= \bigg [ (2 \times 0) + (0) \bigg ]~\mathrm{kJ~mol^{-1}} - \bigg [ (2\times -241.826) \bigg ]~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -483.652~\mathrm{kJ~mol^{-1}} \end{align*}\]

The agreement is very good.